DC Circuits

🧭 Overview

🧠 One-sentence thesis

This section establishes the foundational relationships between charge, current, voltage, resistance, and power in DC circuits, showing how these quantities interact through Ohm's Law and how to measure and calculate them for practical circuit analysis.

📌 Key points (3–5)

• Charge vs. current: Current (Amps) is the rate of charge flow (Coulombs per second); we assume "hole flow" (current flows opposite to electrons).
• Resistance depends on material and geometry: Resistance is calculated from resistivity (ρ), length (L), and cross-sectional area (A); metals have low resistivity and positive temperature coefficients (resistance increases with temperature).
• Ohm's Law relates V, I, R: Voltage = Current × Resistance (V = I·R); for resistors, always use ΔV (voltage difference) to find current, not just voltage at one node.
• Common confusion—voltage vs. current measurement: Voltage is measured across components (two probes on different nodes); current is measured through components (break the circuit and run it through the meter).
• Power balance checks your work: Total power supplied by sources must equal total power dissipated by resistors and other elements; resistors always dissipate power, while sources can supply or dissipate depending on polarity.

⚡ Charge and current fundamentals

⚡ Relationship between charge and current

Current (I) = Charge (Q) / time (t)

• Units: Current in Amps (A), Charge in Coulombs (C), time in seconds (sec).
• Important conversion: 1 Coulomb = 6.242 × 10¹⁸ electrons.
• Current is the rate at which charge flows, not the total amount of charge.
• Example: If a switch causes 2 amps to flow for 4.3 minutes, the total charge is Q = 2 A × (4.3 × 60) sec = 516 C, which equals 3.22 × 10²¹ electrons.

🔄 Hole flow convention

• This course assumes "hole flow" for current direction.
• Current flows in the opposite direction to electron flow.
• This is a standard convention in circuit analysis; the choice does not affect calculations as long as you are consistent.

🧱 Resistance: material properties and geometry

🧱 How resistance is calculated

Resistance (R) = resistivity (ρ) × Length (L) / Area (A)

• Units: If ρ is in Ω·m, then L is in meters and A is in meters².
• Resistance depends on three factors: the material's resistivity, the length of the conductor, and its cross-sectional area.
• Longer conductors or smaller cross-sections increase resistance; larger areas or shorter lengths decrease resistance.
• Example: A 1,000-meter copper wire with 3 mm diameter at 20°C has resistance R = (1.72×10⁻⁸ Ω·m × 1000 m) / (π/4 × (0.003 m)²) = 2.433 Ω.

🌡️ Temperature effects on resistance

R at temperature x = R at 20°C × [1 + α(Tₓ − T₂₀°C)]

• Metals (conductors) have positive temperature coefficients (α > 0): resistance increases as temperature increases.
• Insulators have negative temperature coefficients (α < 0): resistance decreases as temperature increases.
• This positive coefficient in metals is beneficial—excessive current heats the conductor, increasing resistance, which reduces current and prevents thermal runaway.
• Example: The 2.433 Ω copper wire at 20°C becomes 2.529 Ω at 30°C using α = 0.00393.

🔌 Resistivity and conductivity of materials

Material Type	Resistivity (ρ)	Temperature Coefficient (α)	Example
Metals (Conductors)	Low (10⁻⁸ to 10⁻⁷ Ω·m)	Positive	Copper: 1.72×10⁻⁸ Ω·m, α = 0.00393
Insulators	High (10¹⁶ Ω·m or more)	Negative	Air: 1.3×10¹⁶ to 3.3×10¹⁶ Ω·m
Semiconductors	Medium (10⁻¹ to 10² Ω·m)	Negative	Silicon: 6.4×10² Ω·m, α = −0.075

• Conductors have low resistivity and readily allow current flow.
• Insulators (like plastic wire coating) strongly oppose current flow and prevent current from escaping the conductor.

📏 American Wire Gauge (AWG) and ampacity

• AWG is a standard for classifying wire size by the diameter of the metal conductor (not including plastic shield).
• Ampacity is the maximum current a wire can carry before the plastic shield breaks down.
• Larger AWG numbers mean smaller diameter and lower ampacity.
• Example: 12 AWG copper wire has a 2.053 mm diameter and 20 A ampacity; 10 AWG has 2.588 mm diameter and 30 A ampacity.
• Safety tip: Use wire with ampacity well above the expected current; home wiring often uses 12 AWG with 20 A circuit breakers to ensure the breaker trips before the wire is damaged.

🔺 Non-circular cross-sections

• For rectangular or other shapes, calculate the cross-sectional area (e.g., width × thickness for a rectangle) and use the same resistance formula.
• Example: A 1-meter gold bar (5 mm wide, 1 mm thick) at 0°C has area A = 0.005 m × 0.001 m = 0.000005 m², so R₂₀°C = (2.44×10⁻⁸ × 1) / 0.000005 = 0.00488 Ω, then scale to 0°C: R₀°C = 0.00455 Ω.

🔢 Circular mils (CM) and alternative units

• 1 mil = 0.001 inches (milli-inch).
• Circular mils (CM) are used in some AWG tables: 1 CM = (π/4) square mils.
• If resistivity is given in CM·Ω/ft, use area in CM and length in feet.
• Example: A wire with 0.02 inch diameter (20 mils) has area = (π/4)(20 mils)² = 314.16 square mils = 400 CM. For 100 ft of wire with ρ = 5 CM·Ω/ft, R = (5 × 100) / 400 = 1.25 Ω.

🔋 Ohm's Law: voltage, current, and resistance

🔋 Basic Ohm's Law

Voltage (V) = Current (I) × Resistance (R)

• Units: Voltage in Volts (V), Current in Amps (A), Resistance in Ohms (Ω).
• Can also be written in terms of conductance (G = 1/R): Current = Voltage × Conductance (I = V·G).
• Units for conductance: Siemens (S).
• Example: A 3.2 MΩ resistor connected to a 9 V battery has conductance G = 1/(3.2×10⁶) = 0.3125 μS and current I = 9 V × 0.3125 μS = 2.8125 μA.

🧮 Unit shortcuts for Ohm's Law

• Volts and Ohms → Amps: I = V/R with V in Volts and R in Ω gives I in Amps.
• Volts and kΩ → mA: I = V/R with V in Volts and R in kΩ gives I in mA.
• Volts and MΩ → μA: I = V/R with V in Volts and R in MΩ gives I in μA.
• This shortcut reduces calculator errors—just divide the numbers and set the correct current unit.
• Example: 4.5 V / 2 MΩ = 2.25 μA (no need to enter 2×10⁶).

🔌 Resistors and color codes

• Resistor symbol: A zigzag line in circuit diagrams.
• Carbon-film resistors often use a 4-band color code to indicate resistance and tolerance.
• First two bands = digits, third band = multiplier (number of zeros), fourth band = tolerance.
• Example: Brown-Black-Red-Gold = 1-0-×100-5% = 1,000 Ω ± 5%.
• Power rating: Resistors have a maximum power they can dissipate (e.g., ¼ watt) before damage; exceeding this causes overheating and failure.

📐 Measuring resistance with a multimeter

• Set the multimeter to Ohms mode (Ω) at a range higher than the expected resistance.
• Place probes on both ends of the resistor (component must be disconnected from the circuit or powered off).
• Example: A nominal 1,000 Ω resistor with 5% tolerance reads 986 Ω on the meter, which is within tolerance.

🔌 Voltage: sources and measurement

🔌 What is voltage?

Voltage (or potential difference) is created by a separation of charge.

• Voltage is measured across components or between two nodes.
• Battery symbol: Two parallel lines (longer line is +, shorter is −); the + and − signs are often omitted but must be remembered.
• An ideal independent voltage source has no internal resistance and maintains its voltage regardless of the circuit it is connected to.
• A non-ideal voltage source (actual battery) has small internal resistance, causing the terminal voltage to drop slightly under load.
• Example: A 9 V ideal battery connected to 100 Ω delivers 9 V across the resistor; a 9 V battery with 2 Ω internal resistance delivers only 8.82 V across the same resistor.

🔋 Battery types and internal resistance

• Alkaline batteries (AA, AAA, C, D, 9V) are inexpensive and common; internal resistance is typically 0.15–0.3 Ω for 1.5 V cylindrical types.
• Lithium-Ion batteries are increasingly popular for higher energy density.
• 9 V rechargeable NiMH batteries have internal resistance around 1 Ω at full charge, 1.5 Ω at half charge.
• Internal resistance causes voltage drop under load and limits the maximum current the battery can supply.

📏 Measuring voltage with a multimeter

• Set the multimeter to DC Volts (DCV) mode at a range higher than the expected voltage.
• Place the + (red) probe on the higher voltage node and the − (black) probe on the lower voltage node.
• If probes are reversed, the meter displays a negative voltage.
• Example: Measuring a 4.85 V battery pack with the 20 V setting (not 2 V, which would show an error).
• Voltage across components: Place probes on both sides of the component; a positive reading means current flows from + probe to − probe through the component.

🔦 Diodes and LEDs as voltage elements

• A forward-biased diode acts like a small voltage source (constant drop model).
• Silicon diodes (e.g., 1N914) have approximately 0.7 V drop when forward biased.
• LEDs have larger drops: red LEDs ≈ 1.8 V, green LEDs ≈ 2 V (assumed values in this course).
• A diode is a "current valve"—current flows only in one direction (anode to cathode).
• Reverse-biased diode acts as an open circuit (no current flow).
• Example: A green LED in series with a 1 kΩ resistor and 9 V battery has approximately 2 V across it; the constant drop model replaces the LED with a 2 V battery for simplified calculations.
• LED current limit: The LEDs used in this course have a 20 mA maximum current rating; exceeding this causes very bright light, color change (green → yellow), and eventual permanent failure.

🔄 Voltage symbols in circuit diagrams

• Four common DC voltage symbols are used interchangeably: battery symbol, circle with + and −, and variations.
• All represent an independent DC voltage source.

🌐 Nodes and voltage measurement

• Voltage at a node is measured with respect to ground (0 V reference).
• Single subscript notation: Vₐ means voltage at node a relative to ground.
• Double subscript notation: V_AB = Vₐ − V_b (voltage from node A to node B).
• Voltage is the same at all points on a node (a connection shared by 2 or more elements).
• Example: In a circuit, node b is at 6 V everywhere along the blue line representing that node.

⚡ Current: flow and measurement

⚡ Key differences between voltage and current

Voltage	Current
Measured across components (two nodes)	Flows through components
Same at all points on a node	Changes only at branches (splits)
Components change voltage (voltage drop)	Components do not change current (same current in, same current out)
Double subscript notation used (V_AB)	Only single subscript or label (I_R1)

• Don't confuse: Voltage is measured between two points; current is measured by breaking the circuit and inserting the meter in series.

📏 Measuring current with a multimeter

• Set the multimeter to DC Amps (A) mode.
• Break the circuit and insert the meter in series so current flows through it.
• CAUTION: Never place the meter probes across a resistor (in parallel) when in current mode—this bypasses the resistor, causing excessive current that can blow the fuse or damage the meter.
• If the fuse blows, current reads zero at all settings, but voltage readings still work; replace the fuse or check if internal circuitry is damaged.
• Example: To measure current through a resistor, disconnect one end of the resistor from the breadboard and connect the meter in series between the resistor and the rest of the circuit.

🔢 Calculating current through a resistor

I_R = ΔV / R = (Vₐ − V_b) / R

• Always use ΔV (voltage difference across the resistor), not just the voltage at one node.
• This is the most common mistake students make: calculating I = V/R instead of I = ΔV/R.
• The current is positive in the direction from higher voltage (Vₐ) to lower voltage (V_b).
• Example: If Vₐ = 4.84 V and V_b = 2.42 V across a 1,000 Ω resistor, then I = (4.84 − 2.42) / 1000 = 2.42 mA.

🔌 Current sources

An ideal independent current source forces a fixed current through the circuit section it is connected to, regardless of the resistance.

• Symbol: Circle with an arrow inside.
• Current sources have a voltage across them (unlike voltage sources, which have current through them).
• Non-ideal current source has a large parallel resistor; ideal current source assumes infinite parallel resistance (resistor has no effect).
• Current sources are built from multiple transistors and are common in integrated circuits (ICs).
• Example: A 55 mA current source connected to a 100 Ω resistor has voltage V = I × R = 55 mA × 100 Ω = 5.5 V across the source.

🌳 Branches and current flow

• A branch is a path in the circuit where current can split or combine.
• Current does not change as it flows through components in series, but it splits at branches (parallel paths).
• Example: In a circuit with 4 branches and 7 nodes, there are 6 different currents (some branches share the same current if they are in series).

🔍 Determining current from voltage probes

• If voltage probes are placed at nodes, use ΔV across each resistor to find the current through it.
• Example: If node voltages are 30 V and 0 V across a 2 kΩ resistor, then I = (30 − 0) / 2000 = 15 mA.

🔥 Power and energy

🔥 Power equations

Power (P) = Voltage (V) × Current (I)

• Units: Watts (W).
• Two alternative forms:
  • P = V² / R (useful when you know voltage and resistance)
  • P = I² × R (useful when you know current and resistance)
• For sources (voltage or current sources): Use only P = V × I.
• For resistors: Use P = V²/R or P = I²·R (whichever is easier).

⚖️ Supplied vs. dissipated power

• Dissipated (absorbed) power: Resistors and diodes always dissipate power (convert electrical energy to heat or light).
• Supplied (delivered) power: Sources can supply power (negative sign) or dissipate power (positive sign) depending on polarity.
• In this course, only the magnitude of power is required (ignore the sign unless asked specifically for "power of" a source).
• Example: "What is the power supplied by I1?" asks for magnitude only (e.g., 0.1 W); "What is the power of I1?" would include the sign (−0.1 W for supplied power).

🔄 When a source dissipates vs. supplies power

• Voltage source dissipates power if current flows from + to − inside the source (unusual; typically means it is being charged).
• Voltage source supplies power if current flows from − to + inside the source (normal operation).
• Current source dissipates power if the voltage across it is positive in the direction of current flow.
• Current source supplies power if the voltage across it is negative in the direction of current flow.
• Example: In a circuit, if a current source has a positive power reading on a probe, it is dissipating power; if negative, it is supplying power.

⚖️ Conservation of power (power balance)

∑ P_supplied = ∑ P_dissipated

• Total power supplied by all sources must equal total power dissipated by all resistors, diodes, and any sources that dissipate.
• Use this to check your work: If the power does not balance, you made a calculation error or misidentified a source as a supplier/dissipater.
• Resistors and diodes are always on the dissipated side.
• Example: In a circuit, if resistors dissipate 1.119 W total and sources supply 1.119 W total, the power balance is satisfied.

🔋 Battery capacity and milliamp-hours (mAh)

• mAh rating indicates how long a battery will last at a given current draw.
• Calculation: Battery life (hours) = mAh rating / current draw (mA).
• Example: An RC car drawing 70 mA from 6 AAA alkaline batteries (each ≈ 1000 mAh) will last approximately 1000 / 70 = 14.3 hours (in practice, less, because the car stops working before voltage drops to zero).
• Power rating of equipment: If a device runs on 6 V and draws 50 mA, its power rating is 6 V × 0.05 A = 0.3 W.

⚡ Energy

Energy (E) = Power (P) × time (t)

• Units:
  • Watts × seconds = Joules (J)
  • Kilowatts × hours = kilowatt-hours (kWh) (used for electricity bills)
• Example: A 3.6 W TV running for 3.5 hours uses:
  • Energy (Joules) = 3.6 W × (3.5 × 3600) sec = 45,360 J
  • Energy (kWh) = 0.0036 kW × 3.5 hrs = 0.0126 kWh

🔒 Fuses and circuit protection

• Fuses limit current by breaking (opening the circuit) when current exceeds a rated value.
• A thin metallic conductor inside the fuse melts due to excessive heat at the rated current, becoming an open circuit.
• Once blown, the fuse must be replaced.
• AC circuits: Circuit breakers (e.g., 20 A for home wiring) or fuses (e.g., in Christmas lights).
• DC circuits: Fuses are common (e.g., automobile fuses for 12 V battery circuits).
• If a fuse repeatedly blows, there is a problem with the circuit (e.g., short circuit or overload).

🔗 Combining resistors in series and parallel

🔗 Definitions of series and parallel

Configuration	Rule	Equation
Series	Two resistors share one connection, and nothing else is connected to that point	R_total = R₁ + R₂ + R₃ + …
Parallel	Two resistors share two connections (both ends connected together)	R_total = (R₁·R₂)/(R₁+R₂) for 2 resistors; 1/R_total = 1/R₁ + 1/R₂ + … for 3+

• Series: Resistors are in a single path; current is the same through all.
• Parallel: Resistors are in separate paths between the same two nodes; voltage is the same across all.
• Don't confuse: A resistor can be in series with one group and in parallel with another, depending on the circuit topology.

➕ Series resistor equation

R_total = R₁ + R₂ + R₃ + … + R_N

• Simply add all resistances.
• Example: 1 kΩ + 2 kΩ + 3 kΩ = 6 kΩ.

➗ Parallel resistor equations

• For exactly 2 resistors:

  R_total = (R₁ × R₂) / (R₁ + R₂)

  • This equation only works for 2 resistors; do not use for 3 or more.
  • Example: 3 kΩ ∥ 6 kΩ = (3 × 6) / (3 + 6) = 18 / 9 = 2 kΩ.

• For 3 or more resistors:

  1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + … + 1/R_N

  • This involves "fractions of fractions" and is error-prone in calculators.
  • Alternative: Combine resistors two at a time using the 2-resistor formula.
  • Example: For 2 kΩ, 4 kΩ, and 6 kΩ in parallel, first combine 2 kΩ and 4 kΩ: (2×4)/(2+4) = 8/6 = 1.333 kΩ; then combine 1.333 kΩ and 6 kΩ: (1.333×6)/(1.333+6) = 1.09 kΩ.

🎯 When to use series/parallel combining

• Module 1 (this section): Combine all resistors in a circuit down to a single equivalent resistor with one source.
• Module 2 (Thevenin equivalent): Combine resistors to find the Thevenin equivalent resistance.
• Module 3 (AC circuits): Combine inductors and capacitors in series/parallel networks.

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Note: The excerpt includes some personal anecdotes, acknowledgments, and references to external links and course-specific details (e.g., "ENGR 2431," "OU," "Multisim," "project"). These have been included where they provide context for definitions, examples, or warnings, but the focus remains on the technical content as presented in the excerpt.

🧭 Overview

🧠 One-sentence thesis

Combining resistors in series or parallel is a foundational skill that simplifies circuits to a single equivalent resistance, enabling calculation of source currents and power, and this technique will be reused throughout the course for different circuit elements.

📌 Key points (3–5)

• When to combine resistors: Only combine when there is one source (or can be reduced to one source); avoid combining resistors whose specific voltage or current you need to find.
• Series vs parallel definitions: Series means two resistors share one point with nothing else connected; parallel means two resistors share two points.
• Common confusion: Students often mistake resistors for being in series when they share a point but other elements are also connected to that point—this violates the series rule.
• Calculation rules: Series resistances add directly; parallel resistances use the product-over-sum formula (for two resistors) or reciprocal-sum formula (for multiple resistors).
• Why it matters: This skill is used to find total resistance, source current, and power supplied, and will be applied to inductors and capacitors in later modules.

🔌 Series and parallel definitions

🔌 Series connection rule

Rule: R1 and R2 are in Series if both are true: (1) R1 and R2 share one point (or connection), and (2) Nothing else is connected to that point.

• The key is both conditions must be met.
• If a third element connects to the shared point, the two resistors are not in series.
• Example: In Example 1.14, R1 and R2 share a point, but R3 is also connected to that point, so R1 and R2 are not in series.

🔌 Parallel connection rule

Rule: R3 and R4 are in Parallel if: R3 and R4 share two points.

• The resistors must connect at both ends (two shared nodes).
• It is often easier to see parallel connections if you drag resistors along the wire to visualize the shared points.
• Example: In Example 1.14, R2 and R3 share two points, so they are in parallel.

🧮 Combining formulas

🧮 Series formula

• For multiple resistors in series: R_Total = R1 + R2 + R3 + ... + RN
• The total resistance is the sum of all individual resistances.
• Sanity check: The total resistance will be larger than the largest resistor.

🧮 Parallel formula (two resistors)

• For exactly 2 resistors in parallel: R_Total = (R1 times R2) divided by (R1 + R2)
• This is the "product-over-sum" formula.
• Note: This equation ONLY works for 2 resistors.
• Sanity check: The total resistance will be smaller than the smallest resistor; if the two resistors are identical, the result is exactly half of one resistor (e.g., 4kΩ parallel with 4kΩ equals 2kΩ).

🧮 Parallel formula (three or more resistors)

• For 3 or more resistors in parallel: R_Total = 1 divided by (1/R1 + 1/R2 + 1/R3 + ... + 1/RN)
• This is the reciprocal-sum formula.
• Alternative method: Combine multiple parallel resistors two at a time using the two-resistor formula to avoid "fractions of fractions" calculation mistakes.
• Another option: Use conductance (G = 1/R):
  • Step 1: G_Total = G1 + G2 + G3 + ... + GN (or G_Total = 1/R1 + 1/R2 + 1/R3 + ... + 1/RN)
  • Step 2: R_total = 1/G_Total

🛠️ Guidelines for reducing circuits

🛠️ When to find total resistance (Guideline 1 and 2)

• Guideline 1: Only find total resistance if there is one source or the circuit can be reduced to one source.
  • Example: Four 1.5 V batteries in series can be combined into one 6 V battery.
• Guideline 2: If you need to solve for a specific resistor's voltage or current, do not combine that resistor with another resistor—combining it may eliminate your ability to find that specific value.

🛠️ Order and visualization (Guideline 3 and 4)

• Guideline 3: Start combining resistors as far away from the source as possible and work toward the source.
  • This prevents mistakes like trying to combine resistors that are neither in series nor parallel.
• Guideline 4: You can move (drag) resistors anywhere along a wire as long as you don't move them across a branch.
  • This makes it easier to see whether two resistors are in series or parallel.
  • Example: In Example 1.14, dragging R3 along the wire makes it obvious that R2 and R3 are in parallel.

🛠️ Redrawing and sanity checks (Guideline 5 and 6)

• Guideline 5: Redraw the circuit each time you combine two resistors.
  • This takes longer but reduces mistakes, especially for students prone to errors.
• Guideline 6: Always perform a sanity check:
  • Series: Total resistance is larger than the largest resistor.
  • Parallel: Total resistance is smaller than the smallest resistor.
  • Parallel (two close values): Result is fairly close to half the average of the two resistors.

📐 Worked examples

📐 Example 1.14 walkthrough

• Question 1: Can we find total resistance? Yes, there is one source (12 V battery).
• Question 2: Does finding total resistance help? Yes, we need source current and power, not individual resistor values.
• Question 3: Which resistors to combine first? Start far from the source: R2 and R3.
  • Common mistake: Trying to combine R1 and R2 (they share a point, but R3 is also connected, so they are not in series).
• Question 4: Are R2 and R3 in series or parallel? Drag R3 along the wire to see they share two points → parallel.
  • Rx = R2 parallel R3 = (4k times 6k) divided by (4k + 6k) = 2.4 kΩ
  • Sanity check: 2.4 kΩ is less than 4 kΩ (smallest resistor); half the average is 2.5 kΩ, close to 2.4 kΩ.
• Redraw the circuit: Now R1 and Rx are in series (they share a point and nothing else is connected).
  • Rtot = R1 + Rx = 1 kΩ + 2.4 kΩ = 3.4 kΩ
• Find source current: Is = Vs / Rtot = 12 V / 3.4 kΩ = 3.53 mA
• Find power supplied: P_Supplied = Is times Vs = 3.53 mA times 12 V = 42.36 mW

📐 Example 1.15 summary

• By inspection: Rtot = R1 parallel with {R2 + R3 + [R4 parallel (R5 + R6)]} = 10 parallel 45 = 8.182 Ω
• Step-by-step:
  • Step 1: R5 and R6 in series → Ra = 20 + 30 = 50 Ω
  • Step 2: R4 and Ra in parallel → Rb = 50 parallel 50 = 25 Ω
  • Step 3: R2, R3, Rb in series → Rc = 5 + 15 + 25 = 45 Ω
  • Step 4: R1 and Rc in parallel → Rtot = 10 parallel 45 = 8.182 Ω
• Power supplied: Vs = I1 times Rtot = 2.5 A times 8.182 Ω = 20.5 V; P_supplied = I1 times Vs = 2.5 A times 20.5 V = 51.1 W

📐 Example 1.16 and "Reduce and Return" method

• Total resistance: R_T = [(R4 + R5) parallel R3 parallel R2] + R1
• Source current: Calculate using Ohm's Law after finding R_T.
• Voltage across R1: V_R1 = I_S times R1 = 3.99 mA times 3000 Ω = 11.97 V
• Currents through R2 and R3: Voltage at the top node is 15 V minus 11.97 V = 3.03 V; bottom is 0 V (ground).
  • I_R2 = (3.03 V – 0 V) / 4kΩ = 0.7575 mA
  • I_R3 = (3.03 V – 0 V) / 1kΩ = 3.03 mA
• "Reduce and Return" method: Reduce the circuit to one resistor, find source current, then return to the original circuit to find other voltages and currents using Ohm's Law.
• Limitation: To find R4 or R5 current/voltage, new methods are needed (KVL or KCL in Module 1, or matrix methods in Module 2). The excerpt states that the Reduce and Return method becomes cumbersome and inferior to matrix methods when multiple calculation steps are required.

🔍 Common confusions

🔍 Series vs neither

• Don't confuse: Two resistors that share one point are not automatically in series.
• The second rule must also be true: nothing else can be connected to that shared point.
• Example: In Example 1.14, R1 and R2 share a point, but R3 is also connected, so R1 and R2 are neither in series nor parallel and cannot be combined.

🔍 Parallel visualization

• Don't confuse: Parallel resistors may not look parallel in the circuit diagram.
• Use Guideline 4: drag resistors along the wire to see if they share two points.
• Example: In Example 1.14, R2 and R3 do not look parallel until R3 is dragged along the wire.

🔍 Two-resistor vs multi-resistor parallel formulas

• Don't confuse: The product-over-sum formula (R1 times R2 divided by R1 + R2) ONLY works for exactly 2 resistors.
• For 3 or more resistors, use the reciprocal-sum formula or combine two at a time.
• The excerpt warns that the reciprocal-sum formula has "fractions of fractions" and students frequently make calculator mistakes, so combining two at a time is recommended.

🧭 Overview

🧠 One-sentence thesis

KVL states that the sum of voltages around any closed loop equals zero, and the Voltage Divider Rule provides a shortcut for finding voltages across series resistors when only one voltage source is present.

📌 Key points (3–5)

• What KVL states: the sum of all voltages around a closed loop (starting and ending at the same point) equals zero.
• How to apply KVL: label polarity on every element, travel around a loop, write voltage terms with signs based on the polarity you encounter, then solve.
• When VDR can be used: only when all elements are in series (same current flows through all) and there is exactly one voltage source in the loop.
• Common confusion: VDR is the most incorrectly used formula in circuit analysis—it cannot be used when elements are not all in series or when multiple voltage sources exist.
• Why it matters: KVL allows solving for unknown currents and voltages; VDR simplifies calculations when its strict conditions are met.

🔁 Kirchhoff's Voltage Law fundamentals

🔁 The two forms of KVL

KVL is defined as: sum of voltage drops = sum of voltage rises, or equivalently, sum of all voltages around a closed loop = 0.

• The preferred form is "sum of voltages around a closed loop equals zero" because it avoids confusion about whether a voltage source causes a drop or rise.
• "Around a closed loop" means you travel through the circuit and return to the same starting point.
• Any path you choose will yield the same result as long as start and end points are identical.

🔁 Why the zero-sum form is preferred

• The excerpt notes it is "sometimes difficult to know whether a voltage source causes a voltage drop or rise in a circuit."
• Using the zero-sum form sidesteps this ambiguity—you simply record the sign you encounter as you travel.

📝 Step-by-step KVL procedure

📝 Step 1: Label polarity on every element

• Voltage sources: polarity is set by the battery symbol (+ sign on the long line).
• Current sources: the arrowhead side has the + sign.
• Resistors: + and – signs are written in the direction of the assumed current flow.

📝 Step 2: Travel around a closed path

• Start at any point in the circuit.
• Travel around any path until you return to the same starting point.
• As you cross each element, write its voltage (I times R for resistors, Vs for sources) with the sign you encounter:
  • Travel into the + sign → positive term.
  • Travel into the – sign → negative term.

📝 Step 3: Solve the KVL equation

• If only one unknown exists, solve directly.
• If multiple unknowns exist, you need additional KVL and/or KCL equations (or other information).
• Important: if your calculation yields a negative current, the assumed direction was wrong—the magnitude is correct, but reverse the direction. Re-solving is NOT necessary.

📝 Example scenario

Example: A loop with a 12 V source and four resistors (5, 10, 20, 25 ohms) in series.

• KVL equation: –12 + 5·Is + 10·Is + 20·Is + 25·Is = 0
• Solve: Is = 12 / (5+10+20+25) = 0.2 A
• Since the result is positive, the assumed current direction was correct.
• Voltage across the 25-ohm resistor: V = 0.2 · 25 = 5 V

⚡ Voltage Divider Rule (VDR)

⚡ What VDR is and when it applies

Voltage Divider Rule (VDR): Vx = Vs · (Rx / Rtotal)

• Vx: the voltage across the resistor you are finding.
• Vs: the one and only voltage source in the circuit.
• Rtotal: total resistance (sum of all series resistors).
• Strict conditions: VDR can be used if and only if:
  1. All elements are in series (same current flows through all).
  2. Only one voltage source is in the loop.

⚡ Why VDR is a shortcut

• The excerpt states VDR "turns a 4 step problem into a 1 step problem" when conditions are met.
• Example: Using the same circuit as before, VR4 = 12 · (25 / 60) = 5 V—same result, fewer steps.

⚡ Common misuse warning

• The excerpt emphasizes: "VDR is probably the most incorrectly used formula in all of circuit analysis."
• Don't confuse: VDR cannot be used if elements are not all in series or if multiple voltage sources exist.
• Always verify both conditions before applying VDR.

🔌 KVL with multiple voltage sources

🔌 Handling multiple sources

• When multiple voltage sources are present, carefully follow the polarity rules.
• Example: A loop with a 12 V source, five resistors, and additional sources of +2 V, –1.5 V, and +0.7 V.
• KVL: –12 + 1k·Is + 2 + 2k·Is + 3k·Is – 1.5 + 4k·Is + 0.7 + 5k·Is = 0
• Solve: Is = (12 + 1.5 – 2 – 0.7) / (1k + 2k + 3k + 4k + 5k) = 0.72 mA

🔌 Getting the signs correct

• The excerpt notes "the trickiest part about this problem is getting all of the signs correct."
• If you carefully follow the polarity-labeling rules, you will have no problem.
• Some circuits may be unrealistic (e.g., a battery in the wrong polarity), but you must still solve them to practice techniques.

🔋 KVL with current sources

🔋 Polarity of current sources

• The positive sign is placed on the arrowhead of the current source.
• The current source forces a specific current magnitude and direction through all series elements.

🔋 Example scenario

Example: A current source forcing 20 mA clockwise through four resistors (4 kΩ, 3 kΩ, 2 kΩ, 6 kΩ).

• KVL starting at ground, going clockwise: + (20 mA · 4 kΩ) + (20 mA · 3 kΩ) + (20 mA · 2 kΩ) + (20 mA · 6 kΩ) – Vs = 0
• Solve: Vs = 80 + 60 + 40 + 120 = 300 V

🔋 Ground symbol note

• The ground symbol sets the voltage at that node to zero.
• No current flows into the wire where the ground symbol is added—it can be ignored when making calculations, just like wiring to a multimeter.

🔄 Multi-loop circuits and KVL

🔄 Writing multiple KVL equations

• In circuits with multiple loops, you can write KVL equations for each individual loop and for combinations of loops.
• Example: A 3-loop circuit can yield 7 different KVL equations (loop1, loop2, loop3, loop12, loop13, loop23, loop123).

🔄 Using current probes vs assumptions

• The excerpt shows an example where current probes determine the correct direction, so polarity is known to be correct.
• However, using simulation tools to determine current directions is NOT required.
• You can assume any direction; if the answer is negative, reverse the assumed direction without re-solving.

🔄 When to use Module 2 methods instead

• The excerpt mentions the "Reduce and Return" technique is "cumbersome" and "easy to mess up."
• Once a third step requiring KVL or KCL is needed, the Reduce and Return method should be "scrapped for Module 2 methods."
• KVL is useful for practice and understanding, but matrix methods (Module 2) are preferred for complex multi-step calculations.

🧭 Overview

🧠 One-sentence thesis

Kirchhoff's Current Law states that current flowing into any point in a circuit equals current flowing out, and the Current Divider Rule provides a shortcut for calculating how current splits among parallel resistors.

📌 Key points (3–5)

• KCL definition: The sum of currents entering a branch or node equals the sum of currents leaving that branch or node.
• Branch vs node application: KCL at a branch (single point) is easier to visualize; KCL at a node (connecting multiple branches) is more powerful and often requires fewer equations.
• Current Divider Rule (CDR): A shortcut formula for finding current through one resistor when multiple resistors in parallel share a known incoming current.
• Common confusion: In CDR, the positions of R_x and R_total are swapped compared to the Voltage Divider Rule (VDR); CDR uses R_total in the numerator, VDR uses R_x in the numerator.
• Puzzle-solving approach: When resistor values are unknown, use known node voltages, Ohm's Law (I = ΔV/R), and KCL methodically to calculate missing currents, voltages, and resistances.

⚡ Kirchhoff's Current Law fundamentals

⚡ What KCL states

Kirchhoff's Current Law (KCL): The sum of currents entering a branch (or node) equals the sum of currents leaving that branch (or node).

• Mathematically: sum of I_IN = sum of I_OUT.
• Unlike KVL, there are no fixed rules for writing KCL equations—you only need to assume current directions at each branch or node.
• If your calculation yields a negative current, the assumed direction was wrong; you can reverse the direction without reworking the problem.

🔀 Branch version vs node version

Version	Description	Advantage
Branch KCL	Sum of currents into a single point = sum out of that point	Easier to understand; the branch is a single connection point
Node KCL	Sum of currents into a node = sum out of that node	More powerful; often requires fewer equations to solve a circuit

• Branch: A single point where currents meet; straightforward to identify which currents connect.
• Node: A larger region connecting multiple branches; more difficult but reduces the number of equations needed.
• Example: In Example 1.22, two branch equations were needed (at B1 and B2) to find I_R4; in Example 1.23, one node equation at Node X directly solved for I_R4.

🧩 How to apply KCL

• Step 1: Assume directions for all currents at the branch or node.
• Step 2: Write the equation: currents in = currents out.
• Step 3: Solve for the unknown current.
• Step 4: If the result is negative, the actual direction is opposite to your assumption.

Example (branch KCL): At branch B1, if I_R1 flows in and I_R2 and I_x flow out, then I_R1 = I_R2 + I_x. Given I_R1 = 3.99 mA and I_R2 = 0.7575 mA, solve: I_x = 3.99 - 0.7575 = 3.2325 mA.

Example (node KCL): At Node X, if I_R1 flows in and I_R2, I_R3, I_R4 flow out, then I_R1 = I_R2 + I_R3 + I_R4. This single equation directly solves for I_R4 without needing intermediate currents.

🔧 Current Divider Rule (CDR)

🔧 What CDR does

Current Divider Rule (CDR): If you know the current entering a node and have multiple resistors in parallel connected to that node, the current through any one parallel resistor can be calculated directly.

• Formula: I_x = I_s times (R_total divided by R_x).
• I_s: the current coming into the node (from a voltage source or current source).
• R_x: the resistor through which you want to find the current (I_x).
• R_total: the total equivalent resistance of all the parallel resistors at that node.

🔄 CDR vs VDR: key difference

• Voltage Divider Rule (VDR): V_x = V_s times (R_x divided by R_total).
• Current Divider Rule (CDR): I_x = I_s times (R_total divided by R_x).
• Don't confuse: The positions of R_x and R_total are swapped between the two rules.
• In VDR, the resistor of interest (R_x) is in the numerator; in CDR, R_total is in the numerator.

🧮 How to use CDR

1. Calculate R_total: Find the equivalent resistance of all parallel resistors. Use the formula: 1/R_total = 1/R1 + 1/R2 + 1/R3 + ... (for two resistors: R_total = (R1 times R2) divided by (R1 + R2)).
2. Identify I_s: The current entering the node (either from a voltage source through a series resistor or directly from a current source).
3. Apply CDR: I_x = I_s times (R_total divided by R_x).

Example (from Example 1.24): Four resistors (1Ω, 2Ω, 3Ω, 4Ω) in parallel, with a 10 A current source.

• R_total = 1 / (1/1 + 1/2 + 1/3 + 1/4) = 0.48 Ω.
• I_R1 = 10 times (0.48 / 1) = 4.8 A.
• I_R2 = 10 times (0.48 / 2) = 2.4 A.
• I_R3 = 10 times (0.48 / 3) = 1.6 A.
• I_R4 = 10 times (0.48 / 4) = 1.2 A.
• Verify with KCL: 10 A = 4.8 + 2.4 + 1.6 + 1.2 = 10 A ✓

⚠️ When to use CDR

• Use CDR when you have parallel resistors and know the total current entering the node.
• CDR is a shortcut; you can always use KCL and Ohm's Law instead, but CDR is faster.
• CDR works with current sources (I_s is the source current) or with voltage sources (I_s is the current calculated from the voltage source and series resistance).

🧩 Solving complex circuits with KCL and Ohm's Law

🧩 The puzzle-solving approach

• When some resistor values or node voltages are unknown, treat the problem as a puzzle.
• Use what you know (given voltages, known resistances) and methodically calculate new values.
• Combine KCL, Ohm's Law (I = ΔV / R), and power equations to find missing quantities.

🔍 Step-by-step strategy

1. Identify known values: Ground reference (0 V), given node voltages, known resistances, source voltages/currents.
2. Calculate currents from known voltages: Use I = (V_a - V_b) / R for resistors between two known nodes.
3. Apply KCL: At nodes or branches, use current conservation to find unknown currents.
4. Calculate new node voltages: Use V = I times R to find voltage drops, then add/subtract from known node voltages.
5. Find unknown resistances: Use R = ΔV / I once you know both voltage and current.
6. Verify with power balance: Check that power supplied equals power dissipated.

📐 Example walkthrough (from Example 1.25)

Given: Node A = 7.96 V, Node B = 1.29 V, ground = 0 V, 12 V battery, various resistors.

Step 1: Calculate currents using known node voltages:

• I_2k = (7.96 - 0.7) / 2000 = 3.63 mA.
• I_1k = (12 - 7.96) / 1000 = 4.04 mA.

Step 2: Apply KCL to find I_3k:

• At the top node: I_1k = I_2k + I_3k → I_3k = 4.04 - 3.63 = 0.41 mA.

Step 3: Calculate new node voltages:

• V_C = V_A - I_3k times 3k = 7.96 - 0.41 mA times 3k = 6.73 V.
• V_D = V_C - I_3k times 400 = 6.73 - 0.41 mA times 400 = 6.57 V.

Step 4: Find unknown resistor R7:

• V_E = V_D - 5 = 6.57 - 5 = 1.57 V.
• R7 = (V_E - V_B) / I_3k = (1.57 - 1.29) / 0.41 mA = 683 Ω (round-off errors expected due to limited precision in given values).

Step 5: Continue the process for other unknowns (I_5k, I_R6, R6, power supplied).

⚠️ Common pitfalls

• Round-off errors: When given values have limited precision (e.g., 2 digits), expect small discrepancies in calculated results.
• Current source behavior: The current through a current source is fixed and does not change regardless of other components in series with it.
• Direction assumptions: Always assume a direction for unknown currents; a negative result simply means the actual direction is opposite.
• Node voltage notation: Use double-subscript notation (e.g., V_DA = V_D - V_A) to track voltage differences across components clearly.

🔋 Working with current sources

🔋 Current source properties

• A current source sets the current in its branch; this current is fixed.
• The current does not change even if voltage sources or resistors are placed in series with the current source.
• The voltage across a current source adjusts to maintain the specified current.

🧮 Example with multiple current sources (from Example 1.26)

Given: Three current sources (I1 = 30 mA, I2 = 10 mA, I3 = 5 mA), Node A = 84 V, various resistors.

Step 1: Calculate voltage drops across resistors with known currents:

• V_DA = 1k times 30 mA = 30 V.
• V_CD = 2k times 30 mA = 60 V.

Step 2: Calculate node voltages:

• V_D = V_DA + V_A = 30 + 84 = 114 V.
• V_C = V_CD + V_D = 60 + 114 = 174 V.

Step 3: Apply KCL to find unknown currents:

• At Node A: I1 + I3 = I2 + I4 → 30 + 5 = 10 + I4 → I4 = 25 mA.
• At Node B: I2 + I4 = I5 → 10 + 25 = I5 → I5 = 35 mA.

Step 4: Calculate remaining node voltages and unknown resistor:

• V_E = V_A - 2 = 84 - 2 = 82 V.
• V_B = V_E - I4 times 3k = 82 - 25 mA times 3k = 7 V.
• Rx = (V_B - 0) / I5 = 7 / 35 mA = 200 Ω.

🔍 Key insight

• With current sources, you know the current immediately; use this to calculate voltage drops and apply KCL to find other currents.
• The puzzle-solving approach works the same way: start with what you know, calculate step by step, and verify with KCL or power balance.

🧭 Overview

🧠 One-sentence thesis

Source transformations allow you to convert between voltage sources with series resistors (Thevenin form) and current sources with parallel resistors (Norton form), which can greatly simplify circuit analysis and is required for certain matrix methods.

📌 Key points (3–5)

• What source transformations do: convert a voltage source in series with a resistor into a current source in parallel with a resistor, or vice-versa, to simplify circuits.
• When transformations are required: Mesh Matrix analysis requires only voltage sources; Nodal Matrix analysis requires only current sources.
• Key constraint: a resistor must be present in series (Thevenin) or parallel (Norton) for standard transformation; without it, only approximate methods work.
• Common confusion: the voltage across or current through the internal resistor changes after transformation—only voltages and currents at the external connection points ("Nubbies") remain the same.
• How to transform values: use Ohm's Law to convert between voltage and current source values, keeping the resistor value unchanged.

🔄 The two forms and when to use them

🔋 Thevenin Form

Thevenin Form: a voltage source in series with a resistor.

• The voltage source and resistor are connected one after the other in a single path.
• This form is required for Mesh Matrix analysis (covered in section 2.3).
• Example: a 12 V source in series with a 1 kΩ resistor.

⚡ Norton Form

Norton Form: a current source in parallel with a resistor.

• The current source and resistor are connected side-by-side, sharing the same two nodes.
• This form is required for Nodal Matrix analysis (covered in section 2.4).
• Example: a 12 mA source in parallel with a 1 kΩ resistor.

🎯 Why transform

• Transforming can greatly simplify a circuit.
• Example: converting a three-loop mesh problem into a one-node problem makes the analysis much easier.
• Certain analysis methods only accept one form, so transformation is mandatory.

📏 The five transformation rules

📏 Rule 1: Resistor values stay equal

• The Thevenin resistance (Rth) equals the Norton resistance (Rn): Rth = Rn.
• The resistor value does not change during transformation.

🔀 Rule 2: Resistor location changes

• In Thevenin form, the resistor is in series with the voltage source.
• In Norton form, the resistor is in parallel with the current source.
• The resistor "moves" from series to parallel (or vice-versa).

⚠️ Rule 3: A resistor must be present

• If there is no resistor in series with the voltage source (Thevenin) or no resistor in parallel with the current source (Norton), a standard transformation cannot be made.
• In such cases, only an approximate transformation (covered in section 2.2) is possible.
• Don't confuse: an ideal source without a resistor cannot be transformed using the standard method.

🧮 Rule 4: Use Ohm's Law for source values

• To convert the source value, apply Ohm's Law.
• From Thevenin to Norton: current source value = voltage source value divided by resistor value.
• From Norton to Thevenin: voltage source value = current source value multiplied by resistor value.
• Example: a 12 V source with 1 kΩ resistor transforms to a 12 mA current source with 1 kΩ resistor (12 V / 1 kΩ = 12 mA).

🧭 Rule 5: Polarity and direction alignment

• The positive side of the voltage source aligns with the head (arrow tip) of the current source.
• This ensures the transformed source delivers power in the same direction.
• Example: if the voltage source's positive terminal is on the left, the current source arrow should point left.

🔌 The "Nubbies" concept and what stays the same

🔴 What are the "Nubbies"

• The two connection points (marked as red dots in the figures) where external circuit elements attach.
• Any circuit elements connected externally to these two points will have exactly the same voltage and current values whether the source is in Thevenin or Norton form.
• Example: resistors R2, R3, and R4 connected to the Nubbies will have identical voltages in both forms.

⚠️ What changes inside the transformation

• The voltage across the internal resistor (Rth or Rn) is different in Thevenin vs Norton form.
• The current through the internal resistor is also different.
• Don't confuse: only the external behavior at the Nubbies is preserved; internal voltages and currents change.

🚫 When NOT to transform

• If you need to calculate the voltage between the voltage source and the series resistor, do not transform that source.
• If you need to know the current through the series resistor in Thevenin form, do not transform.
• Example: in a circuit with a 12 V source and 3 Ω resistor, the current through the 3 Ω resistor differs between Thevenin and Norton forms, so transformation would lose that information.

🛠️ Practical examples

🛠️ Example: Simplifying a three-loop problem

• Starting circuit: 12 V source in series with 1 kΩ resistor (Thevenin form), creating a three-loop mesh problem.
• Transformation: convert to Norton form with a 12 mA current source in parallel with 1 kΩ resistor.
• Result: the problem becomes a one-node problem, much easier to solve.
• Calculation: current source = 12 V / 1 kΩ = 12 mA.

🛠️ Example: Preparing for Nodal Matrix analysis

• Starting circuit: 12 V source in series with 3 Ω resistor.
• Goal: Nodal Matrix analysis requires only current sources.
• Transformation: convert to Norton form with a 4 A current source in parallel with 3 Ω resistor.
• Calculation: current source = 12 V / 3 Ω = 4 A.
• Result: the circuit now has only three nodes (identified by wire colors), ready for nodal analysis.

🔄 Handling different source locations

• The source and resistor may not always be oriented exactly as shown in the basic diagram.
• As long as the five rules are followed, the transformation works regardless of location or polarity.
• Keep track of the Nubbies to ensure correct external connections.
• Example: the source can be on the left, right, top, or bottom; the transformation process remains the same.

🧭 Overview

🧠 One-sentence thesis

Approximate source transformations allow circuits that are not in standard Thevenin or Norton form to be transformed by adding a virtual resistor sized to minimize error, enabling the use of simpler matrix solution methods.

📌 Key points (3–5)

• Why approximate transformations are needed: some voltage or current sources lack the series or parallel resistor required for standard transformation, but matrix methods require all sources to be the same type (all voltage or all current).
• The virtual resistor is not real: it is added only for mathematical transformation purposes and must be sized to introduce minimal error into the circuit.
• Voltage source rule: add a virtual resistor in series that is 1000 times smaller than the smallest resistor in the circuit.
• Current source rule: add a virtual resistor in parallel that is 1000 times larger than the largest resistor in the circuit.
• Common confusion: the transformed source values can be very large (e.g., 500 A or 400 kV), but these are mathematical constructs, not actual physical sources in the circuit.

🎯 Why approximate transformations exist

🎯 The matrix method limitation

• Matrix techniques (mesh matrix and nodal matrix) are much easier than alternative circuit-solving methods.
• Mesh matrix analysis requires all voltage sources; nodal matrix analysis requires all current sources.
• Sometimes a circuit contains sources that are not in the correct form (Thevenin or Norton), so they cannot be directly transformed.

🔧 What Thevenin and Norton forms require

Thevenin Form: a voltage source with a resistor in series.

Norton Form: a current source with a resistor in parallel.

• If a voltage source has no series resistor, it is not in Thevenin form.
• If a current source has no parallel resistor, it is not in Norton form.
• Approximate transformations solve this by adding a virtual resistor so the source can be transformed.

⚠️ The virtual resistor is not real

• The added resistor is a mathematical tool only; it does not physically exist in the circuit.
• It must be sized carefully to minimize the error it introduces.
• Example: adding a 1 milliohm resistor in a circuit with megaohm resistors would cause round-off errors when combining resistors.

🔌 Voltage source approximate transformation

🔌 The sizing rule

Rule of Thumb: Add a virtual resistor in series with the voltage source that is 1000 times smaller than the smallest resistor in the circuit.

• The goal is to make the virtual resistor as small as possible compared to other resistors.
• A smaller virtual resistor has less impact on the circuit behavior.
• Example: if the smallest resistor in the circuit is 10 Ω, add a 10 mΩ (0.01 Ω) resistor in series.

🧮 How to perform the transformation

1. Identify the smallest resistor value in the circuit.
2. Divide that value by 1000 to get the virtual resistor size.
3. Add the virtual resistor in series with the voltage source.
4. Transform using the standard formula: current source value equals voltage divided by the series resistance.

📐 Example scenario

Example: A 5 V voltage source in a circuit where the smallest resistor is 10 Ω.

• Virtual resistor: 10 Ω / 1000 = 10 mΩ = 0.01 Ω.
• Add 0.01 Ω in series with the 5 V source.
• Transformed current source: I_N = 5 V / 0.01 Ω = 500 A.
• The excerpt shows that node voltages in the circuit with and without the added resistor are nearly the same, confirming minimal error.

⚠️ Don't confuse the transformed value with reality

• The 500 A current source is a mathematical construct, not an actual 500 amp source.
• The transformation is purely for simplifying the solving process.

🔋 Current source approximate transformation

🔋 The sizing rule

Rule of Thumb: Add a virtual resistor in parallel with the current source that is 1000 times larger than the largest resistor in the circuit.

• The goal is to make the virtual resistor as large as possible compared to other resistors.
• A larger virtual resistor has less impact on the circuit because it draws less current.
• Example: if the largest resistor in the circuit is 40 Ω, add a 40 kΩ (40,000 Ω) resistor in parallel.

🧮 How to perform the transformation

1. Identify the largest resistor value in the circuit.
2. Multiply that value by 1000 to get the virtual resistor size.
3. Add the virtual resistor in parallel with the current source.
4. Transform using the standard formula: voltage source value equals current multiplied by the parallel resistance.

📐 Example scenario

Example: A 10 A current source in a circuit where the largest resistor is 40 Ω.

• Virtual resistor: 40 Ω × 1000 = 40 kΩ = 40,000 Ω.
• Add 40 kΩ in parallel with the 10 A source.
• Transformed voltage source: V_th = 10 A × 40 kΩ = 400 kV.
• The excerpt shows that node voltages in the circuit with and without the added resistor are the same, confirming minimal error.

⚠️ Don't confuse the transformed value with reality

• The 400 kV voltage source is a mathematical construct, not an actual 400 kilovolt source.
• The transformation is purely for making the solving process easier.

🧩 Key cautions and context

🧩 When not to transform

The excerpt mentions (in the CAUTION section before Section 2.2) that:

• If you need to find the voltage between a voltage source and its series resistor, do not transform that source—the node voltage cannot be calculated after transformation.
• If you need to find the current through the series resistor, do not transform—the current through that resistor is different in the transformed circuit.

🧩 Why these large values appear

• The excerpt emphasizes that 500 A and 400 kV sources are not actually in the circuit.
• They are simply the result of the mathematical transformation.
• The purpose is to enable the use of matrix methods, which are much easier to solve.

🧩 The role of matrix methods

• The excerpt states that matrix techniques are "so much easier than all of the alternative methods that they should be used if possible."
• Approximate transformations are a tool to make matrix methods applicable when sources are not initially in the correct form.

🧭 Overview

🧠 One-sentence thesis

The Mesh Matrix Method allows you to solve multi-loop circuits by populating matrices directly from circuit inspection, eliminating the tedious algebra and sign errors inherent in the traditional Branch Current Method.

📌 Key points (3–5)

• Branch Current Method: writes KVL equations for each minor loop and KCL equations at branches, then solves the system algebraically—straightforward but error-prone and time-consuming.
• Mesh Matrix Method: populates resistance and voltage matrices by following simple rules (diagonal = sum of resistors in a loop; off-diagonal = negative sum of shared resistors; voltage sources placed by polarity), then solves [R]∙[I]=[V] using matrix inversion.
• Common confusion: "loop current" does not mean the same current flows around the entire loop—only the portion not shared with other loops; shared branches require KCL to find their actual current.
• Source transformation prerequisite: Mesh Matrix requires only voltage sources; any current sources must first be transformed to voltage sources (standard or approximate).
• Why it matters: as circuits grow more complex (more loops), the algebra savings become profound—Mesh Matrix lets you enter values straight into a calculator or MATLAB without writing equations.

🔄 Branch Current Method: the algebraic foundation

🔄 What the Branch Current Method does

• Writes one KVL equation for each minor loop in the circuit.
• Writes KCL equations at branches where loops connect.
• Solves the resulting system of equations for all unknown currents.

Minor loop: a closed circuit path that contains no other elements inside the loop; equivalently, you cannot pass through another circuit element by moving your hand through the loop without leaving it.

🧮 Why it is tedious

• Requires careful sign tracking when writing KVL equations (easy to make mistakes).
• Demands substantial algebra to reduce the system to solvable form (substitution, rearrangement).
• Example: even a 2-loop circuit requires 2 KVL + 1 KCL equation, then substitution to reduce to a 2×2 matrix; a 3-loop circuit becomes "extremely time consuming."

🔍 Example walkthrough (Example 2.5)

Circuit: 2 minor loops, 3 unknown currents (I₁, I₂, I₃).

Steps:

1. Write KVL for minor loop 1: –9 + 1000·I₁ + 2000·I₃ = 0
2. Write KVL for minor loop 2: –2000·I₃ + 3000·I₂ + 0.7 = 0
3. Write KCL at the branch: I₁ = I₂ + I₃
4. Substitute I₃ = I₁ – I₂ into the KVL equations.
5. Rearrange to matrix form: [R]∙[I] = [V], then solve B = A⁻¹·C.

Result: I₁ = 3.946 mA, I₂ = 1.445 mA, I₃ = 2.501 mA.

Don't confuse: the "loop current" I₁ is not the current in every wire of loop 1—the shared 2 kΩ resistor carries I₃, which is found by KCL.

🧩 Mesh Matrix Method: direct matrix population

🧩 Core idea

• Skip writing KVL/KCL equations entirely.
• Populate the [A]∙[B]=[C] matrix by inspection using six rules.
• Solve B = A⁻¹·C to get all loop currents.

Matrix form: [R]∙[I] = [V], where [R] is the resistance matrix (A), [I] is the current matrix (B), and [V] is the voltage matrix (C).

📏 The six Mesh Matrix rules

Rule	What to do	Why
1. Voltage sources only	Transform all current sources to voltage sources first	Mesh Matrix cannot handle current sources directly
2. Minor loops set matrix size	n minor loops → A is n×n, B and C are n×1	Each loop gets one equation
3. Assume all currents clockwise	Always draw loop currents clockwise; negative answer means counter-clockwise	Eliminates confusion about current direction
4. Diagonal of A	A₁₁, A₂₂, A₃₃, etc. = sum of all resistors in that loop	Positive sign; represents total resistance in the loop
5. Off-diagonal of A	A₁₂, A₂₁, etc. = –(sum of resistors shared by those two loops)	Negative sign; matrix is symmetric (A₁₂ = A₂₁)
6. Voltage sources in C	Place in row corresponding to the loop; positive if you hit the – terminal first when traveling clockwise	Sign convention matches KVL but flipped because voltage is on the right side of the equation

✅ Example: verifying the method (Example 2.6)

Same circuit as Example 2.5, now solved by Mesh Matrix rules:

• A₁₁ = 1000 + 2000 = 3000 Ω (resistors in loop 1)
• A₂₂ = 2000 + 3000 = 5000 Ω (resistors in loop 2)
• A₁₂ = A₂₁ = –2000 Ω (2 kΩ resistor shared by loops 1 and 2)
• C₁₁ = +9 V (traveling clockwise in loop 1, you hit the – terminal first)
• C₂₁ = –0.7 V (traveling clockwise in loop 2, you hit the + terminal first)

Result: the matrix in Figure 2.12 is identical to the one derived algebraically in Figure 2.10, but obtained in seconds by inspection.

🔧 Handling shared branches

• The matrix gives you the loop currents (I₁, I₂, I₃, etc.).
• To find the current in a branch shared by two loops, use KCL.
• Example: if a resistor is between loops 1 and 2, its current is I₁ – I₂ (or I₂ – I₁, depending on direction).

🔌 Source transformations: preparing the circuit

🔌 Why transformation is necessary

• Mesh Matrix requires only voltage sources (Rule 1).
• If the circuit contains current sources, you must transform them to voltage sources before applying Mesh Matrix.

🔄 Standard vs approximate transformation

• Standard: current source in parallel with a resistor (Norton form) → voltage source in series with the same resistor (Thévenin form).
• Approximate: current source with no parallel resistor → add a "virtual" resistor 1000× larger than the largest resistor in the circuit, then transform.

Important note from the excerpt:

"Keep in mind there is obviously NOT an actual 500 Amp or a 400 kilovolt source in the last two example circuits. When doing a source transformation (standard or approximate) you are not actually putting in this large valued source into the circuit. It is simply a mathematical transformation for the purpose of making the solving process easier."

🧪 Example with transformation (Example 2.8)

Circuit: contains 3 current sources.

Steps:

1. Transform all 3 current sources to voltage sources (Figure 2.16).
2. Apply Mesh Matrix rules to the transformed circuit (3 minor loops → 3×3 matrix).
3. Solve: I₁ = 11.551 mA, I₂ = 8.122 mA, I₃ = –0.610 mA.

Don't confuse: a negative loop current (I₃ = –0.610 mA) means the actual current flows counter-clockwise, not that there is an error.

🧮 Solving the matrix: tools and techniques

🧮 Matrix inversion

• Once [A] and [C] are populated, solve B = A⁻¹·C.
• Can be done with a calculator, MATLAB, or GNU Octave (open-source alternative).

MATLAB example code:

A = [3000, -2000; -2000, 5000];
C = [9; -0.7];
B = inv(A)*C;
I1 = B(1)
I2 = B(2)
I3 = I1 - I2

📐 Cramer's Rule (manual alternative)

• If matrix inversion tools are unavailable, use Cramer's Rule.
• For each unknown in B, replace the corresponding column in A with C, take the determinant, and divide by the determinant of the original A.
• The excerpt notes this is useful when working with symbolic variables (e.g., Laplace transforms).

🔍 Checking your work

• Use Multisim current probes oriented clockwise (or multimeters with + and – in the clockwise direction).
• If the probe reads positive, the current is clockwise; if negative, it is counter-clockwise.

🧪 Extended examples: more complex circuits

🧪 Three-loop circuit (Example 2.7)

Circuit: 3 minor loops, 4 voltage sources (9 V, 6 V, 4 V, 1 V), 5 resistors.

Matrix setup:

• A₁₁ = 100 + 200 = 300 Ω
• A₂₂ = 200 + 300 + 400 = 900 Ω
• A₃₃ = 400 + 500 = 900 Ω
• A₁₂ = A₂₁ = –200 Ω (200 Ω shared)
• A₁₃ = A₃₁ = 0 (no shared resistors)
• A₂₃ = A₃₂ = –400 Ω (400 Ω shared)
• C₁₁ = +9 – 6 = 3 V (two sources in loop 1)
• C₂₁ = +6 – 4 = 2 V (two sources in loop 2)
• C₃₁ = +4 + 1 = 5 V (two sources in loop 3, both negative terminals first)

Result: I₁ = 17.044 mA, I₂ = 10.566 mA, I₃ = 10.252 mA.

Shared currents: I₄ = I₁ – I₂ = 6.478 mA, I₅ = I₂ – I₃ = 0.314 mA.

🧪 Circuit with current sources (Example 2.8)

• Before: 3 current sources.
• After transformation: all converted to voltage sources (Figure 2.16).
• Result: I₁ = 11.551 mA, I₂ = 8.122 mA, I₃ = –0.610 mA (counter-clockwise).

🧪 Approximate transformation (Example 2.9)

• Problem: 1 mA current source with no parallel resistor.
• Solution: add a virtual 1 MΩ resistor (1000× the largest resistor, 1 kΩ), then transform.
• Result: I₁ = 3 mA, I₂ = 2 mA, I₃ = 1 mA.

Key insight: the virtual resistor does not physically exist; it is a mathematical tool to enable transformation.

🔑 Key advantages and limitations

🔑 Advantages of Mesh Matrix

• Speed: no need to write or simplify equations; populate matrices by inspection.
• Accuracy: fewer opportunities for sign errors or algebraic mistakes.
• Scalability: the time savings grow dramatically as the number of loops increases.

⚠️ Limitations and caveats

• Voltage sources only: must transform current sources first.
• Dependent sources: require an extra step (move dependent terms from C to A matrix; see Appendix A.1 for examples); the A matrix loses symmetry.
• Minor loops only: the method does not use "major loops" (loops with elements inside them).

🔄 Comparison with Nodal Matrix

• The excerpt mentions that Nodal Matrix Analysis (Section 2.4) is the node-based counterpart to Mesh Matrix.
• Table 2.1 (referenced but not shown) provides a side-by-side summary of both methods.

Don't confuse: Mesh Matrix is "loop-based" (focuses on loops and uses KVL); Nodal Matrix is "node-based" (focuses on nodes and uses KCL).

🧭 Overview

🧠 One-sentence thesis

Nodal Matrix analysis simplifies circuit solving by allowing node voltages to be calculated directly through matrix inspection without writing KCL equations or performing algebra, though it requires all sources to be current sources.

📌 Key points (3–5)

• What Nodal Analysis does: focuses on solving for node voltages instead of loop currents, using KCL at each node.
• Traditional vs Matrix method: traditional nodal requires writing KCL equations and algebra; Nodal Matrix populates matrices by inspection with no algebra needed.
• Source requirement: Nodal Matrix can ONLY work with current sources; voltage sources must be transformed first.
• Common confusion: when to use Nodal vs Mesh Matrix—choose based on the resulting A matrix size and whether the circuit has more current or voltage sources; avoid approximate transformations when possible.
• Why it matters: Nodal Matrix directly solves for node voltages, avoiding extra calculations needed in Mesh Matrix when node voltages are the goal.

🔄 Traditional Nodal Analysis Method

📝 How traditional nodal works

• Write a KCL equation at each node (except the reference/ground node).
• Replace current through resistors using the equation: current through resistor = voltage difference divided by resistance.
• Solve the resulting equations to find node voltages.

⚠️ Sign convention to avoid mistakes

Key technique: assume all currents are going OUT of the node when writing the KCL equation.

• This ensures the node voltage at the current node is always the first term in the voltage difference expression.
• Example: at Node A, write current as (V_A - V_ref)/R, not (V_ref - V_A)/R.
• This convention reduces sign errors compared to the traditional "sum of currents in = sum of currents out" approach.

🔧 Handling mixed sources

• Traditional nodal analysis can solve circuits with both current and voltage sources without transformation.
• Voltage sources directly set node voltages (no equation needed for that node).
• Current sources appear as terms in the KCL equations.

🧮 Converting to matrix form

After writing KCL equations and simplifying:

• Collect terms with each node voltage.
• Arrange into matrix form: A·B = C, where B contains the unknown node voltages.
• This matrix can be solved, but traditional nodal requires algebra to get to this form.

🎯 Nodal Matrix Analysis Rules

🔑 Core requirements and setup

Nodal Matrix Method: A systematic way to populate the matrix equation [A]·[B] = C (also written [G]·[V] = I) by inspection, where G is conductance, V is node voltage, and I is current.

Rule 1 - Source restriction: You can ONLY have current sources.

• If voltage sources are present, perform source transformation first (standard or approximate).
• This is the opposite of Mesh Matrix, which requires only voltage sources.

Rule 2 - Matrix dimensions: The number of nodes (excluding reference/ground) determines matrix sizes.

• 1 node: A is [1×1], B and C are [1×1]
• 2 nodes: A is [2×2], B and C are [2×1]
• 3 nodes: A is [3×3], B and C are [3×1]
• Pattern continues for more nodes.

Rule 3 - What B matrix contains: Node voltages referenced to ground.

• To verify in simulation: use voltage probe at the node, or multimeter with + at node and - at ground.

🔢 Populating the A matrix (conductance matrix)

Rule 4 - Diagonal cells (positive values):

• Cells A_11, A_22, A_33, etc. contain the sum of conductance values (1/R) of all resistors touching the corresponding node.
• Example: A_11 = sum of (1/R) for all resistors connected to Node A.
• These are always positive.

Rule 5 - Off-diagonal cells (negative values):

• Always have a negative sign.
• Add the conductance values of resistors touching BOTH nodes corresponding to the cell, then multiply by -1.
• Example: A_12 (the Node A-Node B cell) = -1 × (sum of 1/R for resistors between A and B).
• The A matrix is symmetric: A_12 = A_21, A_13 = A_31, etc.

📥 Populating the C matrix (current source matrix)

Rule 6 - Current source placement and signs:

• Current sources go in the row corresponding to the node they touch.
• If a current source touches two nodes, it appears in two rows.
• Sign rule:
  • Current going INTO the node → positive value
  • Current going OUT of the node → negative value

⚖️ Mesh Matrix vs Nodal Matrix Comparison

📊 Key differences

Aspect	Mesh Matrix	Nodal Matrix
Source requirement	ONLY voltage sources	ONLY current sources
Solves for	Loop currents	Node voltages
B matrix assumption	Currents are clockwise (CW = positive)	Voltages referenced to ground
Diagonal A cells	Sum of resistances in each loop	Sum of conductances touching each node
Off-diagonal A cells	-1 × (sum of resistances touching both loops)	-1 × (sum of conductances touching both nodes)
C matrix sign rule	Positive if going clockwise into - sign of voltage source	Positive if current goes INTO node

🤔 Choosing between methods

When to use Nodal Matrix:

• Circuit has more current sources than voltage sources.
• Transforming to all current sources avoids approximate transformations.
• Goal is to find node voltages directly (Mesh requires extra calculations after solving).

When to use Mesh Matrix:

• Circuit has more voltage sources than current sources.
• Generally preferred when other factors are equal, because conductance values (1/R) can create rounding errors with very small numbers.

Decision factors:

1. Compare the resulting A matrix size for each method (prefer smaller).
2. Count voltage vs current sources (prefer the more abundant type).
3. Check if transformations would be approximate (avoid if possible).
4. Consider what the problem asks for (node voltages favor Nodal; loop currents favor Mesh).

⚠️ Common Pitfalls and Practical Considerations

🔴 Rounding errors in Nodal Matrix

Critical issue: Conductance values (1/R) produce very small numbers that cause significant rounding errors.

• The A matrix fractions must be carried to many decimal places (4th or 5th minimum, sometimes 6th or more).
• Example: 1/(1MΩ) + 1/(1Ω) = 1.000001—if you don't go to 6 decimal places, the 1MΩ resistor is essentially removed.
• Small resistors combined with large resistors are especially problematic.
• Even with 4-5 decimal places, node voltage results can differ noticeably from simulation.

Recommendation: Due to this rounding issue, Mesh Matrix is usually preferable when no other factors make Nodal Matrix more desirable.

🔄 Working with mixed-source circuits

Voltage sources in Nodal analysis:

• A voltage source directly forces a node to a specific voltage.
• No KCL equation is needed for that node—just substitute the known voltage value.
• Example: if a 2V source sets V_C = 2V, plug 2V into other node equations and reduce the number of unknowns.

Approximate transformations:

• When a current source lacks a parallel resistor (not in Norton Form), add a virtual resistor 1000 times larger than the largest resistor in the circuit.
• This is not ideal; avoid if possible by choosing Nodal Matrix when it prevents approximate transformations.

🧮 Solving the matrix equation

Standard method: B = (A)^(-1) · C

• Requires calculator or computer to invert the A matrix.

Cramer's Rule alternative:

• Feasible for 2×2 or 3×3 matrices by hand.
• Not realistic for larger matrices without computer assistance.

🔌 Calculating power from results

After solving for node voltages:

• Use P = V · I for each source (the only power equation for sources).
• Determine if a source supplies or dissipates power:
  • Current source: dissipates power only if the voltage across it is positive in the direction of current flow.
  • Negative power in simulation means the source is a power supplier.

Example: if all current sources show negative power, all are supplying power to the circuit.

🛠️ Step-by-Step Example Workflow

🔍 Example scenario: Simple two-node circuit

Given: Circuit with one current source and three resistors at Node A, reference node at ground.

Traditional nodal steps:

1. Write KCL at Node A assuming all currents go OUT: -5mA + (V_A - 0)/4000 + (V_A - 0)/2000 + (V_A - 0)/6000 = 0
2. Simplify: V_A/4000 + V_A/2000 + V_A/6000 = 0.005
3. Solve: V_A = 5.45V

Nodal Matrix steps:

1. Count nodes (excluding reference): 1 node → A is [1×1], B and C are [1×1]
2. Populate A_11: sum of conductances = 1/4000 + 1/2000 + 1/6000
3. Populate C_11: current source going into node = +5mA = 0.005A
4. Solve: V_A = C_11 / A_11 = 5.45V

Key insight: Nodal Matrix creates the same matrix by inspection, skipping all algebra.

🔧 Example scenario: Three-node circuit with voltage source

Given: Circuit with current source, voltage source, and multiple resistors at Nodes A, B, C.

Handling the voltage source:

• Voltage source forces V_C = 2V (no equation needed for Node C).
• Write KCL equations only for Nodes A and B.
• Substitute V_C = 2V into Node B equation.

Matrix result: 2×2 system for V_A and V_B.

Alternative - Mesh Matrix approach:

• Transform current source to voltage source (if in Norton Form).
• Solve for loop currents I1, I2.
• Calculate node voltages from loop currents: V_A = 40 - I1·4000, V_B = V_A - I1·1000.

Comparison: Nodal Matrix gives node voltages directly; Mesh Matrix requires extra calculations after solving for currents.

🎯 Example scenario: Choosing the method

Given: Circuit with 2 voltage sources and 2 current sources, 3 nodes, 3 loops.

Analysis:

• Both methods yield 3×3 A matrix (tie).
• Equal number of voltage and current sources (tie).
• One current source NOT in Norton Form → approximate transformation needed for Mesh Matrix.

Decision: Use Nodal Matrix to avoid approximate transformation.

Result: Nodal Matrix solves directly for V_A, V_B, V_C; Mesh Matrix would require approximate transformation plus extra calculations to get node voltages from loop currents.

🧭 Overview

🧠 One-sentence thesis

The superposition method solves multi-source circuits by calculating the effect of each independent source separately (with all other sources "zeroed out") and then adding the individual results together, which works because linear electrical systems obey the superposition principle.

📌 Key points (3–5)

• What superposition does: breaks a multi-source circuit into simpler sub-problems—one per independent source—then sums the results.
• When it's useful: preferred when multiple sources cannot be transformed without approximation; usually more time-consuming than Mesh or Nodal Matrix methods.
• How to "zero out" sources: replace voltage sources with short circuits (0 V) and current sources with open circuits (0 A), leaving only internal resistance if sources are non-ideal.
• Common confusion: students often think voltage across an open circuit is zero, but it equals the difference between the node voltages on either side (rarely zero).
• Critical limitation: superposition works only for voltage and current—power cannot be calculated by adding sub-problem powers; you must first find total voltage/current, then compute power.

🔍 Linearity and the superposition principle

🔍 What makes a system linear

A linear electrical component or system obeys: V(I₁ + I₂) = V(I₁) + V(I₂) or I(V₁ + V₂) = I(V₁) + I(V₂).

• Plain language: if you cut the input (current or voltage) in half, the output (voltage or current) also cuts in half.
• Why it matters: only linear systems allow superposition; non-linear components (like diodes) violate this rule.
• Example: a resistor is approximately linear—doubling current doubles voltage (Ohm's law)—though heating effects cause small deviations at high temperatures.

⚠️ Non-linear components

• Diode: extremely non-linear; the current-versus-voltage curve does not satisfy the linearity equation.
• Resistive heating elements (e.g., car rear-window defrosters, incandescent lamp filaments): temperature significantly changes resistance, breaking the linear approximation.
• Why we approximate: simplifies problem-solving; engineers accept small errors when temperature effects are negligible.

🔄 Linear systems vs. linear components

• A circuit with a non-ideal diode is non-linear at the component level.
• If the diode is replaced by a constant-voltage battery (constant-drop model), the system becomes linear even though the battery itself doesn't obey the linearity equation (voltage is constant regardless of current).
• The excerpt emphasizes: "the idea of a linear system is what we will focus on" for superposition.

🛠️ The superposition procedure

🛠️ Step-by-step method

1. Count the sources: number of sub-problems = number of independent sources.
2. For each sub-problem: activate only one source; zero out all others.
3. Solve each sub-problem: find the desired voltage or current due to that single source.
4. Add the results: sum all sub-problem voltages (or currents) to get the total.

🔌 How to zero out sources

Source type	How to zero it	Why
Voltage source	Replace with a short circuit (wire)	Voltage across a wire is always 0 V
Current source	Replace with an open circuit (break)	Current through an open circuit is always 0 A

• For non-ideal sources: replace with their internal resistance (Rᵢₙₜ).
  • Non-ideal voltage source: small resistor in series → zero it out by replacing with that resistor.
  • Non-ideal current source: large resistor in parallel → zero it out by replacing with that resistor.
• For ideal sources (this course): internal resistance is 0 Ω (voltage) or ∞ Ω (current), so just use short/open.

🚨 Common mistake: open-circuit voltage

• Don't confuse: voltage across an open circuit is not zero.
• It equals the difference between the node voltages on the two sides of the break.
• Example: if one side is at 5 V and the other at 2 V, the open-circuit voltage is 3 V.

📐 Worked examples and key techniques

📐 Example with ideal sources (Figure 2.34)

• Circuit has a 2 mA current source and a 0.7 V voltage source.
• Sub-problem 1 (2 mA source only): voltage source → short circuit; current flows through 1 kΩ resistor → V₁ = 2 V, I₁ = 2 mA.
• Sub-problem 2 (0.7 V source only): current source → open circuit; no current flows → V₂ ≈ 0 V, I₂ ≈ 0 A (Multisim shows –700 pV, which is machine epsilon noise).
• Total: V = 2 V + 0 V = 2 V; I = 2 mA + 0 A = 2 mA.

📐 Polarity and sign conventions

• Critical: assume the same current direction and voltage polarity for all sub-problems so results can be added algebraically.
• If a source forces current in the opposite direction, the calculated value will be negative.
• Example: if you assume clockwise current but a source drives it counter-clockwise, the sub-problem result is negative; the final sum accounts for direction.

📐 Simplifying circuits in sub-problems

• You may combine resistors (series/parallel) only if the simplification does not eliminate the voltage or current you need to solve for.
• Example (Figure 2.36): 3 kΩ and 4 kΩ in parallel can be combined because Vₓ is not across them; but don't add the result in series with 2 kΩ or you lose Vₓ.

📐 Tricky situations: shorts and opens

1. Short in parallel with resistors: the resistors are "shorted out" (removed). Reason: 0 Ω ∥ R = 0 Ω.
2. Open in series with resistors: the resistor network has no effect (no current flows). Reason: ∞ Ω + R = ∞ Ω.

• Example (Figure 2.38): when a voltage source is zeroed (short), resistors in parallel with it disappear; when a current source is zeroed (open), resistors in series with it disappear.

⚠️ Critical limitation: power

⚠️ You cannot use superposition for power

CAUTION: You CANNOT calculate power using the superposition technique (only voltage or current).

• Why not: power is proportional to voltage × current (or I² or V²), which is non-linear.
• Example (Figure 2.35): P₅ₖΩ = 230 mW in the full circuit, but P₅ₖΩ₁ + P₅ₖΩ₂ = 193 mW + 1.6 mW = 194.6 mW ≠ 230 mW.
• Correct approach: use superposition to find total voltage and current, then calculate power from those totals.
  • Example: I₅ₖΩ = 6.79 mA (from superposition); P₅ₖΩ = I² R = (6.79 mA)² × 5 kΩ = 230 mW ✓

⚠️ Why this happens

• Superposition is valid only for linear relationships.
• Power = V × I is a product, not a sum, so the linearity equation does not apply.

🎯 When to use superposition

🎯 Advantages

• Turns a complicated multi-source circuit into multiple simpler single-source sub-problems.
• Useful when sources cannot be transformed without approximation.
• Important concept across many engineering disciplines (mechanical, chemical, etc.).

🎯 Disadvantages

• Usually more difficult and time-consuming than Mesh Matrix or Nodal Matrix methods.
• Requires careful tracking of polarity and sign conventions across sub-problems.

🎯 Recognizing superposition problems

• Key phrase: "Due to Only" (e.g., "find voltage due to only the 2 mA source").
• This signals you should solve only one sub-problem, not the entire circuit.
• Demonstrates understanding of the superposition technique.

🎯 Skills developed

• Handling circuits with opens and shorts (also used in Thevenin/Norton equivalents and steady-state RLC analysis in Module 3).
• Understanding linear system behavior.

🔧 Practical notes

🔧 Machine epsilon in simulation

• Multisim (and other programs) have a "machine epsilon" (≈10⁻¹⁰ to 10⁻¹⁶).
• If you see extremely small values (e.g., –700 pV = –700 × 10⁻¹² V), assume it is zero.
• Example: MATLAB command eps returns the estimated machine epsilon.

🔧 Ideal vs. non-ideal sources

• This course uses ideal sources: internal resistance is 0 Ω (voltage) or ∞ Ω (current).
• When zeroing out: voltage source → short (wire); current source → open (break).
• Non-ideal sources would leave Rᵢₙₜ in the circuit, but that is not covered here.

🔧 Rounding and precision

• When combining very different resistor values (e.g., 1 MΩ and 1 Ω), you may need 6+ decimal places to avoid rounding errors.
• This is one reason Mesh/Nodal Matrix methods are often preferred over superposition.

🧭 Overview

🧠 One-sentence thesis

Thevenin and Norton equivalent circuits allow any complicated multi-source, multi-resistor circuit to be replaced by a single independent source and one resistor, simplifying analysis when connecting different load resistors.

📌 Key points (3–5)

• What the technique does: replaces a complex circuit "external to" a specific resistor with one source and one resistor (Thevenin form: voltage source in series; Norton form: current source in parallel).
• How to find it: two sub-problems—(1) find Rth by zeroing all sources and combining resistors; (2) find Vth by calculating the open-circuit voltage where the load resistor was removed.
• Common confusion: when finding Rth, the open circuit created by removing the load resistor is not treated as a normal open circuit—do not remove resistors in series with it; think of it as an ohmmeter connection point.
• Why it matters: any resistor value connected to the equivalent circuit will have the same voltage and current as in the original circuit; enables Maximum Power Transfer Theorem (max power when load resistor equals Rth).
• Multiple methods exist: four methods to find Rth (combining resistors, short-circuit current, test source, source transformations), but Method 1 (combining resistors) is standard for circuits without dependent sources.

🔄 Thevenin and Norton forms

🔄 What they are

Thevenin form: a voltage source with a resistor in series.
Norton form: a current source with a resistor in parallel.

• Both forms were defined in Section 2.1.
• The two forms are interchangeable via source transformation: set I_N = Vth / Rth and R_N = Rth.
• Example: after finding Vth = 5.75 V and Rth = 31.25 Ω, the Norton equivalent has I_N = 5.75 / 31.25 A and R_N = 31.25 Ω.

🎯 Why use them

• A circuit with multiple voltage/current sources and many resistors can be replaced by one source and one resistor.
• Once you have the Thevenin equivalent, transforming to Norton is straightforward using the simple method from Section 2.1.
• Usually no need to solve for Norton separately—just find Thevenin and transform.

🛠️ How to find the Thevenin equivalent (Method 1)

🛠️ Setup step

• Identify the resistor (call it R_L) you are finding the equivalent "external to."
• Remove R_L from the circuit, leaving two connection points (called "Nubbies").
• If the problem says "find the Thevenin equivalent circuit" without mentioning R_L, it means R_L has already been removed and you work at the Nubbies.

🔧 Sub-problem 1: Find Rth

• Zero out all independent sources: replace voltage sources with short circuits, current sources with open circuits.
• Combine resistors starting as far away from the Nubbies as possible, moving toward them, until one resistor remains—that is Rth.
• Example (from Example 2.18): Rth = [(8 + 10) || 30] + 20 = 31.25 Ω.
• Caution: the open circuit where R_L was removed is not a normal open circuit when finding Rth. Do not remove resistors in series with it. Think of it as an ohmmeter connection—an ohmmeter injects current, so the open behaves like a current source, not a true open. This is different from finding Vth, where an open circuit behaves normally.

⚡ Sub-problem 2: Find Vth

• Go back to the original circuit (with R_L removed).
• Calculate the open-circuit voltage across the Nubbies—that is Vth.
• Resistors in series with the open circuit have no effect on Vth (no current flows through them, so no voltage drop).
• Example (from Example 2.18): transform the 1 A current source to a voltage source, remove the 20 Ω resistor (in series with the open), solve the single-loop circuit with KVL to find current I = 0.125 A, then Vth = 8 – 18·0.125 = 5.75 V.
• Caution: do not use Rth when solving for Vth. The two sub-problems are completely separate; no information should be passed between them.

🔗 Verification

• Multisim can measure Vth (voltmeter in Volts mode) and Rth (multimeter in Ohms mode).
• When measuring resistance with an ohmmeter, always remove power from the circuit (zero out sources) or the measurement will be wrong or damage the meter.

🔌 Connecting the load and equivalence

🔌 Connecting R_L

• After finding Rth and Vth, connect R_L to the Nubbies of the Thevenin (or Norton) equivalent circuit.
• The voltage across and current through R_L will be identical to the original circuit, regardless of R_L's value.
• Example (from Figure 2.55): connecting an extremely small resistor or an extremely large resistor to the Nubbies produces the same voltage and current in the original, Thevenin, and Norton circuits.

🔌 No R_L case

• If the original circuit had no R_L (only Nubbies), the Thevenin and Norton equivalents also have no R_L—just an open circuit at the Nubbies.
• Any resistor value can be connected later, and the three circuits (original, Thevenin, Norton) will behave identically.

⚡ Maximum Power Transfer Theorem

⚡ The theorem

Maximum Power Transfer Theorem: connecting a load resistor R_L to a circuit transfers maximum power to R_L when R_L = Rth.

• If asked "what value of R_L allows maximum power transfer to R_L?", solve for Rth (sub-problem 1 only).
• The power dissipated in R_L is P_RL = I² · R_L = [Vth / (Rth + R_L)]² · R_L.
• Graphically, plotting power vs. R_L shows a peak at R_L = Rth.
• Example (from Figure 2.56): when Rth = 1000 Ω, power in R_L is maximum at R_L = 1000 Ω.

⚡ Why it works

• When R_L is very small, current is high but voltage across R_L is low → low power.
• When R_L is very large, voltage is high but current is low → low power.
• The sweet spot (maximum power) occurs when R_L = Rth.

🧮 Alternative methods for finding Rth

🧮 Method 1 (standard for ENGR 2431)

• Zero independent sources and combine resistors.
• Limitation: cannot be used if the circuit has dependent sources.
• This is the only method required for ENGR 2431.

🧮 Method 2 (short-circuit current)

• Find Vth from the open-circuit voltage (as in Method 1).
• Find I_N by placing a short circuit across the Nubbies and calculating the short-circuit current.
• Calculate Rth = Vth / I_N.
• Example (from Figure 2.62): I_N = 2.182 A, Vth = 4.174 V → Rth = 1.913 Ω.

🧮 Method 3 (test source)

• Find Vth from the open-circuit voltage (as in Method 1).
• Zero out independent sources, then connect a test voltage source (e.g., 1 V) or test current source (e.g., 1 A) to the Nubbies.
• Solve for the corresponding current (if using test voltage) or voltage (if using test current).
• Calculate Rth = V_Test / I_Test.
• Example (from Figure 2.63): V_Test = 1 V, I_Test = 0.523 A → Rth = 1.913 Ω.

🧮 Method 4 (transform and combine)

• Perform source transformations starting away from the Nubbies, moving toward them, and combine resistors after each transformation.
• Continue until only one source and one resistor remain—that resistor is Rth.
• If the final source is a voltage source, its value is Vth; if a current source, transform to find Vth.
• Example (from Figure 2.64): successive transformations yield Vth = 4.174 V, Rth = 1.913 Ω.

🔗 Connection to other techniques

🔗 Superposition and Thevenin

• The excerpt notes that solving problems with opens and shorts (as done in superposition) is a valuable skill for Thevenin/Norton problems.
• Removing a resistor and analyzing the circuit is similar to the sub-problems in superposition.
• This skill is also used in Module 3 for steady-state analysis of RLC circuits.

🔗 When to use Thevenin

• Thevenin is useful when you want to analyze how different load resistors affect a circuit without re-solving the entire circuit each time.
• It simplifies repeated calculations for varying R_L values.

🧭 Overview

🧠 One-sentence thesis

Capacitors resist instantaneous voltage changes, making them essential for applications requiring voltage smoothing, surge protection, and energy storage/release cycles.

📌 Key points (3–5)

• Core property: The voltage across a capacitor cannot change instantaneously, which makes capacitors ideal for smoothing filters and surge protection.
• Charging and discharging behavior: Capacitors charge exponentially toward a source voltage and discharge exponentially back to zero, with current flowing during transient periods but dropping to zero at steady state.
• Dielectric materials affect capacitance: Different dielectric materials (mica, ceramic, electrolytic) produce different capacitance ranges and voltage ratings.
• Common confusion: Capacitors behave non-linearly (governed by differential equations), but small time intervals allow linear approximations during transient analysis.
• Practical types and selection: Electrolytic capacitors are polarized and offer high capacitance; ceramic and film capacitors are non-polarized with different performance trade-offs.

⚡ How capacitors resist voltage changes

⚡ The fundamental property

The voltage across a capacitor CANNOT change instantaneously.

• This property arises from the relationship between current and voltage in a capacitor, governed by differential equations.
• Because voltage cannot jump suddenly, capacitors smooth out voltage spikes and prevent sudden changes in circuits.
• Why it matters: This makes capacitors well-suited for smoothing filters, surge protectors, and ripple voltage removal in AC-to-DC rectifier circuits.

🔄 Current-voltage relationship

The excerpt presents the governing equations:

• Current through a capacitor: i_C(t) = C · (dv/dt), or numerically i_C = C · (Δv/Δt)
• Voltage across a capacitor: v_C(t) = (1/C) · integral of i_C(t) from t_0 to t, plus initial voltage v_C(t_0)

In plain language:

• Current depends on how fast voltage is changing (the rate of change).
• If voltage changes quickly, current is large; if voltage is constant (Δv = 0), current is zero.
• At DC steady state, voltage is constant, so current through a capacitor is zero.

Don't confuse: Capacitors do not behave like resistors (linear Ohm's Law). Their behavior is exponential, though small time steps allow linear approximation.

🔋 Charging and discharging cycles

🔋 Store phase (charging)

• A capacitor is charged by connecting it to a voltage source (e.g., a 12 V battery) through a resistor.
• Initially (at t = 0+, right after the switch is pressed), the capacitor voltage is still near zero, so maximum current flows.
• Over time, the capacitor voltage rises exponentially toward the source voltage (12 V in the example).
• At steady state (t = ∞), the capacitor voltage equals the source voltage, and current drops to zero.

Example from the excerpt:

• At t = 0+: capacitor voltage is ~1.3 V, current is ~10.7 mA.
• At t = ∞: capacitor voltage is 12 V, current is 0 mA.

🔓 Release phase (discharging)

• When the charging switch is opened and a discharge switch is pressed, the capacitor acts like a battery with rapidly declining voltage.
• The capacitor discharges through a resistor (and possibly an LED in the example), with current flowing in the opposite direction.
• Initially (at t = 0+), the capacitor voltage is still at its charged value (e.g., 12 V), so current is maximum.
• Over time, the capacitor voltage drops exponentially toward zero, and current decreases to zero.

Example from the excerpt:

• At t = 0+: capacitor voltage is 12 V, current is ~10 mA (accounting for ~2 V drop across a green LED).
• At t = ∞: capacitor voltage is 0 V, current is 0 mA (fully discharged).

🕰️ Transient vs steady state

Region	Voltage behavior	Current behavior
Transient (0 < t < ∞)	Changing exponentially	Non-zero, decreasing exponentially
Steady state (t = ∞)	Constant (equals source or zero)	Zero

• The excerpt emphasizes that during transient periods, voltage and current are constantly changing, but at any single "frozen" moment in time, the circuit can be analyzed as a simple DC problem using Ohm's Law, KVL, and KCL.

🧱 Dielectric materials and capacitor types

🧱 Dielectric constant and capacitance

The excerpt provides a table of dielectric constants (relative permittivity, ε_r) for various materials:

• Mica (muscovite): 5.0 to 8.7
• Ceramic materials: vary widely (e.g., alumina 8 to 10)
• Distilled water: 80

How it affects capacitance:

• Capacitance formula: C = (ε_r · ε_0 · A) / d, where ε_0 is the permittivity of free space, A is plate area, and d is separation.
• Higher dielectric constant → higher capacitance for the same geometry.

Example from the excerpt:

• A 20 mm² capacitor with 1 mm mica dielectric has a capacitance range of 0.884 pF (ε_r = 5) to 1.538 pF (ε_r = 8.7).

🔌 Common capacitor types

🔌 Electrolytic capacitors (aluminum or tantalum)

• Polarized: have a positive (anode, longer leg) and negative (cathode, shorter leg) side.
• Must be connected in the correct direction (+ to – in the direction of current flow).
• Capacitance range: typically up to millifarads (mF), much larger than other types.
• Failure mode: when breakdown voltage is exceeded, they "pop" and are damaged (bulging or exploding).
• Don't confuse: These are the only common capacitors that are polarized; reversing polarity will damage them.

🔹 Ceramic disk capacitors

• Non-polarized: can be connected in either direction.
• Capacitance range: low picofarads (pF) to low microfarads (μF).
• Cost: usually cheaper than other ceramic types (e.g., MLCC).
• Voltage: typically only for low voltage ranges.
• Labeling example: "22 4" means 22 followed by 4 zeros in pF, i.e., 220,000 pF.

🎞️ Film capacitors (plastic capacitors)

• Non-polarized: can be connected in either direction.
• Dielectric types: polyester, polystyrene, Teflon, polypropylene, Mylar (PET).
• Cost and performance: more expensive than ceramic, but better performance and reliability.
• Applications: high current, high power, and/or high temperature; precision applications (smaller tolerances).

🛒 Purchasing considerations

The excerpt lists the most important specifications when purchasing capacitors:

1. Capacitance value: the amount of charge storage.
2. Tolerance: how much the actual capacitance may vary from the rated value.
3. Voltage rating: maximum voltage before damage (breakdown voltage).

Common failure causes: operating beyond rated voltage, power surges, excessive heat.

🧮 Numerical approximation and verification

🧮 Linear approximation for small time steps

• The exact relationship is exponential, but if Δt is small, the change in voltage or current can be approximated as linear.
• This allows simpler calculations during transient analysis.

Example from the excerpt:

• If voltage increases by 2 V over 1 msec in a 1 μF capacitor:
  • i_C(0.001 sec) = C · Δv/Δt = 1E-6 F · 2 V / 0.001 sec = 2 mA

🧮 Verification in simulation

The excerpt describes using Multisim to verify equation i_C = C · (Δv/Δt):

• Rearrange to solve for Δt: Δt = C · |v_C(t_f) - v_C(t_0)| / i_C(t_f)
• Compare calculated Δt from probe voltages and currents to the simulation time stamp.
• Results show close agreement, with small errors due to limited simulation resolution and the exponential function being modeled as linear.

Example results from Table 3.3:

Phase	Calculated Δt	Multisim Δt	Close match?
Store (charging)	0.01386 s	0.015 s	Yes, ~8% error
Release (discharging)	0.02086 s	0.018 s	Yes, ~14% error

Don't confuse: The error does not mean the equation is wrong; it reflects the approximation of an exponential curve as linear over small intervals and simulation resolution limits.

🔬 Energy storage and practical examples

🔬 Energy stored in a capacitor

Energy formula: E = (1/2) · C · V²

Example from the excerpt:

• For the mica capacitor with 12 V applied:
  • Minimum capacitance (0.884 pF): E_min = 6.365E-11 J
  • Maximum capacitance (1.538 pF): E_max = 1.1074E-10 J

📸 Practical application: camera flash

The excerpt describes a "store and release" circuit:

• Store phase: When a switch is pressed (e.g., to take a picture), a capacitor charges from a battery.
• Release phase: When the switch is released, the capacitor's stored energy is discharged to power a flash.

This is the same principle as the RC circuits shown in the figures, where:

• One switch (S1) charges the capacitor through a resistor.
• Another switch (S2) discharges the capacitor through a resistor and LED.

🛠️ Practical tips from the excerpt

• Discharging before measurements: Manually discharge capacitors (e.g., by shorting the legs with a screwdriver) to ensure initial voltage v_C(t_0) = 0 for accurate calculations.
• Damaged capacitors: Bulging electrolytic capacitors indicate failure (shown in photos from a refrigerator circuit board). Replacing them fixed the problem.
• Simulation quirks: In Multisim, use a toggle switch (DIPSW1) to reset capacitor voltage to zero between simulations; most other switches do not reset the capacitor due to a Multisim oddity.

🧭 Overview

🧠 One-sentence thesis

Inductors resist changes in current through magnetic effects and behave oppositely to capacitors—current (not voltage) cannot change instantaneously, and they reach steady state with maximum current and zero voltage.

📌 Key points (3–5)

• What inductance is: an effect from the magnetic field around a current-carrying conductor that resists changes in current by generating an opposing electromotive force (EMF).
• Key difference from capacitors: inductors prevent instantaneous current changes (capacitors prevent instantaneous voltage changes); inductors charge by increasing current to a steady state (capacitors charge by increasing voltage).
• Voltage-current relationships: voltage across an inductor equals L times the rate of current change; current is the integral of voltage over time.
• Common confusion: at DC steady state, inductor voltage is zero (because current is constant, so rate of change is zero), but during transient charging/discharging the voltage is not zero.
• Practical hazard: the (L di/dt) term causes dangerous voltage spikes when switching motors or inductive loads, especially damaging to Field Effect Transistors (FETs).

🔌 What inductors are and why they matter

🔌 Usage and prevalence

• Inductors are used less often than capacitors because they are larger, more expensive, and have fewer applications.
• Primary use: filtering applications, especially at high frequencies where op-amp active filters with only capacitors are difficult to achieve.
• Capacitors are often used together with inductors in these applications.

⚙️ Components with inductance

• Even though the component called an "inductor" is not used frequently, many other components have significant inductance: sensors, motors, transformers.
• Understanding inductance is important for working with these components.

🧲 How inductance works

🧲 Definition and physical mechanism

Inductance (unit: Henry, H): an effect resulting from the magnetic field that forms around a current-carrying conductor that tends to resist changes in the current.

• Electric current through a conductor creates a magnetic flux proportional to the current.
• A change in current creates a change in magnetic flux.
• By Faraday's law, this changing flux generates an electromotive force (EMF) that opposes the change in current.
• Inductance (L) is a measure of the amount of EMF generated for a unit change in current.

🧮 What determines inductance

Inductance depends on:

• N: number of turns of wire
• l: average length of each coil
• A: area of each coil
• μ_r: relative permeability of the core material (the material the coil is wrapped around)
• μ_0: permeability of a vacuum (constant = 1.2566370614×10^-6 kg·m/A²·s²)

The formula is: L = (N² · μ_r · μ_0 · A) / l

• Wrapping wire around a material with high relative permeability (e.g., iron μ_r ≈ 100, nickel μ_r ≈ 200) increases the magnetic flux linking the turns, which increases inductance.

⚡ Energy stored in an inductor

The energy stored is: E = (1/2) · L · i²

• Energy depends on inductance and the square of the current.

📐 Voltage-current relationships

📐 Voltage as a function of current change

The voltage across an inductor is: v(t) = L · (di_L / dt)

Numerically approximated as: V = L · (Δi_L / Δt)

• Voltage is proportional to the rate of change of current.
• If current is constant (Δi = 0), then voltage is zero.
• This is why at DC steady state, inductor voltage is zero.

📐 Current as a function of voltage

The current through an inductor is: i_L(t) = (1/L) · ∫[t_0 to t] v_L · dt + i_L(t_0)

• The i_L(t_0) term is the initial current (initial condition), usually assumed to be zero.
• Current is the integral (accumulated area) of voltage over time.
• Numerically, if initial conditions are zero, i_L(t) can be determined by summing the area under the v_L(t) curve.

🔄 Linear approximation in transient mode

• Both i_L(t) and v_L(t) are exponential functions of time, not linear.
• However, if a small Δt is selected during the transient (charging or discharging) phase, a linear approximation can be made.
• Example: if current increases by 1 mA over 2 msec and L = 4 mH, then v_L(0.002 sec) = 0.004 H · 0.001 A / 0.002 sec = 2 mV.

🔁 Inductor behavior vs capacitor behavior

🔁 Key contrasts

Property	Capacitor	Inductor
Cannot change instantaneously	Voltage	Current
Charging increases	Voltage (current → 0)	Current (voltage → 0)
At steady state	Voltage = supply voltage, current = 0	Current = max, voltage = 0

🔁 Current continuity rule

The current through an inductor CANNOT change instantaneously.

• When a switch changes position (e.g., from charging to discharging), the current direction continues in the same direction.
• Example: if current flows clockwise during charging, it will continue to flow (now counter-clockwise in the new loop) during discharging.
• This affects component placement: an LED must be placed in the opposite direction in an RL circuit compared to an RC circuit.

🔁 Voltage polarity during charging and discharging

• During charging (store circuit): inductor voltage is positive (starts at supply voltage, decreases toward zero).
• During discharging (release circuit): inductor voltage is negative (starts at a large negative value, increases toward zero).
• At steady state (both modes): inductor voltage is zero because current is constant.

⚠️ Practical hazards and protection

⚠️ Voltage spikes from (L di/dt)

• The (L di/dt) term causes a lot of damage to electric circuit components.
• Most motors have internal inductance; when a switch turns a motor on or off, the derivative of current becomes very large.
• This results in a voltage spike with little to no current.

⚠️ Damage to transistor switches

• Bipolar Junction Transistors (BJTs) require current to activate, so they are not very susceptible to the (L di/dt) voltage spike.
• Field Effect Transistors (FETs) only require voltage to activate (current is not needed), so the voltage spike is especially damaging to them.
• Example FETs: BS170 (for small DC motors < 500 mA), IRF540 (power MOSFET for higher current).

⚠️ Protection diodes

• A protection diode (also called fly-back, commutating, or snubber diode) can be used to protect a circuit from the (L di/dt) voltage spike.
• FETs often have internal protection diodes, but it is still a good idea to use an additional external protection diode when using a FET in a motor switching application because the internal diode is often not sufficient.

🔧 Circuit design considerations

🔧 Switch type for store/release circuits

• Unlike capacitors, inductors do not need to be manually discharged; when the voltage source is removed, current dissipates quickly.
• Using two separate SPST (single-pole single-throw) switches cannot work for inductors because as soon as the first switch opens, the inductor current would dissipate.
• Therefore, a SPDT (single-pole double-throw) switch is used to immediately switch between the store (charging) and release (discharging) circuits.

🔧 Steady state in DC circuits

• At DC steady state, current through any component is constant (Δi = 0).
• By the voltage equation v_L = L · Δi/Δt, the voltage across an inductor is forced to be zero.
• However, in the transient region (before reaching steady state), the voltage across the inductor is not zero.

🔧 Calculating steady-state current

• Example from the excerpt: when the SPDT switch engages the store circuit for a long time, the inductor reaches steady state with current = 1.2 A and voltage = 0 V.
• The current can be determined from the resistor: I_R = ΔV/R = (12 - 0)/1000 = 1.2 A.

🔧 Voltage at time = 0+ in release circuit

• At the instant the switch moves to the release position, the inductor voltage can be calculated via Kirchhoff's Voltage Law (KVL).
• Example: going clockwise around the loop: -V_L1 - V_R2 - V_LED = 0 → V_L1 = -12 - 2 = -14 V.
• The negative sign occurs because current flows from the negative to the positive terminal as labeled.

🛠️ Physical construction and measurement

🛠️ Types and appearance

• Many inductors appear homemade; making your own inductor is a common practice.
• Toroidal inductors: types wrapped around a circular ring.
• An LCR meter (measures inductance, capacitance, resistance) is useful for verifying homemade inductors.

🛠️ Example component

• The excerpt references a 47 mH inductor used in "project infinity."

🧭 Overview

🧠 One-sentence thesis

Inductors combine using the same rules as resistors—series inductances add directly, parallel inductances combine reciprocally—making it possible to build custom inductance values from available components.

📌 Key points (3–5)

• Why combine inductors: Often difficult to build or purchase the exact inductance value needed, so combining multiple inductors solves the problem.
• How inductors combine: Inductors follow the exact same combination rules as resistors (series add directly, parallel use reciprocal formula).
• Common confusion: Capacitors combine in the opposite fashion—don't mix up inductor and capacitor combination rules.
• Practical calculation: Total inductance can be found by systematically reducing series and parallel groups step-by-step.
• Energy storage depends on total inductance: At steady state, energy stored equals one-half times total inductance times current squared.

🔧 Why and when to combine inductors

🔧 Practical motivations

The excerpt identifies two main scenarios:

• Building custom inductors: When you wind your own inductor, it may be difficult to achieve the exact inductance value you need.
• Using purchased stock: You may have a set of standard inductors on hand but lack the specific value required for your circuit.

In both cases, combining inductors in series and/or parallel lets you reach the target inductance.

🔄 Relationship to resistors

The theory behind combining inductors in series and parallel is done before capacitors in this document because inductors are combined in the exact same process as combining resistors.

• If you already know how to combine resistors, you can apply the same method to inductors.
• Series inductors add: L_total = L1 + L2 + L3 + ...
• Parallel inductors combine reciprocally: 1/L_total = 1/L1 + 1/L2 + 1/L3 + ...

Don't confuse: Capacitors use the opposite rules (the excerpt explicitly states "capacitors combine in opposite fashion").

🧮 Worked example: calculating total inductance

🧮 Individual inductor value

Example 3.2 starts with identical inductors, each wound on a circular iron bar:

• Diameter: 30 mm → cross-sectional area A = π × (0.03 m)² / 4 = 7.0685×10⁻⁴ m²
• Number of turns N = 100
• Average coil length (magnetic path length) l = 100 mm = 0.1 m
• Relative permeability of iron μ_r = 100
• Permeability of free space μ₀ = 1.2566370614×10⁻⁶ (in SI units)

Using the inductance formula (equation 3.6 in the source):

• L = (N² × μ_r × μ₀ × A) / l
• L = (100² × 100 × 1.2566370614×10⁻⁶ × 7.0685×10⁻⁴) / 0.1
• L = 8.88262 mH per inductor

🔗 Combining the network

The circuit in Figure 3.12 shows inductors L1 through L7 arranged in a mix of series and parallel.

The total inductance is calculated as:

• L_total = L1 + (L2 || L3 || [L4 + L5 + L6 || L7])
• Result: L_total = 12.5837 mH

How to read this notation:

• + means series (add directly)
• || means parallel (use reciprocal formula)
• Brackets [ ] group sub-networks that must be simplified first

Example: The innermost group [L4 + L5 + L6 || L7] means "add L4, L5, L6 in series, then combine that result in parallel with L7."

⚡ Energy stored in the inductor network

⚡ Energy formula and steady-state current

Example 3.3 adds a 1 kΩ resistor in series with the inductor network and a 12 V battery.

At steady state:

• The voltage across the total inductance goes to zero (as shown in Section 3.2 for inductor charging circuits).
• All 12 V appears across the resistor.
• Current through the inductor (and resistor) is I_L = ΔV / R = (12 V – 0 V) / 1 kΩ = 12 mA.

Energy stored in an inductor:

• E = (1/2) × L_total × i_L²
• E = (1/2) × 12.5837 mH × (12 mA)²
• E = 0.906 μJ (microjoules)

🔍 Why steady state matters

• During charging, the inductor voltage is non-zero and current is changing.
• At steady state, the inductor acts like a short circuit (zero voltage), so all the supply voltage drops across the resistor.
• The current becomes constant, and the energy stored in the magnetic field reaches its maximum value.

Don't confuse: The energy formula uses the total inductance of the combined network, not individual inductor values.

📋 Summary table: inductor vs capacitor combination rules

Component	Series combination	Parallel combination	Same as...
Inductors	Add directly: L_total = L1 + L2 + ...	Reciprocal: 1/L_total = 1/L1 + 1/L2 + ...	Resistors
Capacitors	Reciprocal (opposite of inductors)	Add directly (opposite of inductors)	Opposite of resistors

The excerpt emphasizes this contrast to prevent confusion when working with mixed RLC circuits.

🧭 Overview

🧠 One-sentence thesis

Capacitors combine in the opposite fashion to resistors and inductors—parallel capacitors add directly while series capacitors combine using the reciprocal formula—and this total capacitance determines energy storage in the network.

📌 Key points (3–5)

• Opposite combination rules: capacitors in parallel add directly (like resistors in series), while capacitors in series use the reciprocal formula (like resistors in parallel).
• Parallel formula: total capacitance equals the sum of all individual capacitances.
• Series formula: for two capacitors use the product-over-sum; for three or more use the reciprocal sum.
• Common confusion: the combination rules are reversed compared to resistors and inductors—don't apply resistor rules directly to capacitors.
• Energy calculation: total energy stored requires both the total capacitance and the steady-state voltage across the network.

🔄 How capacitor combination differs from resistors and inductors

🔄 The opposite-fashion rule

When combining capacitors, the equation for combining parallel resistors is used for capacitors in series and the equation for combining resistors in series (simply adding them together) is used for capacitors in parallel.

• This is the core principle: capacitor combination formulas are "flipped" relative to resistors and inductors.
• Resistors/inductors in series add directly; capacitors in series use reciprocals.
• Resistors/inductors in parallel use reciprocals; capacitors in parallel add directly.
• Don't confuse: always remember that capacitors behave oppositely to resistors when combining.

🧩 Why this matters

• If you purchased some capacitors but don't have the exact size you need, you can combine them to achieve the desired total capacitance.
• The excerpt notes that inductors combine exactly like resistors, so capacitors are the exception.

➕ Combining capacitors in parallel

➕ Parallel capacitor formula

The equation for multiple capacitors in parallel is:

C_Total = C1 + C2 + C3 + ... + CN

• Simply add all capacitance values together.
• This is the same process as combining resistors in series.
• Example: two capacitors of 100 microfarads each in parallel give 200 microfarads total.

🔌 Recognizing parallel capacitors

• The excerpt emphasizes that two components are in parallel if "there are 2 connections (or they share 2 nodes)."
• The circuit drawing can be tricky; sometimes you need to redraw the circuit to see that capacitors share the same two nodes.
• Example: in Example 3.4, the 6 microfarad capacitor is in parallel with a network of other capacitors "due to the tricky way it is drawn," but they share two nodes.
• Tip: as long as you connect the two ends of any component to the same nodes, it doesn't matter where on those nodes you connect them.

➗ Combining capacitors in series

➗ Series capacitor formulas

For two capacitors in series:

C_Total = (C1 · C2) / (C1 + C2)

Note: This equation ONLY works with 2 capacitors.

For three or more capacitors in series:

C_Total = 1 / (1/C1 + 1/C2 + 1/C3 + ... + 1/CN)

• The two-capacitor formula is a shortcut (product over sum).
• For three or more, you must use the reciprocal sum.
• This is the same process as combining resistors in parallel.

🧮 Step-by-step combination strategy

The excerpt provides a systematic approach:

1. Start as far away from the terminals (Nubbies) as possible.
2. Combine capacitors step by step, working inward.
3. Use the + symbol for parallel and the || symbol for series to write the combination expression clearly.
4. Leave off the prefix (e.g., micro) during calculations to save time and minimize errors, then add it back at the end.

Example from the excerpt: in Example 3.3a, C1 and C2 (both on the far left) are in parallel and add to 200 microfarads; then all three resulting capacitors are in series, giving C_Total = 40 microfarads.

🔍 Verification in simulation

• The total capacitance cannot be measured directly in Multisim.
• To verify: build two circuits—one with the capacitor network, one with a single capacitor equal to C_Total—and add a large resistor so voltage changes slowly.
• If the voltage readings are the same at the same moments in time as they charge, the combination was done correctly.
• The excerpt shows this method in Figure 3.14 and Figure 3.18.

⚡ Energy storage in capacitor networks

⚡ Energy formula and required values

The energy stored in a capacitor network is:

E = (1/2) · C_total · V_c²

• Two unknowns must be found: C_total (from combining capacitors) and V_c (the voltage across the network).
• The excerpt calls these "great add-on questions"—instead of just asking for total capacitance, a problem may ask for total energy.

🔋 Finding steady-state voltage

• At steady state (as time approaches infinity), the voltage across the capacitor network equals the source voltage in simple circuits.
• The excerpt references capacitor charging characteristics from Section 3.1 and Table 3.2.
• Example: in Example 3.4, V_c = 12 V at steady state, so E = (1/2) · 4.6023×10⁻⁶ · 12² = 0.331 millijoules.
• Don't confuse: more complicated circuits (like those in Section 3.6) require more extensive calculations to find V_c.

📐 Important companion equations

The excerpt reminds you to always remember:

• Q = C · V (charge equals capacitance times voltage)
• E = (1/2) · C · V² (energy stored)

These are essential for solving any capacitor problem beyond just finding total capacitance.

🧪 Worked example walkthrough

🧪 Example 3.4 summary

The problem asks for the total energy stored in a capacitor network at steady state.

Given circuit: multiple capacitors (1, 2, 3, 4, 5, 6, 7, 8 microfarads) arranged in a complex series-parallel combination with a 12 V source.

Step 1: Write the combination expression using the opposite-fashion rules:

C_total = 1 + 6 || {8 + 2||(3+4)||5||7}

(Remember: + means parallel, || means series for capacitors.)

Step 2: Combine step by step, starting from the innermost grouping:

• 3 + 4 = 7 microfarads (parallel)
• 2 || 7 || 5 || 7 (series combination)
• Add 8 microfarads (parallel)
• 6 || (result) (series)
• Add 1 microfarad (parallel)
• Final result: C_total = 4.6023 microfarads

Step 3: Find V_c at steady state = 12 V (equals source voltage).

Step 4: Calculate energy: E = (1/2) · 4.6023×10⁻⁶ · 12² = 0.331 millijoules.

🎨 Redrawing circuits for clarity

• The most difficult part is visualizing connections, especially when the 6 microfarad capacitor is in parallel with a network "due to the tricky way it is drawn."
• The excerpt suggests redrawing the circuit to make parallel connections clearer.
• Example: in Figure 3.17, the circuit is redrawn with color-coded nodes (Blue Node) to show that moving a wire from one point on a node to another point on the same node doesn't change the circuit.
• This helps you see that components share two nodes and are therefore in parallel.

🧭 Overview

🧠 One-sentence thesis

Single-loop RC and RL circuits with one voltage source network follow exponential charging and discharging equations governed by a time constant (tau), which determines how quickly capacitors store voltage or inductors store current during transient behavior.

📌 Key points (3–5)

• Time constant (tau): determines charging/discharging speed; tau = R·C for RC circuits, tau = L/R for RL circuits.
• Exponential behavior: capacitor voltage in charging has a (1 - e^(-t/τ)) term (starts at zero, rises); inductor voltage in charging has an e^(-t/τ) term (starts high, falls to zero).
• Common confusion: resistor voltage exponential terms are opposite for RC vs RL circuits—RC resistors decay (e^(-t/τ)), RL resistors rise (1 - e^(-t/τ))—so always solve for loop current first, then use Ohm's Law to find resistor voltage.
• Single-loop requirement: all equations apply only when one voltage source network exists and all components share the same current (single loop); multiple resistor/capacitor/inductor networks can be combined if they are in series or parallel without other component types between them.
• Discharging circuits: capacitor voltage and inductor current both decay exponentially (e^(-t/τ)) from their steady-state charging values.

⏱️ Time constant fundamentals

⏱️ What tau measures

Time constant (tau or τ): the parameter that determines how fast a capacitor or inductor will be charged or discharged.

• For RC circuits: tau = R_eq · C_total (resistance times capacitance).
• For RL circuits: tau = L_total / R_eq (inductance divided by resistance).
• R_eq is the total resistance "seen" by the capacitor or inductor.
• If only one resistor exists, R_eq equals that resistor's value; multiple resistors in a single loop are summed.
• C_total and L_total are the combined capacitance or inductance; if only one capacitor or inductor exists, use its value directly.
• Example: a circuit with R = 20Ω and C = 0.01F has tau = 0.2 seconds; a circuit with L = 1H and R = 5Ω has tau = 0.2 seconds.

⏱️ Why tau matters

• Tau sets the timescale for exponential growth or decay.
• At t = tau (one time constant), an RC capacitor reaches approximately 63.2% of its final voltage (1 - e^(-1) ≈ 0.632).
• At t = tau, an RL inductor voltage has decayed to approximately 36.8% of its initial value (e^(-1) ≈ 0.368).
• Larger tau means slower charging/discharging; smaller tau means faster response.

🔋 Charging circuits (storing energy)

🔋 Capacitor voltage during charging

• Equation: v_C(t) = V_S (1 - e^(-t/τ))
• V_S is the source voltage.
• The capacitor starts at zero volts and exponentially approaches V_S as time increases.
• The (1 - e^(-t/τ)) term reflects "starting low, rising to steady state."
• To solve for time given voltage: t = -ln(1 - v_C(t)/V_S) · τ

🔋 Inductor voltage during charging

• Equation: v_L(t) = V_S (e^(-t/τ))
• The inductor starts with voltage equal to V_S and exponentially decays toward zero.
• The e^(-t/τ) term reflects "starting high, falling to zero."
• To solve for time given voltage: t = -ln(v_L(t)/V_S) · τ
• Don't confuse: inductor voltage falls while capacitor voltage rises during charging.

🔋 Loop current in charging circuits

• RC circuit current: i(t) = V_S (e^(-t/τ)) / (sum of all resistors)
• RL circuit current: i(t) = V_S (1 - e^(-t/τ)) / (sum of all resistors)
• The excerpt emphasizes: always calculate loop current first, then use Ohm's Law (v = i·R) to find any resistor voltage.
• Why this approach: resistor exponential terms are opposite for RC vs RL, causing confusion if you try to write resistor voltage equations directly.
• Example: in an RC circuit, resistors start at maximum voltage and decay (e^(-t/τ)); in an RL circuit, resistors start at zero and rise (1 - e^(-t/τ)).

🔋 Multiple networks in series

• If multiple resistor, capacitor, or inductor networks exist but are separated by other component types, you can rearrange the order (for voltage-across-component calculations, not node voltages).
• Combine networks of the same type into single equivalent values before applying the equations.
• Inductor voltage distribution: use Voltage Divider Rule (VDR)—split total inductor voltage proportionally by inductance, just like resistors.
• Capacitor voltage distribution: split total capacitor voltage inversely proportional to capacitance (similar to Current Divider Rule style)—v_C1 = v_C_total · (C_total / C1).
• Example: two inductors in series (3H and 1.714H) with 12V total across them → the 3H inductor gets 12 · (3 / 4.714) ≈ 7.636V.

🔌 Discharging circuits (releasing energy)

🔌 Capacitor voltage during discharging

• Equation: v_C(t) = v_C_charge_ss (e^(-t/τ))
• v_C_charge_ss is the steady-state voltage the capacitor was charged to (equals V_S if only one capacitor network and one voltage source in the charging circuit).
• The capacitor voltage cannot change instantaneously, so it starts at v_C_charge_ss and decays exponentially.
• Use the time constant from the discharge circuit (tau = R_eq · C, where R_eq is the total resistance in the discharge path).
• Example: a capacitor charged to 12V with discharge tau = 0.1s → at t = 0.05s, v_C = 12 · e^(-0.05/0.1) ≈ 7.278V.

🔌 Inductor current during discharging

• Equation: i_L(t) = i_L_charge_ss (e^(-t/τ))
• i_L_charge_ss is the steady-state current the inductor was charged to (equals V_S / R_total if only one inductor network and resistors in series in the charging circuit).
• The inductor current cannot change instantaneously, so it starts at i_L_charge_ss and decays exponentially.
• Use the time constant from the discharge circuit (tau = L / R_eq, where R_eq is the total resistance in the discharge path).
• Example: an inductor charged to 2.4A with discharge tau = 0.1s → at t = 0.05s, i_L = 2.4 · e^(-0.05/0.1) ≈ 1.456A.

🔌 Capacitor current and inductor voltage in discharging

• Capacitor current: i_C(t) = (v_C_charge_ss / R_total) (e^(-t/τ))
• Inductor voltage: v_L(t) = i_L(t) · R_total (use the current from the exponential decay equation and multiply by total resistance).
• The excerpt notes that inductor voltage can be negative during discharge (e.g., -24V at t=0 in the example), reflecting the direction of the induced voltage opposing the current change.
• Don't confuse: both capacitor voltage and inductor current decay with the same exponential form (e^(-t/τ)) during discharge.

🔌 Limitation: one voltage source network only

• All discharging equations (3.21–3.24) require only resistors and one voltage source network in the discharge circuit.
• Adding a second voltage source (e.g., an LED modeled as a 2V battery) violates this requirement and invalidates the equations.
• The capacitor acts as a dissipating voltage source; the inductor acts as a dissipating current source.
• Example: adding an LED to the discharge circuit in Figure 3.25 causes significant differences in measured values (Figure 3.26), and calculating those values is beyond the scope of the excerpt.

📊 Measurement and verification

📊 Using Multisim oscilloscope

• The excerpt emphasizes that oscilloscope plots (voltage or current vs. time) are more accurate than instantaneous probe values for transient analysis.
• Current clamps (XCP1, XCP2) measure current; the oscilloscope displays in Volts, but with a 1mV/mA conversion, the reading directly equals Amps.
• Voltage measurements: connect oscilloscope terminals directly to nodes; the negative input can be left unconnected only if one side of the component is grounded.
• Cursors on the oscilloscope allow precise time and value measurements (e.g., T1 at t=0, T2 at a later time, delta time T2-T1 is the elapsed time t in the equations).

📊 Comparing calculations to simulation

• The excerpt provides multiple examples (Tables 3.6, Examples 3.5–3.9) showing that transient equation calculations closely match Multisim oscilloscope measurements.
• Example workflow: measure tau from component values → use exponential equations to calculate voltage or current at a specific time → compare to oscilloscope cursor readings.
• Typical agreement: calculated values match Multisim within a few percent (e.g., 268.47 mA calculated vs. 268.385 mA measured).
• Don't confuse: absolute simulation time (e.g., 84.615 ms) is not the same as the time variable t in the equations; t is the elapsed time since the switch moved (delta time T2-T1).

📊 Steady state vs. transient

• Steady state: the value when time approaches infinity; the circuit has stabilized and stopped changing.
• Transient region: the time period between t=0 (switch action) and steady state, where exponential changes occur.
• Example: in an oscilloscope plot, the transient region is approximately between two cursors; after the second cursor, values stabilize (e.g., inductor voltage ≈ 0V, capacitor voltage nearly constant).
• The excerpt notes that DC steady state analysis for more complicated circuits (multiple loops, multiple sources, RLC combinations) will be covered in Section 3.6, but that section's content is not included in this excerpt.

🧭 Overview

🧠 One-sentence thesis

DC steady state analysis simplifies complex circuits containing capacitors and inductors by replacing capacitors with open circuits and inductors with short circuits, allowing engineers to calculate voltages, currents, energy, and charge after transients have settled.

📌 Key points (3–5)

• What steady state means: the DC value when time approaches infinity, after the circuit stabilizes and stops changing at the end of the transient region.
• The two replacement rules: at DC steady state, capacitors act like open circuits and inductors act like short circuits.
• Common confusion: remember the "tricky situations"—a short in parallel removes resistors from the circuit; an open in series makes the resistor network have no effect.
• How to extract information: after solving the simplified circuit, use the voltage across the open (capacitor location) or current through the short (inductor location) to calculate energy, charge, and power.
• Filter classification: DC steady state analysis reveals whether a circuit is a Low Pass Filter (LPF, passes DC through) or High Pass Filter (HPF, blocks DC).

🔄 What DC steady state means

⏱️ Steady state definition

The "steady state" value for a DC circuit is the value when time approaches infinity. In other words, it is the DC value after it stabilizes and stops changing at the end of the transient region.

• The transient region is the period during which voltages and currents are still changing.
• After the transient region ends, values stabilize to constant DC levels.
• Example: In an oscilloscope plot, the inductor voltage stabilizes to approximately zero after the second cursor—0 V is its steady state value; the capacitor voltage is still changing slightly but approaching its steady state.

🎯 Why this matters

• Steady state analysis lets you solve complicated circuits with multiple loops, multiple voltage sources, current sources, and networks of capacitors and inductors.
• The method builds on superposition techniques from earlier sections, but now reactive components are replaced by simpler equivalents.

🔧 The two replacement rules

🔌 Capacitors become open circuits

• At DC steady state, replace every capacitor with an open circuit.
• An open circuit means no current flows through that branch.
• Voltage can still exist across the open terminals.

🔗 Inductors become short circuits

• At DC steady state, replace every inductor with a short circuit.
• A short circuit means the voltage across it is forced to zero.
• Current can still flow through the short.

🧩 Why these rules work

• Capacitors block DC current after charging is complete (no further change in voltage → no current).
• Inductors allow DC current to flow freely once the magnetic field is established (no further change in current → no voltage drop).

⚠️ Tricky situations when simplifying circuits

⚡ Short in parallel with resistors

• A short circuit in parallel with one or more resistors causes the resistors to be removed (shorted out) from the circuit.
• Reasoning: a short is a wire (0 Ω resistor); combining 0 Ω in parallel with R gives total resistance = (0·R)/(0+R) = 0 Ω.
• Example: If an inductor becomes a short and is in parallel with resistors R5 and R6, those resistors are removed from the circuit.

🚫 Open in series with resistors

• An open circuit in series with one or more resistors results in the resistor network having no effect on the circuit.
• Reasoning: an open is an infinite-ohm resistor; ∞ + R = ∞, so the entire branch is open.
• Example: If a capacitor becomes an open and is in series with resistor R4, R4 is removed because no current flows through that branch.

📊 Extracting circuit information after simplification

🔋 Voltage and current at component locations

After solving the simplified circuit (using nodal analysis, mesh analysis, or other methods):

Component	Voltage	Current
Capacitor	Voltage across the open circuit terminals where the capacitor was	Forced to be 0 (open circuit)
Inductor	Forced to be 0 (short circuit)	Current through the short circuit where the inductor was

⚡ Energy stored in reactive components

• Capacitor energy: Calculate the voltage across the capacitor location, then plug into E = ½ C · V²
• Inductor energy: Calculate the current through the inductor location, then plug into E = ½ L · I²
• Example (from Example 3.10): After finding node voltages V_A = 20 V and V_B = 12 V, capacitor voltage V_C = V_B – V_A = –8 V, so E_C1 = ½ (300 μF) · (–8)² = 9.6 mJ.

🔌 Charge on capacitors

• Capacitor charge: Calculate voltage, then plug into Q = C · V
• Inductor charge: Not applicable
• Example (from Example 3.11): V_C1 = 6.91 V across a 1 μF capacitor → Q1 = 1 μF · 6.91 V = 6.91 μC.

🔥 Power analysis

• Capacitors and inductors have zero real power (P = 0 W) at steady state; they only have reactive power (Q in VARs), covered in AC circuits.
• Only resistors and sources have real power in DC steady state.
• Verify the power balance: sum of power supplied = sum of power dissipated.
• Example (from Example 3.11): The 12 V battery supplies 61.1 mW; resistors and the 2 V battery dissipate a total of 61.1 mW, confirming the balance.

🎛️ Filter classification using DC steady state

🔽 Low Pass Filter (LPF)

A circuit that "passes through" DC to the output is called a Low Pass Filter (LPF).

• At DC (0 Hz), the output voltage is non-zero.
• When you perform steady state analysis and the output voltage is approximately equal to (or a fraction of) the input, the circuit is an LPF.
• Capacitor-based LPF: Place the capacitor in parallel with the output resistor. The capacitor becomes an open, so current flows through the output resistor and voltage appears across it.
• Inductor-based LPF: Place the inductor in series with the input. The inductor becomes a short, allowing DC to pass through to the output.
• Example: In Figure 3.33, the first two circuits use inductors in series; the third and fourth use capacitors in parallel with the output path.

🔼 High Pass Filter (HPF)

If the DC signal is not passed through to the output it is called a High Pass Filter (HPF).

• At DC (0 Hz), the output voltage is zero (or very close to zero).
• When you perform steady state analysis and the output voltage is 0 V, the circuit is an HPF.
• Capacitor-based HPF: Place the capacitor in series with the output. The capacitor becomes an open, blocking current to the output resistor → V_out = 0 V.
• Inductor-based HPF: Place the inductor in parallel with the output path. The inductor becomes a short, diverting all current away from the output → V_out = 0 V. Alternatively, take the output directly across the inductor; the short forces V_out = 0 V.
• Example: In Figure 3.32, all four circuits show 0 V at the output, confirming they are HPF circuits.

🔄 Converting between HPF and LPF

• Capacitors and inductors are opposites when it comes to filtering.
• You can switch a capacitor for an inductor (or vice versa) to convert a circuit between HPF and LPF.
• Example: The first two circuits in Figure 3.33 were created by replacing the capacitor in the corresponding HPF circuits with an inductor.

🔌 Passive vs active filters

• Passive filters: Created from capacitors, inductors, and resistors; no external power required (the input source provides all power).
• Active filters: Created with transistors or Operational Amplifiers (Op Amps); require external power to function.
• Passive filters are the type discussed in this section.
• Active filters have many advantages and are covered in specialized resources (e.g., chapter 20 of "Opamps for Everyone").

🧮 Step-by-step problem solving

📝 General procedure

1. Replace components: Capacitors → open circuits; inductors → short circuits.
2. Simplify the circuit: Remove resistors shorted out by inductor shorts; remove resistor branches opened by capacitor opens.
3. Solve the simplified circuit: Use nodal analysis, mesh analysis, or other DC circuit techniques (similar to superposition problems from section 2.5).
4. Extract component values: Use the solved voltages and currents to find capacitor voltages, inductor currents, energy, charge, and power.

🧪 Example 3.10 walkthrough

• Goal: Determine the energy in every reactive component.
• Step 1: Replace capacitor C1 with an open, inductors L1 and L2 with shorts.
• Step 2: Redraw the circuit; identify it as a 2-node nodal matrix problem (nodes V_A and V_B).
• Step 3: Solve the matrix → V_A = 20 V, V_B = 12 V.
• Step 4: Calculate capacitor voltage: V_C = V_B – V_A = –8 V → E_C1 = ½ (300 μF) · (–8)² = 9.6 mJ.
• Step 5: Calculate inductor currents using Ohm's law and KCL: I_L2 = 1.6 A, I_L1 = 2.4 A.
• Step 6: Calculate inductor energies: E_L1 = ½ (1 mH) · (2.4)² = 2.88 mJ; E_L2 = ½ (2 mH) · (1.6)² = 2.56 mJ.

🧪 Example 3.11 walkthrough

• Goal: Determine capacitor charges and verify power balance.
• Step 1: Replace capacitors with opens, inductors with shorts; simplify the circuit (remove R4 in series with open; remove R5 and R6 shorted out by L2).
• Step 2: Solve as a 2-loop mesh matrix problem → I1 = 5.091 mA, I2 = 1.636 mA.
• Step 3: Calculate capacitor voltages: V_C1 = (I1 – I2) · R2 = 6.91 V; V_C2 = 0 V (both sides of the open are at the same node).
• Step 4: Calculate charges: Q1 = 1 μF · 6.91 V = 6.91 μC; Q2 = 2 μF · 0 V = 0 C.
• Step 5: Perform power analysis: sum of power supplied = 61.1 mW (from 12 V battery); sum of power dissipated = 3.272 mW (2 V battery) + 25.92 mW (R1) + 8.092 mW (R2) + 23.87 mW (R3) = 61.1 mW.
• Step 6: Verify with Multisim (power is zero for all capacitors and inductors; only resistors and sources have real power).

📋 Summary table

📊 DC Steady State Summary

Aspect	Capacitor in DC steady state	Inductor in DC steady state
Model	Replace with open circuit	Replace with short circuit
Voltage	Voltage across open circuit terminals	Forced to be 0 (short circuit)
Current	Forced to be 0 (open circuit)	Current through the short circuit
Energy	Calculate voltage, plug into E = ½ C · V²	Calculate current, plug into E = ½ L · I²
Charge	Calculate voltage, plug into Q = C · V	Not Applicable

🔍 Don't confuse

• Open vs short: An open blocks current but can have voltage; a short allows current but forces voltage to zero.
• Transient vs steady state: Transient is the changing period; steady state is the final, stable value.
• Real power vs reactive power: At DC steady state, capacitors and inductors have zero real power (P = 0 W); resistors and sources have real power.

🧭 Overview

🧠 One-sentence thesis

Passive filters require no external power and rely solely on the input source, whereas active filters use transistors or operational amplifiers and require external power, with active filters offering many advantages despite some situations favoring passive designs.

📌 Key points (3–5)

• What passive filters are: filters built from electrical components that need no external power—the input source provides all power needed.
• What active filters are: filters created with transistors or Op Amps that require external power to function.
• Common confusion: voltage reduction in circuits—reduction may be due to multiple resistors splitting current into different paths, not necessarily due to the capacitor or inductor.
• When to use each type: while passive filters are preferred in some situations, active filters have many advantages overall.

🔌 Passive vs Active Filters

🔌 Passive filters defined

Passive filters: analog filters created from electrical components that do not require external power to function; the input source provides all the power needed.

• The term "passive" means no external power is required to make the devices work.
• All energy comes from the input signal itself.
• Example: a filter built only with resistors, capacitors, and inductors is passive.

⚡ Active filters defined

Active filters: filters created with transistors or Operational Amplifiers (Op Amps) that require external power for them to function.

• These filters need a separate power supply beyond the input signal.
• They use active components (transistors, Op Amps) rather than only resistors, capacitors, and inductors.
• The excerpt mentions that a good place to learn about active filters is chapter 20 of the eBook "Opamps for Everyone."

📊 Comparison table

Filter Type	Components	External Power Needed	Advantages
Passive	Resistors, capacitors, inductors	No—input source provides all power	Preferred in some situations
Active	Transistors or Op Amps (plus other components)	Yes—requires external power supply	Many advantages (not detailed in excerpt)

⚠️ Common Confusion About Voltage Reduction

⚠️ What causes voltage reduction

• The excerpt emphasizes that voltage reduction in circuits is not always due to capacitors or inductors.
• In the examples given (second and fourth circuits), voltage is reduced, but the reduction is due to multiple resistors in the circuit that cause current to split into different paths.
• Don't confuse: a capacitor or inductor being present does not automatically mean it is causing the voltage drop.

🔍 Circuit behavior examples

The excerpt describes four circuit examples (Figure 3.33, Passive Low Pass Filter examples):

• 3rd circuit: output voltage is taken directly across the battery because no current flows through the resistor, so there is no voltage drop.
• 4th circuit: the capacitor acts as an open circuit, causing it to have no effect on the output.
• 2nd and 4th circuits: voltage is reduced, but this is due to resistor configuration (current splitting), not the reactive components themselves.

🎯 When to Choose Each Type

🎯 Practical considerations

• The excerpt states that "there are some situations where passive filters are preferred."
• However, it also notes that "active filters have many advantages."
• The specific advantages and situations are not detailed in this excerpt.
• For further study on active filters, the excerpt recommends chapter 20 of "Opamps for Everyone."