Apply superposition theorem: Worked examples — Hardware deep dive

This is a child of the parent topic. Everything here leans on Linearity and homogeneity, Kirchhoff's Voltage and Current Laws, and Voltage divider and current divider.

The scenario matrix

Every superposition problem is one (or a mix) of these case classes. The examples below are labelled with the cell they cover, and together they fill every row.

#	Case class	What is special	Covered by
A	Two voltage sources, same polarity	contributions add	Ex 1
B	Two voltage sources, opposing polarity	one contribution is negative → partial cancel	Ex 2
C	Mixed V + I source	one divider, one parallel-R step	Ex 3
D	A source contributing exactly zero	balanced bridge / no path	Ex 4
E	Degenerate input: killed source shorts out a resistor	topology collapses	Ex 5
F	Limiting value: a resistor → 0 or → ∞	check the formula stays sane	Ex 6
G	Word problem (real world)	translate words → circuit	Ex 7
H	Exam twist: a dependent source present	must NOT be killed	Ex 8

Ex 1 — Two voltage sources, same polarity (Cell A)

Forecast: both sources push node $A$ up, so both contributions are positive and $V_{A}$ lands somewhere between the two source-driven pulls. Guess a number before continuing.

Kill $V_{2}$ (short it). Now $V_{1}$ drives $R_{1}$ into the parallel combination $R_{2} ∥ R_{3}$ . Why this step? With $V_{2} = 0$ , its terminal is a wire to ground, so $R_{2}$ now runs from node $A$ straight to ground, in parallel with $R_{3}$ . $R_{2} ∥ R_{3} = \frac{2 \cdot 4}{2 + 4} = \frac{8}{6} = 1.333 \dots Ω$ $V_{A, 1} = V_{1} \cdot \frac{R _{2} ∥ R _{3}}{R _{1} + R _{2} ∥ R _{3}} = 10 \cdot \frac{1.3333}{2 + 1.3333} = 10 \cdot 0.4 = 4 V$ Why the divider? $R_{1}$ and the parallel block form a simple two-resistor divider from $10 V$ to ground (Voltage divider and current divider).
Kill $V_{1}$ (short it). By identical structure ( $R_{1} ∥ R_{3}$ from node to ground, driven through $R_{2}$ ): $R_{1} ∥ R_{3} = \frac{2 \cdot 4}{6} = 1.3333 Ω, V_{A, 2} = 4 \cdot \frac{1.3333}{2 + 1.3333} = 4 \cdot 0.4 = 1.6 V$
Add algebraically. Both push $A$ up → same sign: $V_{A} = 4 + 1.6 = 5.6 V$

Worked example Verify

Direct KCL at $A$ : $\frac{V _{A} - 10}{2} + \frac{V _{A} - 4}{2} + \frac{V _{A}}{4} = 0$ . Multiply by 4: $2 (V_{A} - 10) + 2 (V_{A} - 4) + V_{A} = 0 \Rightarrow 5 V_{A} = 28 \Rightarrow V_{A} = 5.6 V$ ✓ Units: volts, correct.

Ex 2 — Opposing voltage sources (Cell B)

Forecast: $V_{1}$ still pulls the node up; the reversed $V_{2}$ now pulls it down. So one contribution should be negative and the total is less than in Ex 1. Guess.

$V_{1}$ alone — unchanged from Ex 1: $V_{A, 1} = + 4 V$ . Why unchanged? Killing sources ignores their polarity; a shorted source is a wire either way. Only the active source's polarity matters.
$V_{2}$ alone. The magnitude of the divider is the same as Ex 1 ( $1.6 V$ ), but because $V_{2}$ 's minus faces the node, it drives node $A$ negative: $V_{A, 2} = - 1.6 V$ Why the sign? Reversing the battery flips the sign of the whole contribution — this is exactly Linearity and homogeneity: scaling a source by $- 1$ scales its response by $- 1$ .
Add algebraically. $V_{A} = 4 + (- 1.6) = 2.4 V$

Worked example Verify

KCL with reversed source: $\frac{V _{A} - 10}{2} + \frac{V _{A} + 4}{2} + \frac{V _{A}}{4} = 0 \Rightarrow 2 (V_{A} - 10) + 2 (V_{A} + 4) + V_{A} = 0 \Rightarrow 5 V_{A} = 12 \Rightarrow V_{A} = 2.4 V$ ✓

Ex 3 — Mixed voltage + current source (Cell C)

Forecast: the voltage source gives a divider; the current source gives $I \times R_{∥}$ . You already know the parent answer is $14.4 V$ — predict the current before computing.

$12 V$ alone (open the 3 A). Divider: $V_{A, 1} = 12 \cdot \frac{6}{4 + 6} = 7.2 V$
$3 A$ alone (short the 12 V). Now $R_{1} ∥ R_{2}$ from node to ground: $R_{1} ∥ R_{2} = \frac{4 \cdot 6}{10} = 2.4 Ω, V_{A, 2} = 3 \cdot 2.4 = 7.2 V$
Add: $V_{A} = 7.2 + 7.2 = 14.4 V$ .
Current down $R_{2}$ : superpose the current directly. $I_{2} = \frac{V _{A, 1}}{6} + \frac{V _{A, 2}}{6} = 1.2 + 1.2 = 2.4 A$

Worked example Verify

$V_{A} / R_{2} = 14.4/6 = 2.4 A$ ✓ — computing the current from the total voltage agrees with adding the two current pieces, because current (unlike power) is linear.

Ex 4 — A source contributing exactly zero (Cell D)

Forecast: the bridge is symmetric. Guess whether the bridge current is zero, and why.

Find the two midpoint potentials without the galvanometer. Why? If no current flows in $R_{g}$ , the potentials of $B$ and $C$ are set purely by the two dividers. $V_{B} = 8 \cdot \frac{R _{2}}{R _{1} + R _{2}} = 8 \cdot \frac{100}{200} = 4 V, V_{C} = 8 \cdot \frac{R _{4}}{R _{3} + R _{4}} = 4 V$
Compare. $V_{B} - V_{C} = 0 V$ , so no voltage drives $R_{g}$ : $I_{g} = \frac{V _{B} - V _{C}}{R _{g}} = 0 A$ Why this matters for superposition: a source's contribution to a particular target can be exactly zero even though the source is fully active. Zero is a legitimate term in the algebraic sum — don't skip it, record it.

Worked example Verify

Balance condition $R_{1} R_{4} = R_{2} R_{3}$ : $100 \cdot 100 = 100 \cdot 100$ ✓, so $I_{g} = 0$ regardless of $R_{g}$ . If we perturb $R_{4} \to 110$ : $V_{C} = 8 \cdot 110/210 = 4.190 V$ , $V_{B} - V_{C} = - 0.190 V \neq = 0$ — confirming the zero was due to balance, not luck.

Ex 5 — Degenerate case: killed source shorts a resistor (Cell E)

Forecast: killing the voltage source doesn't delete $R_{1}$ — it leaves $R_{1}$ as a wire-connected resistor. Predict whether $R_{1}$ still matters in the $I_{s}$ -alone case.

$V_{1}$ alone (open the 2 A). With $I_{s}$ open, current flows $V_{1} \to R_{1} \to R_{L}$ : $V_{L, 1} = 6 \cdot \frac{R _{L}}{R _{1} + R _{L}} = 6 \cdot \frac{6}{9} = 4 V$
$I_{s}$ alone (short $V_{1}$ ). Shorting $V_{1}$ turns the $V_{1}$ – $R_{1}$ branch into just $R_{1}$ from the top node to ground. So the $2 A$ sees $R_{1} ∥ R_{L}$ : $R_{1} ∥ R_{L} = \frac{3 \cdot 6}{9} = 2 Ω, V_{L, 2} = 2 \cdot 2 = 4 V$ Why $R_{1}$ still counts: killing $V_{1}$ removed the source, not the resistor. The resistor stays and reshapes the topology — this is the "degenerate" trap.
Add: $V_{L} = 4 + 4 = 8 V$ .

Worked example Verify

KCL at load node $V_{L}$ (source-transform check): current from $V_{1}$ branch $\frac{6 - V _{L}}{3}$ plus injected $2$ minus load current $\frac{V _{L}}{6} = 0$ . $\Rightarrow \frac{6 - 8}{3} + 2 - \frac{8}{6} = - 0.667 + 2 - 1.333 = 0$ ✓

Ex 6 — Limiting behaviour (Cell F)

Forecast: if $R_{2} \to 0$ (node hard-tied to ground) then $V_{A} \to 0$ . If $R_{2} \to \infty$ , the node floats — predict its voltage from the 12 V source and the 3 A source separately.

$R_{2} \to 0$ . Divider term $12 \cdot \frac{R _{2}}{R _{1} + R _{2}} \to 12 \cdot \frac{0}{4} = 0$ . Parallel term $3 \cdot (R_{1} ∥0) = 3 \cdot 0 = 0$ . $V_{A} \to 0 + 0 = 0 V$ Why sane? A dead short to ground pins $V_{A} = 0$ no matter what the sources do — both contributions vanish, as they must.
$R_{2} \to \infty$ . Divider term $12 \cdot \frac{R _{2}}{R _{1} + R _{2}} \to 12 \cdot 1 = 12 V$ (no current through $R_{1}$ , so no drop). Parallel term: $R_{1} ∥ R_{2} \to R_{1} = 4 Ω$ , so $3 \cdot 4 = 12 V$ . $V_{A} \to 12 + 12 = 24 V$ Why sane? With no path to ground except back through $R_{1}$ , all $3 A$ flows through $R_{1}$ ( $3 \cdot 4 = 12 V$ ) on top of the source's own $12 V$ .

Worked example Verify

At $R_{2} = 6$ we had $14.4 V$ ; the limits $0$ and $24$ bracket it, and the general formula $V_{A} = 12 \cdot \frac{R _{2}}{4 + R _{2}} + 3 \cdot \frac{4 R _{2}}{4 + R _{2}}$ is monotonic increasing in $R_{2}$ from $0$ to $24$ . Check midpoint $R_{2} = 6$ : $12 \cdot 0.6 + 3 \cdot 2.4 = 7.2 + 7.2 = 14.4$ ✓

Ex 7 — Word problem (Cell G)

Forecast: the panel gives a divider (small, because $10 ≫ 2$ ), the charger gives $1 A \times (2∥10)$ . Guess which dominates.

Panel alone (open the charger). Divider $2 Ω$ (wiring) to $10 Ω$ (device): $V_{D, panel} = 5 \cdot \frac{10}{2 + 10} = 5 \cdot 0.8333 = 4.1667 V$
Charger alone (short the panel's ideal source). Charger sees $2∥10$ : $2∥10 = \frac{20}{12} = 1.6667 Ω, V_{D, charger} = 1 \cdot 1.6667 = 1.6667 V$
Add: $V_{D} = 4.1667 + 1.6667 = 5.8333 V$ . Panel contributes $\approx 71%$ , charger $\approx 29%$ . Why translate this way: "ideal source behind resistance" is a Thevenin and Norton equivalents model; superposition then treats each real-world supply as one killable source.

Worked example Verify

KCL at $D$ : $\frac{V _{D} - 5}{2} + \frac{V _{D}}{10} - 1 = 0$ . $\times 10$ : $5 (V_{D} - 5) + V_{D} - 10 = 0 \Rightarrow 6 V_{D} = 35 \Rightarrow V_{D} = 5.8333 V$ ✓ Units: volts.

Ex 8 — Exam twist: a dependent source is present (Cell H)

Forecast: with a single independent source there's only one sub-circuit. But you may not open the dependent source — predict whether $V_{A}$ is larger or smaller than the plain divider value $10 \cdot \frac{4}{6} = 6.667 V$ .

Only one independent source, so superposition is a single term — but keep $g V_{A}$ active. Why keep it? Dependent sources encode real feedback in the circuit; killing one deletes physics and gives a wrong answer. Superposition toggles independent sources only.
Write KCL at $A$ , dependent current into the node counts as negative on the "leaving" side: $\frac{V _{A} - 10}{2} + \frac{V _{A}}{4} - g V_{A} = 0$ Why this equation? KCL (Kirchhoff's Voltage and Current Laws): current toward source + current to ground − injected dependent current $= 0$ .
Solve with $g = 0.25$ : multiply by 4: $2 (V_{A} - 10) + V_{A} - 4 (0.25) V_{A} = 0 \Rightarrow 2 V_{A} - 20 + V_{A} - V_{A} = 0 \Rightarrow 2 V_{A} = 20 \Rightarrow V_{A} = 10 V$ The dependent source raised $V_{A}$ from $6.667$ to $10 V$ — proof it cannot be ignored.

Worked example Verify

Plug $V_{A} = 10$ : $\frac{10 - 10}{2} + \frac{10}{4} - 0.25 \cdot 10 = 0 + 2.5 - 2.5 = 0$ ✓

Connections

Parent: Apply superposition theorem
Linearity and homogeneity
Kirchhoff's Voltage and Current Laws
Voltage divider and current divider
Thevenin and Norton equivalents
Nodal and mesh analysis
Why power does not superpose