Worked examples — Kirchhoff's current law (KCL), Kirchhoff's voltage law (KVL)

Before we start, one sentence to earn every symbol:

Every worked example below leans on one global rule. State it once, obey it everywhere:

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The scenario matrix

Every problem Kirchhoff can throw at you falls into one of these cells. The examples below are labelled with the cell(s) they cover.

Cell	What makes it tricky	Example
A. Single loop, one battery	baseline — sign discipline only	Ex 1
B. Node, in vs out signs	which currents are $+$, which $-$	Ex 2
C. Assumed direction WRONG	answer comes out negative	Ex 3
D. Two batteries aiding	EMFs add	Ex 4a
E. Two batteries opposing	EMFs subtract, current can reverse	Ex 4b
F. Degenerate: $R=0$ (short)	limiting behaviour, $I\to \infty$?	Ex 5
G. Degenerate: $R=\infty$ (open)	no current, all voltage across the gap	Ex 6
H. Dead battery: $\epsilon =0$	battery becomes a plain wire/resistor	Ex 6
I. Two loops (KCL + KVL together)	simultaneous equations	Ex 7
J. Real-world word problem	translate words → circuit	Ex 8
K. Exam twist: "find $V$ across the open gap"	voltage where no current flows	Ex 9

Cells D–E share one figure; F–H share one; the two-loop cell gets its own.

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Ex 1 — Cell A: the baseline single loop

Forecast: guess the current before reading on — is it more or less than $9$ A?

Figure s01 — a rectangular loop: the battery ($\epsilon =9$ V) sits on the left, resistors $R_1$ and $R_2$ line the top, and the red arrow shows the assumed clockwise current $I$.

1. Assume $I$ clockwise (red arrow in the figure). Why this step? KVL is an equation in $I$; we must name a direction so "with the current" and "against it" have meaning.
2. Walk the loop clockwise from the battery's $-$ terminal, writing signed changes: $+\epsilon -I{R}_1-I{R}_2=0.$ Why this step? $-$ to $+$ through the battery is a rise ($+\epsilon$); each resistor traversed with the current is a drop ($-IR$).
3. Solve for $I$: $I=\frac{\epsilon }{R_1+R_2}=\frac{9}{1+2}=3\text{ A}.$ Why this step? Pure algebra; series resistances add, so this is the same as $I=\epsilon /R_{\text{eq}}$.

Verify: total drop $=I{R}_1+I{R}_2=3(1)+3(2)=9\text{ V}=\epsilon$. ✓ Units: $\text{V}/\Omega =\text{A}$. ✓

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Ex 2 — Cell B: node signs

Forecast: is $I_3$ positive (really in) or negative (secretly out)?

1. Write KCL with in $=+$, out $=-$: $I_1+I_2+I_3-I_4=0.$ Why this step? A node stores no charge, so what enters must leave: $\sum I_k=0$.
2. Plug numbers: $4+1+I_3-6=0\Rightarrow I_3=1\text{ A}.$ Why this step? Solve the linear equation. Positive ⇒ $I_3$ genuinely flows in, as assumed.

Verify: in $=4+1+1=6\text{ A}$, out $=6\text{ A}$. Balanced. ✓

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Ex 3 — Cell C: the wrong-guess (negative answer)

Forecast: what sign will $I$ have this time?

1. Assume $I$ counter-clockwise. Now traversing the loop counter-clockwise, the battery is entered $+$ to $-$ (a drop) and resistors are traversed with this new current. $-\epsilon -I{R}_1-I{R}_2=0.$ Why this step? The physics is unchanged; only our labelling flipped, so the signs flip.
2. Solve: $I=\frac{-\epsilon }{R_1+R_2}=\frac{-9}{3}=-3\text{ A}.$ Why this step? The minus sign is the tell: the real current is $3\text{ A}$ in the opposite (clockwise) direction — exactly Ex 1.

Verify: magnitude $|I|=3\text{ A}$ matches Ex 1; direction reversed as expected. ✓

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Ex 4 — Cells D & E: two batteries, aiding then opposing

Figure s02 — two side-by-side panels. Left: both EMF arrows point the same way, so they add to $10$ V (shown in red). Right: the EMF arrows point opposite ways, so they subtract to $2$ V (shown in red).

Ex 4a — Cell D: batteries aiding

Forecast: should the effective push be $6+4$ or $6-4$?

1. Assume $I$ clockwise; walk clockwise. Both batteries are climbed $-\to +$ (both rises): $+{\epsilon }_1+{\epsilon }_2-IR=0.$ Why this step? Aiding EMFs both lift you, so they add.
2. Solve: $I=\frac{{\epsilon }_1+{\epsilon }_2}{R}=\frac{6+4}{5}=2\text{ A}.$

Verify: drop across $R$ = $IR=2(5)=10\text{ V}={\epsilon }_1+{\epsilon }_2$. ✓

Ex 4b — Cell E: batteries opposing

Forecast: which battery wins, and does the current reverse?

1. Walk clockwise. ${\epsilon }_1$ is a rise ($-\to +$), but ${\epsilon }_2$ is now climbed $+\to -$ (a drop): $+{\epsilon }_1-{\epsilon }_2-IR=0.$ Why this step? Opposing EMFs subtract — one pushes, the other pushes back.
2. Solve: $I=\frac{{\epsilon }_1-{\epsilon }_2}{R}=\frac{6-4}{5}=0.4\text{ A}.$ Why this step? Positive ⇒ the stronger battery (${\epsilon }_1$) sets the direction; the weaker one is being charged.

Verify: if the batteries were equal (${\epsilon }_2=6$), $I=0$ — a perfect deadlock. Sanity check passes. ✓

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Ex 5 — Cell F: the short circuit ($R\to 0$)

Forecast: does the current blow up to infinity?

Figure s03 — the same rectangular loop, but the external short wire is drawn thick and red ($R=0$) while the battery's internal resistance $r=0.5\Omega$ sits inside the source. The point: $r$ is the only thing limiting the current.

1. KVL around the loop (only $r$ resists): $\epsilon -Ir-IR=0,R=0.$ Why this step? Even a "perfect" wire loop still contains the battery's own $r$. There is no such thing as truly zero total resistance.
2. Solve with $R=0$: $I=\frac{\epsilon }{r+R}=\frac{12}{0.5+0}=24\text{ A}.$ Why this step? The limiting value: as external $R\to 0$, current climbs to $\epsilon /r$, its maximum — large but finite, because $r$ saves us.
3. Terminal voltage (voltage across the external short): $V_{\text{term}}=\epsilon -Ir=12-24(0.5)=0\text{ V}.$ Why this step? All the EMF is eaten inside the battery, so the outside sees $0\text{ V}$ — the classic dead-short signature.

Verify: power in $r$ = $I^{2}r=24^2\times 0.5=288\text{ W}$; power from EMF = $\epsilon I=12\times 24=288\text{ W}$. Energy balances. ✓ (This is why shorting a battery makes it hot.)

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Ex 6 — Cells G & H: open circuit and dead battery

Forecast: guess both currents before reading.

1. Open circuit (G): with a gap, no closed path exists, so $I={\lim }_{R\to \infty }\frac{\epsilon }{R}=0\text{ A}.$ Why this step? Charge cannot cross the gap; KCL at the gap forces $I=0$.
2. Dead battery (H): set $\epsilon =0$ in KVL for a loop with resistor $R=3\Omega$: $0-IR=0\Rightarrow I=0\text{ A}.$ Why this step? With no EMF to lift charge, nothing drives it — the battery is just a piece of wire (or resistor). No source, no current.

Verify: both give $I=0$, but for different reasons: (G) no path, (H) no push. Cross-check: in (H), power delivered $=\epsilon I=0$. ✓

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Ex 7 — Cell I: two loops, KCL + KVL together

Forecast: which branch carries more current — the $4\Omega$ or the $6\Omega$?

Figure s04 — the battery on the left feeds a node that splits into two parallel branches. The red branch is $R_1=4\Omega$, carrying the larger current $I_1=3$ A; the black branch is $R_2=6\Omega$ with $I_2=2$ A. The main current $I=5$ A is labelled on the top rail.

1. KVL, left loop (battery + $R_1$): $\epsilon -I_1{R}_1=0\Rightarrow I_1=12/4=3\text{ A}$. Why this step? $R_1$ alone spans the full EMF (parallel branches share the same voltage).
2. KVL, right loop (battery + $R_2$): $\epsilon -I_2{R}_2=0\Rightarrow I_2=12/6=2\text{ A}$. Why this step? Same reasoning; $R_2$ also sees the full $12\text{ V}$.
3. KCL at the node: $I=I_1+I_2=3+2=5\text{ A}$. Why this step? The main current splits; nothing is lost at the junction.

Verify: parallel equivalent $R_{\text{eq}}=\frac{4\cdot 6}{4+6}=2.4\Omega$, so $I=12/2.4=5\text{ A}$. ✓ The lower resistance ($4\Omega$) carries more current — as predicted.

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Ex 8 — Cell J: real-world word problem

Forecast: total draw — under or over $10\text{ A}$?

1. Translate: parallel across $12\text{ V}$ ⇒ each sees $12\text{ V}$ (same as Ex 7's logic). Why this step? "In parallel across the battery" is the word-cue for equal branch voltages.
2. Branch currents via Ohm's Law: $I_{\text{head}}=\frac{12}{3}=4\text{ A},I_{\text{tail}}=\frac{12}{9}\approx 1.33\text{ A}.$ Why this step? Each branch's current is set by its own resistance under the shared $12\text{ V}$; Ohm's law $I=V/R$ turns each known voltage-and-resistance into a current.
3. KCL — total draw: $I_{\text{fuse}}=4+\frac{4}{3}=\frac{16}{3}\approx 5.33\text{ A}.$ Why this step? The fuse carries the sum of both branches.

Verify: $5.33\text{ A}<10\text{ A}$ ⇒ fuse holds. ✓ Cross-check: $R_{\text{eq}}=\frac{3\cdot 9}{12}=2.25\Omega$, $I=12/2.25=5.33\text{ A}$. ✓

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Ex 9 — Cell K: exam twist, voltage across an open gap

Forecast: if no current flows, is the reading $0$ or $10\text{ V}$?

1. Current first: an ideal voltmeter is an open circuit, so KCL forces $I=0$ (Cell G again). Why this step? No closed path for charge ⇒ zero current everywhere.
2. With $I=0$, no resistor drops any voltage: $I{R}_1=I{R}_2=0$. Why this step? Ohm's Law: a drop needs current; $I=0$ ⇒ zero drop.
3. KVL to find the gap voltage $V_g$: walking the loop, the only non-zero term is the battery: $\epsilon -I{R}_1-V_g-I{R}_2=0\Rightarrow V_g=\epsilon =10\text{ V}.$ Why this step? All of the EMF appears across the one place no current flows — the open gap. This is the exam trap: "no current" does not mean "no voltage."

Verify: the whole $10\text{ V}$ must appear somewhere; with zero resistor drops, it lands entirely on the gap. ✓

Recall Quick self-test

Ex 5 short-circuit current with $\epsilon =12$, $r=0.5$? ::: $24\text{ A}$ Ex 9 voltmeter reading across the open gap? ::: $10\text{ V}$ (the full EMF) Ex 4b current when two equal EMFs oppose? ::: $0\text{ A}$ (deadlock)

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Connections

• Ohm's Law — every "drop" above is $V=IR$.
• Series and Parallel Resistors — used as the sanity check in Ex 1, 7, 8.
• Conservation of Charge — the reason KCL forces $I=0$ across a gap (Ex 6, 9).
• Conservation of Energy — the power balances in Ex 5.
• Electric Potential — the gap voltage in Ex 9 is a potential difference.
• Wheatstone Bridge — combines many of these cells at once.
• Mesh and Nodal Analysis — automates Ex 7's two-loop bookkeeping.