Kirchhoff's Voltage Law

DC Circuits · 10 min read

Your phone charger contains two voltage sources — the mains supply and an internal regulation stage — wired in the same loop.

Series-parallel reduction cannot handle two EMFs in one loop. Kirchhoff's Voltage Law can — it works on every closed loop, no matter how many sources it contains.

The statement

Start at any node, walk all the way around a closed loop, and return to the same node. The voltage changes you pick up along the way must cancel out:

\sum_{\text{loop}} V_k = 0

A battery pushes you up in potential; a resistor with current flowing through it pulls you down (if you're walking in the current's direction).

Add the lifts, subtract the drops, and the sum is zero — because you ended where you started and a node can only have one potential.

A simple KVL loop

One battery drives current through two series resistors. Walking clockwise gives one voltage rise (battery) and two voltage drops (resistors).

Clockwise walk: +V from the battery, −IR₁ across R₁, −IR₂ across R₂. Sum = 0.

Sign convention

The tricky part of KVL isn't the arithmetic — it's the signs. Pick a walking direction around the loop (clockwise or anticlockwise) and commit to it for every element.

If you guess the current direction wrong, the algebra corrects you — the solved $I$ comes out negative.

Walk direction	Element	Signed term
− → + (into + terminal)	Battery	$+ \mathcal{E}$
+ → − (into − terminal)	Battery	$- \mathcal{E}$
With assumed current	Resistor	$-IR$
Against assumed current	Resistor	$+IR$

The four-step method

Pick a loop direction — clockwise or anticlockwise. Mark it on the circuit and keep it for every element.
Assign a sign to each element. Battery walked − → + gives $+ \mathcal{E}$ ; resistor walked with the assumed current gives $-IR$ .
Write $\sum V = 0$ — sum all the signed terms and set the result equal to zero.
Solve for the unknown. A negative answer means your assumed current direction was backwards — accept it, don't restart.

Why it works — energy conservation

Voltage is energy per unit charge. Walking a closed loop returns you to your starting node, so the net energy per charge must be zero.

Otherwise you'd be creating or destroying energy by walking in circles. KVL is that conservation statement written in loop form.

KVL is the tool you reach for when the circuit stops collapsing. Pick a loop direction, keep your signs consistent, write

\sum V = 0

, and solve.

Worked example — simple loop

12 V battery, R₁ = 4 Ω, R₂ = 8 Ω in a single series loop. Find the current $I$ .

Clockwise walk: +12 V from the battery (− → +), −4I across R₁, −8I across R₂. Sum = 0.

Walk clockwise; assume $I$ flows clockwise.
Assign signs: battery (walked − → +): $+ 12\text{ V}$ . R₁ (walked with current): $-4I$ . R₂ (walked with current): $-8I$ .
Write KVL: $12 - 4I - 8I = 0$ .
Solve: $12 = 12I \Rightarrow I = 1\text{ A}$ . Verify: $V_{R1} = 4\text{ V}$ , $V_{R2} = 8\text{ V}$ , sum = 12 V ✓

Worked example — two opposing sources

12 V source (left) and 4 V source (right, opposing) in the same loop, with R₁ = 6 Ω and R₂ = 2 Ω.

Series-parallel reduction cannot handle two EMFs in one loop. KVL writes one equation and solves directly.

Walk clockwise: +12 V (left battery, − → +), −6I (R₁), −4 V (right battery entered at +, so + → −), −2I (R₂). Net EMF = 8 V drives I = 1 A.

Walk clockwise; assume $I$ clockwise.
12 V battery (walked − → +): $+12$ . 4 V battery (walked + → −, opposing): $-4$ . R₁: $-6I$ . R₂: $-2I$ .
$12 - 4 - 6I - 2I = 0 \Rightarrow 8 = 8I$ .
$I = 1\text{ A}$ , flowing clockwise. A negative result would mean the current is actually anticlockwise — the algebra corrects your arrow.

When to use KVL vs. reduction

For clean series-parallel networks, reduction is faster — there's no loop equation to write. KVL earns its keep when:

Two or more sources appear in the same loop.
The network is a bridge or has a diagonal branch that doesn't fit the series/parallel template.
You need a specific branch quantity and the expansion bookkeeping of reduction feels uglier than one loop equation.

In bigger circuits you'll write one KVL equation per independent loop and solve the resulting system (mesh analysis). That generalisation comes later — for this topic, one loop at a time is enough.

KVL loop explorer

Drag the sliders — watch the loop equation update live and see the voltage drops as proportional bars. The two bars always sum to V: that's KVL in action.

Source voltage12.0 V

1 V24 V

R₁4 Ω

R₂8 Ω

KVL loop equation

12 - 1.000 \times 4 - 1.000 \times 8 = 0

I = 1.000 A

Voltage drops —

V_{R1} + V_{R2}

= 12.0 V (KVL)

Think before calculating

Predict the answer, then reveal it to check.

Scenario 1: 9 V battery, R₁ = 1 Ω, R₂ = 2 Ω in series. Find I and the voltage across R₂.

Scenario 2: Two sources 10 V and 6 V opposing in one loop, one resistor R = 4 Ω. Find I and its direction.

Scenario 3: You write the loop equation and get I = −0.5 A. What does the negative sign mean?

Common mistakes

✅ KVL: the algebraic sum of voltages around any closed loop = $0$
✅ Pick one loop direction and keep it consistent for every element
✅ Battery walked − → + gives $+ \mathcal{E}$ ; resistor walked with current gives $-IR$
✅ A negative $I$ means your assumed direction was backwards — accept it
✅ Use KVL when two or more EMFs share a loop, or the circuit can't be reduced