Kirchhoff's rules

When a circuit is too tangled for series/parallel, two simple rules — one for junctions, one for loops — always crack it.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §28.3.

Before you start

What you need first

Topic 16 — current $I$.
Topic 17 — Ohm's law $V = IR$.
Topic 19 — a battery's EMF.

What you'll be able to do

Apply the junction rule: $\sum I_{\text{in}} = \sum I_{\text{out}}$.
Apply the loop rule: $\sum V = 0$ around any loop.
See both as conservation of charge and of energy.
Find an unknown current or voltage in a circuit.

When reduction fails

When series and parallel aren't enough

Some circuits (with several batteries, or bridged branches) can't be boiled down to simple series/parallel. Kirchhoff's two rules always work — and both are just conservation laws you already trust:

Junction rule — about current (charge is never lost).
Loop rule — about voltage (energy is never lost).

Rule 1 — the junction rule (current)

At any junction (where wires meet), the current flowing in equals the current flowing out — charge can't pile up or vanish:

$$ \sum I_{\text{in}} = \sum I_{\text{out}} $$

5 A + 3 A in → 8 A must flow out.

📐 Worked example 1

Junction rule

At a junction, $I_1 = 5$ A and $I_2 = 3$ A flow in. One wire carries $I_3 = 6$ A out. Find the current $I_4$ in the other outgoing wire.

1In = out:

$$ 5 + 3 = 6 + I_4 $$

2Solve:

$$ I_4 = 2\ \text{A} $$

Rule 2 — the loop rule (voltage)

Go all the way around any loop and add up the voltage changes — they total zero (you end where you started):

$$ \sum V = 0 \quad\text{(around a loop)} $$

The battery lifts the voltage by $\varepsilon$; the resistors drop it by $IR$ each. So the lift equals the sum of the drops.

Lift by $\varepsilon$, drop across $R_1$ and $R_2$ — back to zero.

📐 Worked example 2

Loop rule

A single loop has a 12 V battery and two resistors $R_1 = 4\ \Omega$, $R_2 = 8\ \Omega$ in series, carrying current $I = 1$ A. Find each voltage drop and check the loop rule.

1Drops from Ohm's law:

$$ V_1 = IR_1 = (1)(4) = 4\ \text{V}, \qquad V_2 = IR_2 = (1)(8) = 8\ \text{V} $$

2Loop rule: lift $=$ sum of drops:

$$ \varepsilon = V_1 + V_2 = 4 + 8 = 12\ \text{V} \;\checkmark $$

The two drops add up to the battery's lift — the loop rule in action.

✏️ Try it yourself

A single loop has a 9.0 V battery and three resistors $1.0\ \Omega$, $2.0\ \Omega$, and $6.0\ \Omega$ in series.

(a) Use the loop rule to find the current.
(b) Find the voltage across the $6.0\ \Omega$ resistor.

(a) Loop rule: $9.0 = I(1.0+2.0+6.0) = 9.0\,I$, so $I = 1.0\ \text{A}.$ (b) $V_6 = IR = (1.0)(6.0) = 6.0\ \text{V}.$

Common mistakes

Mistake	Fix
Getting signs wrong in the loop rule.	A battery from $-$ to $+$ is a $+\varepsilon$ lift; crossing a resistor with the current is a $-IR$ drop.
Forgetting charge is conserved at a junction.	Total current in must equal total current out.
Reaching for Kirchhoff when series/parallel would do.	Reduce first; use Kirchhoff only when the circuit won't simplify.
Mixing up which rule is which.	Junctions → current (charge); loops → voltage (energy).

Recap — the whole topic on one screen

Rule	Statement	Conserves
Junction	$\sum I_{\text{in}} = \sum I_{\text{out}}$	charge
Loop	$\sum V = 0$ around a loop	energy

Next topic

RC circuits

Put a resistor and a capacitor together and the charge builds up — or drains away — over time, set by the time constant $\tau = RC$. This is the last topic before the midterm.

→ Topic 22 · RC circuits

Mistake	Fix
Getting signs wrong in the loop rule.	A battery from \(-\) to \(+\) is a \(+\varepsilon\) lift; crossing a resistor with the current is a \(-IR\) drop.
Forgetting charge is conserved at a junction.	Total current in must equal total current out.
Reaching for Kirchhoff when series/parallel would do.	Reduce first; use Kirchhoff only when the circuit won't simplify.
Mixing up which rule is which.	Junctions → current (charge); loops → voltage (energy).

Rule	Statement	Conserves
Junction	\(\sum I_{\text{in}} = \sum I_{\text{out}}\)	charge
Loop	\(\sum V = 0\) around a loop	energy