Kirchhoff's Laws

Kirchhoff's first law: the sum of currents into a junction equals the sum out (conservation of charge). Kirchhoff's second law: round any closed loop, the sum of e.m.f.s equals the sum of p.d.s (conservation of energy). From these: series $R=R_1+R_2+\dots$ ; parallel $\dfrac1R=\dfrac1{R_1}+\dfrac1{R_2}+\dots$ .

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH12AQA 7408 Paper 1≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

State Kirchhoff's laws and link them to conservation of charge and energy.
Derive and use the series and parallel resistance formulae.
Apply Kirchhoff's laws to circuits with more than one source.
Find currents and p.d.s in combined series–parallel networks.

Pause and think

You connect a $1$ Ω resistor in parallel with a $1000$ Ω resistor. Is the combined resistance nearer $1$ Ω, $500$ Ω or $1001$ Ω?

Section 2 of 14

Core idea

Start hereTwo batteries fighting

Connect a 12 V and a 6 V battery in parallel to a lamp. Does the 6 V battery help? Surprisingly, it gets charged by the other one. Kirchhoff's laws, which are just conservation of charge and energy, reveal what really happens.

The big idea

1st law: $\sum I_{\text{in}}=\sum I_{\text{out}}$ (charge). 2nd law: round any loop, $\sum E=\sum IR$ (energy).

1. First law

Currents at a junction

5=3+2

. Charge cannot pile up at a junction.

2. Series and parallel from the laws

Combining resistors

$V=IR_1+IR_2\Rightarrow R=R_1+R_2$
Series: same current (1st law), p.d.s add (2nd law).
$I=\frac V{R_1}+\frac V{R_2}$
Parallel: same p.d., currents add.
$\frac1R=\frac1{R_1}+\frac1{R_2}$
Divide by $V$ .

Combined resistance of 6 Ω in parallel with R

Adding a parallel branch always lowers the total resistance, however large the branch.

3. Loops

Worked exampleThe two batteries

12 V (2 Ω) and 6 V (1 Ω) in parallel across 4 Ω.

Goal: Branch currents.
Strategy: Unknowns $I_1$ , $I_2$ ; load current $I_1+I_2$ ; two loops.

$12=2I_1+4(I_1+I_2)\Rightarrow6I_1+4I_2=12$
Loop through the 12 V cell and load.
$6=I_2+4(I_1+I_2)\Rightarrow4I_1+5I_2=6$
Loop through the 6 V cell and load.
$I_1=\tfrac{18}7=2.57\text{ A},\ I_2=-\tfrac67=-0.86\text{ A}$
Solve simultaneously.

Answer

I_2

negative: the 6 V cell is being charged.

✓ CheckLoad p.d.

4(1.71)=6.86

>6

V, so current is pushed into the 6 V cell ✓.

Key takeawayNegative answers A negative current just means the real direction is opposite to your guess.

Pause and think

3 Ω ∥ 6 Ω, in series with 4 Ω?

Summary

Junctions: charge.
Loops: energy.

Junction: Point where currents split or join.

HistoryA student's laws. Gustav Kirchhoff published his circuit laws in

1845

while still a student at Königsberg, aged

21

. He later worked with Bunsen to invent spectroscopy, discovering caesium and rubidium from the coloured lines in their flames.

Fun factBirds on a wire. A bird on a single power line is safe because both its feet are at nearly the same potential: the short length of wire between them has tiny resistance, so the bird, in parallel with it, carries almost no current. Touch a second wire, though, and the loop is completed.

Keep reading for free

Create a free account to read the rest of this lesson: key formulae, worked examples, explore the graph, common mistakes, exam tip, guided practice and more. Your answers are checked and your progress is saved.

Create free account Sign in