Kirchhoff's laws

The idea

Two rules let you analyse any circuit. Both are just conservation laws:

• KCL (current law): charge can’t pile up at a junction, so current in = current out.
• KVL (voltage law): energy is conserved round a loop, so the voltage rises equal the voltage drops.

$\text{KCL: }\sum I=0\text{KVL: }\sum V=0\text{ round any closed loop}$

Sign conventions: count currents into a node as + and out as −. If a current you solve for comes out negative, it simply flows the opposite way to the arrow you drew. Keep the sign and carry on.

Sources in a loop: a source that pushes the same way as the others adds; one turned round subtracts. Three cells of 20, 10 and 20 V with the 10 V reversed give $20-10+20=30$ V.

Series and parallel

	Series	Parallel
Same for each part	Current	Voltage
Total resistance	$R_T=R_1+R_2+\dots$	$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\dots$
Two resistors		$R_T=\frac{R_1{R}_2}{R_1+R_2}$ (product over sum)

Voltage divider (two resistors in series): $V_2=V\frac{R_2}{R_1+R_2}$ Current divider (two in parallel): $I_2=I\frac{R_3}{R_2+R_3}$, the current through one branch uses the other branch’s resistance on top.

The method for any series–parallel circuit

1. Reduce the network to one total resistance $R_T$.
2. Total current $I_T=V/R_T$.
3. Work back out: the voltage across each block, then each branch current.
4. Check KVL (the drops add up to the supply) and KCL (the branches add up to the total).
5. Check the power balance: the power dissipated in the resistors, $\sum I^{2}R$, must equal the power from the sources, $\sum VI$.

The power balance is the best check there is. If it doesn’t balance, something upstream is wrong.

Worked example

A 12 V supply feeds $R_1=4\Omega$ in series with $R_2=6\Omega$ and $R_3=12\Omega$ in parallel. Find every current and check the power balance.

Answer

$R_2\parallel R_3=\frac{6\times 12}{18}=4\Omega$, so $R_T=4+4=8\Omega$ and $I_T=12/8=1.5$ A. $V_{R1}=1.5\times 4=6$ V, so the parallel pair gets $12-6=6$ V. $I_2=6/6=1$ A, $I_3=6/12=0.5$ A. KCL: $1+0.5=1.5$ ✓ Powers: $1.5^2\times 4=9$ W, $1^2\times 6=6$ W, $0.5^2\times 12=3$ W → total 18 W $=12\times 1.5$ ✓

Common mistakes

• Adding parallel resistances directly. Parallel always gives less than the smallest resistor.
• In a current divider, putting the branch’s own resistance on top.
• Throwing away a negative answer. It’s information: the current flows the other way.

Check yourself

1. 10 Ω and 15 Ω in parallel?

$\frac{150}{25}=6\Omega$.

2. A 9 V supply across 1 kΩ and 2 kΩ in series. Voltage across the 2 kΩ?

$9\times \frac{2}{3}=6$ V.