Kirchhoff's laws and combining resistorsEdexcel International A Level Physics: Revision notes

Charge cannot be created or destroyed, and it does not build up at a point in a circuit. So at any junction the total current entering equals the total current leaving. This is Kirchhoff's first law:

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

In a series circuit there are no junctions, so the current is the same everywhere. In a parallel circuit the current splits at a junction and the branch currents add up to the total: $I = I_1 + I_2 + \dots$

Each coulomb of charge gains energy from the source and transfers it to components. Around any closed loop the energy gained per unit charge equals the energy transferred per unit charge. This is Kirchhoff's second law: in a closed loop, the sum of the e.m.f.s equals the sum of the potential differences, $\sum \varepsilon = \sum IR$.

For a single supply of p.d. $V$ with components in series, $V = V_1 + V_2 + \dots$. Components in parallel are connected across the same two points, so they have the same p.d.

Start from the two conservation laws. The current $I$ is the same through each resistor (charge). The supply p.d. is shared: $V = V_1 + V_2$ (energy). With $V = IR$:

$IR = IR_1 + IR_2 \Rightarrow R = R_1 + R_2$

So for $n$ resistors $R = R_1 + R_2 + \dots + R_n$, and the total is always larger than the largest resistor.

The total current splits: $I = I_1 + I_2$ (charge). Each resistor has the same p.d. $V$ (energy). Using $I = V/R$:

$\frac{V}{R} = \frac{V}{R_1} + \frac{V}{R_2} \Rightarrow \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

The total resistance is always smaller than the smallest resistor, because adding a branch gives the charge carriers another route. For two resistors, $R = R_1R_2/(R_1 + R_2)$.

A 12 V battery (negligible internal resistance) is connected to a 10 Ω resistor in series with a 30 Ω and 60 Ω resistor in parallel.

• Parallel pair: $1/R = 1/30 + 1/60 = 1/20$, so $R = 20$ Ω
• Total: $10 + 20 = 30$ Ω; $I = 12/30 = 0.40$ A
• p.d. across the 10 Ω: $4.0$ V, so across the pair: $12 - 4.0 = 8.0$ V
• Branch currents: $8.0/30 = 0.27$ A and $8.0/60 = 0.13$ A, which add to 0.40 A