Series and parallel circuits and powerAQA A-Level Physics: Revision notes

In a series circuit there is a single path, so the resistances add: $R_T = R_1 + R_2 + R_3 + \dots$. The total is always larger than the largest individual resistor.

In a parallel circuit the charge has several paths, so the reciprocals add: $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$. The total is always smaller than the smallest individual resistor, because each extra branch gives the charge another route.

For two resistors in parallel, $R_T = \frac{R_1 R_2}{R_1 + R_2}$ is a quick check. Two equal resistors $R$ in parallel give $R/2$.

Two conservation laws explain every dc circuit.

Conservation of charge: charge is not created or destroyed, so the current into a junction equals the current out. In series the current is the same everywhere. In parallel the branch currents add up to the supply current.

Conservation of energy: the energy a unit charge gains from the supply equals the energy it transfers round any complete loop. So the potential differences across components in series add up to the supply pd, and each parallel branch has the same pd as the supply.

For identical resistors in parallel the current splits equally. For different resistors the larger current goes through the smaller resistance, since $I = V/R$ with the same $V$.

Cells in series: the emfs add, so $n$ identical cells of emf $\varepsilon$ give $n\varepsilon$. The same current passes through every cell.

Identical cells in parallel: the emf stays at $\varepsilon$, the same as one cell, but the current is shared equally, so each cell supplies only $1/n$ of the total current. The combination can supply a given current for longer, because each cell is drained more slowly.

Series cells give a larger pd (brighter lamp, more power). Parallel cells give a longer-lasting supply at the same pd.

Electrical energy transferred is $E = IVt$, where $I$ is current, $V$ is the pd and $t$ is time in seconds. Power is the rate of energy transfer:

$P = IV = I^2 R = \frac{V^2}{R}$

Choose the form that uses the quantity you know. In series the current is common, so $P = I^2R$ shows the largest resistor dissipates the most power. In parallel the pd is common, so $P = V^2/R$ shows the smallest resistor dissipates the most power.

The total power dissipated in all the resistors equals the power delivered by the supply, which is conservation of energy.

A 9.0 V supply is connected to a 6.0 Ω resistor in series with two 12 Ω resistors in parallel.

1. Parallel pair: $1/R = 1/12 + 1/12$, so $R = 6.0$ Ω.
2. Total: $R_T = 6.0 + 6.0 = 12$ Ω.
3. Current: $I = 9.0/12 = 0.75$ A.
4. Power in the 6.0 Ω resistor: $I^2R = 0.75^2 \times 6.0 = 3.4$ W.
5. Each 12 Ω carries 0.375 A and dissipates 1.7 W.
6. Check: $3.4 + 1.7 + 1.7 = 6.8$ W $= IV = 0.75 \times 9.0$.