The circuit shown contains two batteries, each with an emf and an internal resistance, and two resistors. Find

- the current in the circuit,
- the potential difference $V_{ab}$, and
- the power output of the emf of each battery.

There is only 1 loop in the circuit.

$$\begin{aligned} – 4 + 12 &= 4 I + 7 I + 2 I + 3 I \\ 16 I &= 8 \\ I &= 0.5 \, \text{A} \end{aligned}$$

To find the potential difference $V_{ab}$: The current have two ways to flow from a to b – which we shall termed top and bottom circuit. You can use either top or bottom circuit to find the potential difference. Both will yield the same answer.

Using the bottom circuit:

$$ \begin{aligned} V_{b} &= V_{a} – 4I – 4 – 7I \\ V_{a} – V_{b} &= 11I + 4 \\ V_{ab} &= 9.5 \, \text{V} \end{aligned}$$

Using the top circuit:

$$ \begin{aligned} V_{a} &= V_{b} + 12 – 2I – 3I \\ V_{a} – V_{b} &= 12 – 5I \\ V_{ab} &= 9.5 \, \text{V} \end{aligned}$$

Finding the power output is easy. You just have to use $P = IV$. Hence,

$$P_{12} = 6 \, \text{W}$$

$$P_{4} = 4 \, \text{W}$$