In the circuit shown, a 12 V power supply with unknown internal resistance r is connected to a run-down rechargeable battery with unknown emf $\epsilon$ and internal resistance 1 $\Omega$, and to an indicator light bulb of resistance 3 $\Omega$ carrying a current of 2 A. The current through the run down battery is 1 A in the direction shown. Find the unknown current I, the internal r, and the emf $\epsilon$.

Find the power delivered by the 12 V power supply and by the battery being recharged, and find the power dissipated in each resistor.

Loop 1:

$$\begin{aligned} 12 &= Ir + 6 \\ Ir &= 6 \, \text{V} \end{aligned}$$

Loop 2:

$$\begin{aligned} – \epsilon &= -1 + 6 \\ \epsilon &= -5 \, \text{V} \end{aligned}$$

Loop 3:

$$\begin{aligned} 12 + \epsilon &= Ir + 1 \\ Ir &= 6 \, \text{V} \end{aligned}$$

Using the current information in the diagram,

$$\begin{aligned} I &= 1 + 2 \\ &= 3 \, \text{A} \end{aligned}$$

$$r = 2 \, \Omega$$

$$\begin{aligned}P_{12} &= IV \\ &= 3 (12) \\ &= 36 \, \text{W} \end{aligned}$$

$$P_{\epsilon} = -5 \, \text{W}$$

$$\begin{aligned} P_{r} &= I^{2}r \\ &= 18 \, \text{W} \end{aligned}$$

$$P_{1 \Omega} = 1 \, \text{W}$$

$$P_{3 \Omega} = 12 \, \text{W}$$

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