In the given figure, when switch S is open, the voltmeter V reads 3.08 V. When the switch S is closed, the voltmeter reading drops to 2.97V, and the ammeter A reads 1.65 A. Find the emf, the internal resistance of the battery and the resistor R. Assume that the two meters are ideal.

In the given figure, when switch S is open, the voltmeter V reads 3.08 V. When the switch S is closed, the voltmeter reading drops to 2.97V, and the ammeter A reads 1.65 A. Find the emf, the internal resistance of the battery and the resistor R. Assume that the two meters are ideal.

Solution:
We have the relation, $$V=\mathcal{E}-Ir$$ When switch S is open, no current flows through the circuit and thus $I=0$.
Hence, $$\mathcal{E}=V=3.08V$$ When switch S is closed,
$V$ = 2.97 V
$I$ = 1.65 A
$r$ = ?
$R$ = ?

Now, $$\begin{array}{rl}r& =\frac{\mathcal{E}-V}{I}\\ & =\frac{3.08-2.97}{1.65}\\ \therefore r& =0.067\mathrm{\Omega }\end{array}$$ Again, $$\begin{array}{rl}V& =IR\\ 2.97& =1.65\times R\\ R& =1.8\mathrm{\Omega }\end{array}$$ Hence, the emf is 3.08 V, the internal resistance is 0.067 $\mathrm{\Omega }$ and resistance of resistor R is 1.8 $\mathrm{\Omega }$.