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, \[\mathcal{E} = V =3.08 \hspace{0.1cm} V\] When switch S is closed,
\(V\) = 2.97 V
\(I\) = 1.65 A
\(r\) = ?
\(R\) = ?

Now, \[\begin{align*} r&=\frac{\mathcal{E}-V}{I}\\ &=\frac{3.08-2.97}{1.65}\\ \therefore r&=0.067 \hspace{0.1cm} \Omega\\ \end{align*}\] Again, \[\begin{align*} V&=IR\\ 2.97 & =1.65\times R\\ R&=1.8 \hspace{0.1cm} \Omega\\ \end{align*}\] Hence, the emf is 3.08 V, the internal resistance is 0.067 \(\Omega\) and resistance of resistor R is 1.8 \(\Omega\).