Question

In the following circuit, the battery has an emf of \(2 \, \text{V}\) and an internal resistance of \(\frac{2}{3} \, \Omega\). The power consumption in the entire circuit is ______ W.

Answer 1

Correct Answer: 3

First, we calculate the equivalent resistance \( R_{eq} \) of the circuit.

The two branches with \( 2 \, \Omega \) resistors are in parallel. The equivalent resistance for each pair of resistors in parallel is:

\[ \frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{2} = 1 \, \Omega. \]

Since there are two such parallel branches in series, the total resistance \( R_{eq} \) of the circuit is:

\[ R_{eq} = 1 + 1 + \frac{2}{3} = \frac{4}{3} \, \Omega. \]

The power \( P \) consumed in the circuit is given by:

\[ P = \frac{V^2}{R_{eq}}. \]

Substitute \( V = 2 \, V \) and \( R_{eq} = \frac{4}{3} \, \Omega \):

\[ P = \frac{2^2}{\frac{4}{3}} = \frac{4}{\frac{4}{3}} = 3 \, W. \]

Thus, the power consumption in the entire circuit is:

\[ 3 \, W. \]