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

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:

$3 \, W.$