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.
Correct Answer: 3
Solution and Explanation
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:
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