In the following circuit,the battery has an e | vedclass

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(A) First,we calculate the equivalent resistance of the external circuit. The circuit consists of five $2 \ \Omega$ resistors. By symmetry,the potenti

Vedclass · Accepted Answer

$1.5$

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(A) First,we calculate the equivalent resistance of the external circuit. The circuit consists of five $2 \ \Omega$ resistors. By symmetry,the potential at the two middle nodes is equal,so no current flows through the diagonal resistors. The circuit simplifies to two parallel branches,each containing two $2 \ \Omega$ resistors in series.
Each branch has a resistance of $2 \ \Omega + 2 \ \Omega = 4 \ \Omega$.
Since there are two such branches in parallel,the external equivalent resistance $R_{	ext{ext}}$ is $\frac{4 \ \Omega}{2} = 2 \ \Omega$.
The total resistance of the circuit is $R_{	ext{total}} = R_{	ext{ext}} + r = 2 \ \Omega + \frac{2}{3} \ \Omega = \frac{8}{3} \ \Omega$.
The total power consumption in the circuit is given by $P = \frac{E^2}{R_{	ext{total}}}$.
Substituting the values,$P = \frac{2^2}{8/3} = \frac{4}{8/3} = \frac{12}{8} = 1.5 \ W$.

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

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