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(C) The power dissipated in a circuit with a constant voltage $V$ is given by $P = \frac{V^2}{R}$. Since $V = 3\,V$ is constant for all three circuits

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$P_2 > P_1 > P_3$

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(C) The power dissipated in a circuit with a constant voltage $V$ is given by $P = \frac{V^2}{R}$. Since $V = 3\,V$ is constant for all three circuits,$P \propto \frac{1}{R}$. Thus,the circuit with the smallest equivalent resistance will dissipate the most power.$1$. For $R_1$: The circuit consists of several $1\,\Omega$ resistors. Analyzing the symmetry and series-parallel combinations,the equivalent resistance is $R_1 = 1\,\Omega$.$2$. For $R_2$: This is a balanced Wheatstone bridge with a central resistor. Due to symmetry,the potential at the nodes connected to the central resistor is equal,so no current flows through the central resistor. The equivalent resistance is $R_2 = 0.5\,\Omega$.$3$. For $R_3$: This is a bridge circuit in series with another resistor. The equivalent resistance is $R_3 = 2\,\Omega$.Comparing the resistances: $R_2 (0.5\,\Omega) Since $P \propto \frac{1}{R}$,the power dissipation follows the inverse order: $P_2 > P_1 > P_3$.

The figure shows three resistor configurations $R_1, R_2$,and $R_3$ connected to a $3\,V$ battery. If the power dissipated by the configurations $R_1, R_2$,and $R_3$ is $P_1, P_2$,and $P_3$ respectively,then:

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