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(C) For a constant voltage $V$,the power dissipated is given by $P = \frac{V^2}{R}$. Thus,$P \propto \frac{1}{R}$.$1$. For $R_1$: The circuit consists

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$P_2 > P_1 > P_3$

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(C) For a constant voltage $V$,the power dissipated is given by $P = \frac{V^2}{R}$. Thus,$P \propto \frac{1}{R}$.$1$. For $R_1$: The circuit consists of three $1 \, \Omega$ resistors in parallel,which are in series with another $1 \, \Omega$ resistor. The equivalent resistance is $R_1 = (1 \, \Omega \parallel 1 \, \Omega \parallel 1 \, \Omega) + 1 \, \Omega = \frac{1}{3} \, \Omega + 1 \, \Omega = \frac{4}{3} \, \Omega \approx 1.33 \, \Omega$.$2$. For $R_2$: This is a balanced Wheatstone bridge with five $1 \, \Omega$ resistors. The middle resistor carries no current. The equivalent resistance is $R_2 = (1 \, \Omega + 1 \, \Omega) \parallel (1 \, \Omega + 1 \, \Omega) = 2 \, \Omega \parallel 2 \, \Omega = 1 \, \Omega$.$3$. For $R_3$: This is a Wheatstone bridge with four $1 \, \Omega$ resistors in series with a $3 \, \Omega$ resistor. The bridge part has an equivalent resistance of $1 \, \Omega$. Thus,$R_3 = 1 \, \Omega + 3 \, \Omega = 4 \, \Omega$.Comparing the resistances: $R_3 > R_1 > R_2$.Since $P \propto \frac{1}{R}$,we have $P_2 > P_1 > P_3$.

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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