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(A) The power dissipated in a circuit is given by $P = \frac{V^2}{R}$. Since the voltage $V = 3 	ext{ V}$ is constant for all configurations, the pow

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$P_1 > P_2 > P_3$

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(A) The power dissipated in a circuit is given by $P = \frac{V^2}{R}$. Since the voltage $V = 3 	ext{ V}$ is constant for all configurations, the power $P$ is inversely proportional to the equivalent resistance $R$ $(P \propto \frac{1}{R})$.$1$. For configuration $R_1$: The circuit consists of three $1 	ext{ }\Omega$ resistors in parallel. Thus, $R_1 = \frac{1 	ext{ }\Omega}{3} = 0.33 	ext{ }\Omega$.$2$. For configuration $R_2$: This is a Wheatstone bridge circuit. Since all resistors are $1 	ext{ }\Omega$, it is a balanced bridge. The middle resistor carries no current. The equivalent resistance is two parallel branches of $2 	ext{ }\Omega$ each, so $R_2 = \frac{2 	ext{ }\Omega}{2} = 1 	ext{ }\Omega$.$3$. For configuration $R_3$: This is a series-parallel combination. Two $1 	ext{ }\Omega$ resistors in parallel give $0.5 	ext{ }\Omega$, which is in series with another $1 	ext{ }\Omega$ resistor, giving $1.5 	ext{ }\Omega$. This is in parallel with another $1 	ext{ }\Omega$ resistor. Calculating the total, $R_3 = 1.5 	ext{ }\Omega$ (approx). Wait, looking at the diagram for $R_3$: The top two resistors are in parallel $(0.5 	ext{ }\Omega)$, the next two are in parallel $(0.5 	ext{ }\Omega)$, and these are in series with the bottom $1 	ext{ }\Omega$ resistor. Total $R_3 = 0.5 + 0.5 + 1 = 2 	ext{ }\Omega$.Comparing resistances: $R_1 = 0.33 	ext{ }\Omega$, $R_2 = 1 	ext{ }\Omega$, $R_3 = 2 	ext{ }\Omega$.Since $P \propto \frac{1}{R}$, the order of power is $P_1 > P_2 > P_3$.

The figure shows three resistor configurations $R_1, R_2$, and $R_3$ connected to a $3 \text{ V}$ 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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