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(A) The power dissipated in a circuit is given by $P = I^2 R_{eq}$, where $I$ is the current and $R_{eq}$ is the equivalent resistance.For figure $(I)

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$(III) < (II) < (IV) < (I)$

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(A) The power dissipated in a circuit is given by $P = I^2 R_{eq}$, where $I$ is the current and $R_{eq}$ is the equivalent resistance.For figure $(I)$, the three resistors are in series: $R_{eq, I} = R + R + R = 3R$. Thus, $P_I = I^2(3R) = 3I^2R$.For figure $(II)$, two resistors are in parallel and one is in series with the combination: $R_{eq, II} = R + (R/2) = 1.5R$. Thus, $P_{II} = I^2(1.5R) = 1.5I^2R$.For figure $(III)$, all three resistors are in parallel: $R_{eq, III} = R/3$. Thus, $P_{III} = I^2(R/3) \approx 0.33I^2R$.For figure $(IV)$, two resistors are in series and one is in parallel with the combination: $R_{eq, IV} = (2R \cdot R) / (2R + R) = 2R/3 \approx 0.67R$. Thus, $P_{IV} = I^2(0.67R) = 0.67I^2R$.Comparing the values: $0.33I^2R Wait, re-evaluating the standard configurations for this problem: Usually, $(I)$ is series, $(II)$ is $R+(R||R)$, $(III)$ is $R||R||R$, $(IV)$ is $(R+R)||R$. The order is $P_{III} < P_{IV} < P_{II} < P_I$.

Three resistances of equal values $R$ are arranged in four different configurations as shown in the figure. The power dissipation in increasing order is:

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