To compare the powers, we use the fact that all three configurations are connected across the same battery voltage $V=3\text{ V}$.

For a fixed voltage source, $P=\frac{V^2}{R_{\text{eq}}}$ So, smaller equivalent resistance means larger power.

1. Find equivalent resistance of each circuit

From the figure, each circuit is made of identical resistors of resistance $R$.

Circuit I

This is the series combination of two resistors: $R_1 = R+R = 2R$

Hence, $P_1 = \frac{V^2}{2R}$

Circuit II

This is a parallel combination of two identical resistors: $R_2 = \frac{R\cdot R}{R+R} = \frac{R}{2}$

Hence, $P_2 = \frac{V^2}{R/2} = \frac{2V^2}{R}$

Circuit III

This is a single resistor of resistance $R$ connected across the battery, so $R_3 = R$

Hence, $P_3 = \frac{V^2}{R}$

2. Compare the powers

We have $P_2 = \frac{2V^2}{R}, \qquad P_3 = \frac{V^2}{R}, \qquad P_1 = \frac{V^2}{2R}$

Therefore, $P_2 > P_3 > P_1$

3. Check the options

• A: $P_1 > P_2 > P_3$ ❌
• B: $P_1 > P_3 > P_2$ ❌
• C: $P_2 > P_1 > P_3$ ❌
• D: $P_3 > P_2 > P_1$ ❌

The correct order is $P_2 > P_3 > P_1$

So none of the listed options matches this order.

4. Comparison with stored correct answer

Stored correct answer: C

But our derived result is: $P_2 > P_3 > P_1$ which does not match option C.

Therefore, I disagree with the stored answer.