Three resistors, each of resistance R , are arranged in the following…

The correct answer is Option (2) → (B), (D), (C), (A)

The same source drives every combination, so the power dissipated is \( P = E^{2}/R_{\mathrm{eq}} \): the smaller the equivalent resistance, the larger the power.

    (A) Three resistors in series — \( \displaystyle R_{A} = R + R + R = 3R \)

    (B) Three resistors in parallel — \( \displaystyle R_{B} = \frac{R}{3} \)

    (C) One resistor in series with two resistors in parallel — \( \displaystyle R_{C} = R + \frac{R}{2} = \frac{3R}{2} \)

    (D) One resistor in parallel with two resistors in series — \( \displaystyle R_{D} = \frac{R \times 2R}{R + 2R} = \frac{2R}{3} \)

Putting these in \( P = E^{2}/R_{\mathrm{eq}} \):

    \( \displaystyle P_{A} = \frac{E^{2}}{3R} \approx 0.33\,\frac{E^{2}}{R} \qquad P_{B} = \frac{3E^{2}}{R} = 3\,\frac{E^{2}}{R} \)

    \( \displaystyle P_{C} = \frac{2E^{2}}{3R} \approx 0.67\,\frac{E^{2}}{R} \qquad P_{D} = \frac{3E^{2}}{2R} = 1.5\,\frac{E^{2}}{R} \)

So the decreasing order of power is \( P_{B} \gt P_{D} \gt P_{C} \gt P_{A} \), that is (B), (D), (C), (A).