Answer
$P_b < P_d < P_a < P_c$
Work Step by Step
Recall that $P = I\Delta V = I^2R = \displaystyle \frac{\Delta V^2}{R}$
Plugging in the given values to find the power of each resistor, we get:
a.) $\displaystyle P_a = \frac{\Delta V^2}{R}$
b.) $\displaystyle P_b = \frac{(2\Delta V^2)}{R} = \frac{4\Delta V^2}{R}$
c.) $\displaystyle P_c = \frac{\Delta V^2}{(2R)} = \frac{\Delta V^2}{2R}$
d.) $\displaystyle P_d = \frac{\Delta V^2}{\frac{1}{2}R} = \frac{2\Delta V^2}{R}$
So in order of largest to smallest power, we get:
$P_b < P_d < P_a < P_c$
