Let the same potential difference $V$ be applied across each combination.

Total power dissipated by a combination is $P=\frac{V^2}{R_{\text{eq}}}$ So, smaller equivalent resistance means larger power.

Now compute equivalent resistance for each arrangement (standard 3-resistor combinations of equal resistance $R$):

• A: All three in series $R_A=3R$ $P_A=\frac{V^2}{3R}$

• B: Two in parallel, then in series with one resistor $R_B=R+\frac{R\cdot R}{R+R}=R+\frac{R}{2}=\frac{3R}{2}$ $P_B=\frac{V^2}{(3R/2)}=\frac{2V^2}{3R}$

• C: Two in series, then in parallel with one resistor $R_C=\frac{(2R)R}{2R+R}=\frac{2R}{3}$ $P_C=\frac{V^2}{(2R/3)}=\frac{3V^2}{2R}$

• D: All three in parallel $R_D=\frac{R}{3}$ $P_D=\frac{V^2}{R/3}=\frac{3V^2}{R}$

Compare powers: $P_A=\frac{V^2}{3R},\quad P_B=\frac{2V^2}{3R},\quad P_C=\frac{3V^2}{2R},\quad P_D=\frac{3V^2}{R}$

Hence the increasing order is $P_A<P_B<P_C<P_D$

Checking options: This matches Option D.

Comparison with stored answer: Stored answer is B, but the correct order from equivalent resistances is D. So the stored answer appears incorrect.