Question

The above figure shows two resistors X and Y connected in series to a battery. The power dissipated for this combination is P₁. When these resistors are connected in parallel to the same battery, then the power dissipated is given by P₂. Find out the ratio `P_1/P_2`.

Answer 1

For series, total resistance is R + 2R = 3R.

`P_1 = V^2/(3R)`

For parallel, the total resistance is `(2R)/3`.

`P_2 = V^2/((2R)/3) = (3V^2)/(2R)`

`P_1/P_2 = (V^2/(3R))/((3V^2)/(2R))`

`P_1/P_2 = 2/9`