Electricity — Question 4

1:4

Reason — Let R_s and R_p be the equivalent resistance of the wires when connected in series and parallel respectively.

R_s = R + R = 2R

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{R}} + \dfrac{1}{\text{R}} = \dfrac{2}{\text{R}}$

R_p = $\dfrac{\text{R}}{2}$

Let P_s and P_p be the power consumed in series and parallel circuits, respectively.

Power (P) = $\dfrac{\text{V}^2}{\text{R}}$

P_s = $\dfrac{\text{V}^2}{\text{2R}}$

and

P_p = $\dfrac{\text{V}^2}{\dfrac{\text{R}}{2}} = \dfrac{2\text{V}^2}{\text{R}}$

For the same potential difference V, the ratio of the heat produced in the circuit is given by ratio of P_s and P_p hence

$\text{H}_s : \text{H}_p = \dfrac{\dfrac{\text{V}^2}{\text{2R}}}{\dfrac{2\text{V}^2}{\text{R}}} \\[1em] = \dfrac{\text{V}^2 \times \text{R}}{\text{2R} \times 2\text{V}^2} \\[1em] = \dfrac{1}{4}$

Hence, the ratio of the heat produced is 1:4.