First slide

Electrical power

Question

Masses of three wires are in the ratio 1:3:5. Their lengths are in the ratio 5:3:1. When connected in series with a battery, the ratio of heat produced in them will be

Moderate

A

125:15: 1
B

1: 15: 125
C

15:1:125
D

5: 9 : 5

Solution

We know that , $H = i^{2} R$
$∴ H = i^{2} (ρl / A) = \frac{i^{2} {ρl}^{2} d}{m}$
or $H \propto (l^{2} / m)$
or $H_{1} : H_{2} : H_{3} = \frac{(5)^{2}}{1} : \frac{(3)^{2}}{3} : \frac{(1)^{2}}{5}$
or $H_{1} : H_{2} : H_{3} = 125 : 15 : 1$
Note. when they are connected in parallel, then
$\begin{aligned} H = V^{2} / R or H \propto m / l^{2} \\ ∴ H_{1} : H_{2} : H_{3} = 1 : 15 : 125 \end{aligned}$

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