Question

A $1000W$ heating unit is designed to operate on a $120V$ line. If line voltage drops to $110V$, then the percentage of heat output drops by:

• A. $4\%$
• B. $8\%$
• C. $12\%$
• D. $19\%$

Answer 1

The correct option is D $19\%$

Given:

When voltage $V_1=120V$, power $P_1=1000W$

Need to find:

When voltage $V_2=110V$, power $P_2=?$

By joule's law, power consumed is $P=\frac{V^2}{R}$

Hence, $\frac{P_2}{P_1}=\frac{V_2^2}{V_1^2}$, when resistance $R$ is constant.

Subtracting variable $1$ from both sides and simplifying,

$\frac{P_2-P_1}{P_1}=\frac{V_2^2-V_1^2}{V_1^2}$

Percentage drop is:

$\frac{\Delta P}{P_1}\times 100=\frac{P_2-P_1}{P_1}\times 100=\frac{{110}^2-{120}^2}{120}\times 100=-19\%$ (The -ve sign indicates a power drop).