Question 51: Current Electricity

Question

A $500 \mathrm{~W}$ heating unit is designed to operate on a $400 \mathrm{~V}$ line. If line voltage drops to $160 \mathrm{~V}$, the percentage drop in heat output will be:

Accepted Answer

To determine the percentage drop in heat output, we first need to understand that the power output $ P $ of an electrical device is proportional to the square of the voltage $ V $. Mathematically, this relationship is represented as:
$ P \propto V^2 \quad 	ext{or} \quad P = \frac{V^2}{R} $
where $ R $ is the resistance of the heating unit. In this problem, we need to compare the power output at two different voltages, $ 400 \mathrm{~V} $ and $ 160 \mathrm{~V} $.
Let's denote the initial power output at $ 400 \mathrm{~V} $ as $ P_1 $ and the power output at $ 160 \mathrm{~V} $ as $ P_2 $. Given that the initial power output is $ 500 \mathrm{~W} $, we can write:
$ P_1 = 500 \mathrm{~W} $
Since the power is proportional to the square of the voltage:
$ P_1 = k \cdot V_1^2 $
and
$ P_2 = k \cdot V_2^2 $
where $ k $ is a proportionality constant. We can divide these equations to find the ratio of $ P_2 $ to $ P_1 $:
$ \frac{P_2}{P_1} = \frac{V_2^2}{V_1^2} $
Substituting the given voltage values:
$ \frac{P_2}{P_1} = \frac{(160)^2}{(400)^2} = \frac{25600}{160000} = \frac{16}{100} = 0.16 $
This means that the power output at $ 160 \mathrm{~V} $ is $ 0.16 $ times the power output at $ 400 \mathrm{~V} $. To find $ P_2 $, we multiply $ P_1 $ by $ 0.16 $:
$ P_2 = 0.16 \cdot 500 \mathrm{~W} = 80 \mathrm{~W} $
The percentage drop in heat output is then given by:
$ 	ext{Percentage Drop} = \left( \frac{P_1 - P_2}{P_1} ight) 	imes 100 $
Substituting the values:
$ 	ext{Percentage Drop} = \left( \frac{500 \mathrm{~W} - 80 \mathrm{~W}}{500 \mathrm{~W}} ight) 	imes 100 = \left( \frac{420 \mathrm{~W}}{500 \mathrm{~W}} ight) 	imes 100 = 84\% $
Thus, the percentage drop in heat output is 84%. The correct answer is:
Option C - 84%.

Answer

74%

Answer

85%

Answer

84%

Answer

75%

Current Electricity

A 500 \mathrm{~W} heating unit is designed to operate on a 400 \mathrm{~V} line. If line voltage drops to 160 \mathrm{~V}, the percentage drop in heat output will be:

Choose the correct answer: