Question 42: Current Electricity

Question

If voltage across a bulb rated $220 \mathrm{~V}, 50 \mathrm{~W}$ drops by $5 \%$ of its rated value, the percentage of the rated value by which the power would decrease is

Accepted Answer

When the voltage across a resistive bulb changes, its power varies as the square of the voltage. HereÃ¢ÂÂs how to see that:
Rated voltage and power:
$$V_0 = 220	ext{ V},\quad P_0 = 50	ext{ W}.$$
New voltage after a 5 % drop:
$$V = 0.95\,V_0.$$
Power scales like $V^2$:
$$P = P_0\left(\frac{V}{V_0}ight)^2 = P_0\,(0.95)^2 = 50\cdot0.9025 = 45.125	ext{ W}.$$
Percentage decrease in power:
$$\frac{P_0 - P}{P_0}	imes100\% = (1 - 0.9025)	imes100\% = 9.75\%\approx10\%. $$
So the power drops by about
10 %
(Option A).

Answer

$10 \%$

Answer

$2.5 \%$

Answer

$15 \%$

Answer

$5 \%$

Current Electricity

If voltage across a bulb rated 220 \mathrm{~V}, 50 \mathrm{~W} drops by 5 \% of its rated value, the percentage of the rated value by which the power would decrease is

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