During a period of high power demand, the voltage output of the power company is reduced by 5.0%. By what percentage is the power in a resistor decreased?
A) 10%
B) 5.0%
C) 90%
D) 15%
E) 2.5%

Question

Quizplus · Accepted Answer

The power dissipated in a resistor is given by $P = \frac{V^2}{R}$, where $P$ is the power, $V$ is the voltage across the resistor, and $R$ is the resistance of the resistor. If the voltage is reduced by 5%, the new voltage is $0.95V$. The new power is $P' = \frac{(0.95V)^2}{R} = \frac{0.9025V^2}{R}$. The percentage decrease in power is $\frac{P - P'}{P} \times 100 = \frac{V^2/R - 0.9025V^2/R}{V^2/R} \times 100 = (1 - 0.9025) \times 100 = 9.75%$, which rounds to approximately 10%.

During a Period of High Power Demand, the Voltage Output