Question

Suppose that the voltage is reduced by 10 percent (to 90 VV). By what percentage is the power reduced? Assume that the resistance remains constant.

Answer 1

Answer:

The power is reduced by 19 percent.

Explanation:

The formula of power is given by:

$P = \frac{V^{2}}{R}$

In which V is the voltage, and R is the resistance.

I am going to use R = 1 in both cases.

With the original voltage, V = 1, we have

$P = \frac{V^{2}}{R} = \frac{1}{1} = 1$

With the modified voltage, V = 0.9, we have:

$P = \frac{V^{2}}{R} = \frac{0.9^{2}}{1} = 0.81$

So the power is reduced by 1-0.81 = 0.19 = 19 percent.