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

The power is reduced by 19 percent.

Step-by-step explanation:

The formula of power is given by:

`P = (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 = (V^(2))/(R) = (1)/(1) = 1`

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

`P = (V^(2))/(R) = (0.9^(2))/(1) = 0.81`

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