Chapter 25.1, Problem 7PP

To determine

To Calculate:Peak power of the electric light.

Answer to Problem 7PP

  $1.5\times {10}^2\text{W}$

Explanation of Solution

Given:

Average power, $P=75\text{W}$

Formula Used:

  $P=\frac{1}{2}{P}_{\max }$

Where,

  $P:$ average power

  $P_{\max }:$ peak power

Calculation:

For the maximum power of electric light,

  $P=\frac{1}{2}{P}_{\max }$

Rearranging the equation,

  $P_{\max }=\left(2\right)P$

Substituting the values in the above formula,

  $\begin{array}{c}{P}_{\max }=\left(\text{2}\right)\left(\text{75W}\right)\\ =1.5\times {10}^2\text{W}\end{array}$

Conclusion:

Hence, peak power of the electric light is $1.5\times {10}^2\text{W}$ .