Chapter 7.2, Problem 40E

a.

To determine

To find:an expression for power in terms of ${\cos }^{2}2\pi ft$ .

Answer to Problem 40E

  $P=I_0^{2}R\left(1-{\cos }^{2}2\pi ft\right)$

Explanation of Solution

Given information:

  $P=I_0^{2}R{\sin }^{2}2\pi ft$

Calculation:

The expression for power in terms of ${\cos }^{2}2\pi ft$ can be written as follows:

  $\begin{array}{c}P=I_0^{2}R{\sin }^{2}2\pi ft\left[\text{Let }x=2\pi f\right]\\ =I_0^{2}R\left(1-{\cos }^{2}2\pi ft\right)\left[\text{Use }{\sin }^{2}x=1-{\cos }^{2}x\right]\end{array}$

b.

To determine

To find: an expression for power in terms of ${\csc }^{2}2\pi ft$ .

Answer to Problem 40E

  $P=\frac{I_0^{2}R}{{\csc }^{2}2\pi ft}$

Explanation of Solution

Given information:

  $P=I_0^{2}R{\sin }^{2}2\pi ft$

Calculation:

The expression for power in terms of ${\csc }^{2}2\pi ft$ can be written as follows:

  $\begin{array}{c}P=I_0^{2}R{\sin }^{2}2\pi ft\left[\text{Let }\text{}x=2\pi f\right]\\ =I_0^{2}R\left(\frac{1}{{\csc }^{2}2\pi f}\right)\left[\text{Use }\sin x=\frac{1}{\csc x}\right]\\ =\frac{I_0^{2}R}{{\csc }^{2}2\pi f}\end{array}$