Chapter 6.7, Problem 47E

a.

To determine

To find: The amplitude of the current.

Answer to Problem 47E

The amplitude of the currentis $220$ .

Explanation of Solution

Given information: The equation is $I=220\sin \left(60\pi t-\frac{\pi }{6}\right)$ .

Calculation:

Consider the equation.

  $I=220\sin \left(60\pi t-\frac{\pi }{6}\right)$

Consider the equation.

  $I=A\sin \left(kt+c\right)+h$

Compare the above equation.

  $A=220$

Thus, the amplitude of the current is $220$ .

b.

To determine

To find: The period of the current.

Answer to Problem 47E

The period of the current is $\frac{1}{30}$ .

Explanation of Solution

Given information: The equation is $I=220\sin \left(60\pi t-\frac{\pi }{6}\right)$ .

Calculation:

Consider the equation.

  $I=220\sin \left(60\pi t-\frac{\pi }{6}\right)$

Consider the equation.

  $I=A\sin \left(kt+c\right)+h$

Compare the above equation.

  $\begin{array}{c}\frac{2\pi }{k}=\frac{2\pi }{60\pi }\\ =\frac{1}{30}\end{array}$

Thus, the period of the current is $\frac{1}{30}$ .

c.

To determine

To find: The phase shift of the current.

Answer to Problem 47E

The phase shift of the current is $\frac{1}{360}$ .

Explanation of Solution

Given information: The equation is $I=220\sin \left(60\pi t-\frac{\pi }{6}\right)$ .

Calculation:

Consider the equation.

  $I=220\sin \left(60\pi t-\frac{\pi }{6}\right)$

Consider the equation.

  $I=A\sin \left(kt+c\right)+h$

Compare the above equation.

  $\begin{array}{l}-\frac{c}{k}=-\frac{\left(-\frac{\pi }{6}\right)}{60\pi }\\ =\frac{1}{360}\end{array}$

Thus, the phase shift of the current is $\frac{1}{360}$ .

d.

To determine

To find: The current when $t=60$ .

Answer to Problem 47E

The current when $t=60$ is $110$ .

Explanation of Solution

Given information: The equation is $I=220\sin \left(60\pi t-\frac{\pi }{6}\right)$ .

Calculation:

Consider the equation.

  $\begin{array}{c}I=220\sin \left(60\pi t-\frac{\pi }{6}\right)\\ I=220\sin \left(60\pi \left(60\right)-\frac{\pi }{6}\right)\\ =220\sin \left(\frac{5129\pi }{6}\right)\\ =220\left(-0.5\right)\\ =110\end{array}$

Thus, the current when $t=60$ is $110$ .