Chapter 6.4, Problem 89AYU

a.

To determine

What is amplitude and period?

Answer to Problem 89AYU

  $\boxed{Amplitude=220}$ , $\boxed{Period=\frac{1}{60}}$

Explanation of Solution

Given information:

The voltage $V$ in volts produced by an ac generator at time $t$ , in seconds, is

  $V\left(t\right)=220\sin \left(120\pi t\right)$

What is amplitude and period?

Calculation:

Consider the given function.

  $V\left(t\right)=220\sin \left(120\pi t\right)$

  $\begin{array}{l}Amplitude=220\\ w=2\pi f\\ 120\pi =2\pi f\\ f=120/2=60\\ Period=1/f\\ Period=\frac{1}{60}\end{array}$

Hence, $\boxed{Amplitude=220}$ , $\boxed{Period=\frac{1}{60}}$

b.

To determine

Graph $V$ over two periods, beginning at $t=0$ .

Answer to Problem 89AYU

  

Explanation of Solution

Given information:

The voltage $V$ in volts produced by an ac generator at time $t$ , in seconds, is

  $V\left(t\right)=220\sin \left(120\pi t\right)$

Graph $V$ over two periods, beginning at $t=0$ .

Calculation:

use $\boxed{Amplitude=220}$ and $\boxed{Period=\frac{1}{60}}$ to graph the function $V\left(t\right)=220\sin \left(120\pi t\right)$ .

  

Hence, the graph is drawn.

c.

To determine

What is the current $I$ ?

Answer to Problem 89AYU

  $\boxed{I=22\sin \left(120\pi t\right)}$

Explanation of Solution

Given information:

The voltage $V$ in volts produced by an ac generator at time $t$ , in seconds, is

  $V\left(t\right)=220\sin \left(120\pi t\right)$

If a resistance of $R=10ohms$ is present, what is the current $I$ ?

[Use Ohm’s Law, $V=IR$ ]

Calculation:

  $V\left(t\right)=220\sin \left(120\pi t\right)$

Let $V=IR$

put $R=10ohms$ and equate the function.

  $\begin{array}{l}10\cdot I=220\sin \left(120\pi t\right)\\ I=\frac{220\sin \left(120\pi t\right)}{10}\\ I=22\sin \left(120\pi t\right)\end{array}$

Hence, current is, $\boxed{I=22\sin \left(120\pi t\right)}$ .

d.

To determine

What is amplitude and period of current $I$ ?

Answer to Problem 89AYU

  $\boxed{A=22}$ , $\boxed{T=\frac{1}{60}}$

Explanation of Solution

Given information:

The voltage $V$ in volts produced by an ac generator at time $t$ , in seconds, is

  $V\left(t\right)=220\sin \left(120\pi t\right)$

What is amplitude and period of current $I$ ?

Calculation:

We know that,

  $I=22\sin \left(120\pi t\right)$

  $\begin{array}{l}A=22\\ \omega =120\pi \\ T=\frac{2\pi }{120\pi }=\frac{1}{60}\end{array}$

Hence amplitude and time period is $\boxed{A=22}$ , $\boxed{T=\frac{1}{60}}$

e.

To determine

Graph current $I$ up to two period of time.

Answer to Problem 89AYU

  

Explanation of Solution

Given information:

The voltage $V$ in volts produced by an ac generator at time $t$ , in seconds, is

  $V\left(t\right)=220\sin \left(120\pi t\right)$

Graph current $I$ up to two period of time.

Calculation:

We know that,

  $I=22\sin \left(120\pi t\right)$ let graph it.

  

Hence the graph is drawn.