Q. 89 - Precalculus Enhanced with Graphing Utilities

Question

Alternating Current (ac) Generators The voltage V, in volts, produced by an ac generator at time t, in seconds, is

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

(a) What is the amplitude? What is the period?
(b) Graph V over two periods, beginning at $t=0$.
(c) If a resistance of $R=10ohms$ is present, what is the current I ?
[Hint: Use Ohm's Law, $[V=IR]$
(d) What is the amplitude and period of the current I ?
(e) Graph I over two periods, beginning at t=0. 

Step-by-Step Solution

Answer

(a)  Amplitude and period of the given function $220 and \frac{1}{6}$ respectively

(b) Graph V over two periods, beginning at $t=0$

(c) Current $I=22\sin \left(120\mathrm{\pi t}\right)$

(d) current amplitude and period $22 and \frac{1}{60}$respectively

(e)  Graph of the function

Step 1.Given information

The given function $V=220\sin \left(120\mathrm{\pi t}\right)$

Step 2.(a) What is the amplitude? What is the period?

For the given current function $V\left(t\right)=220\sin \left(120\mathrm{\pi t}\right)$, we can find the amplitude and period by comparing it with $y=A\sin \left(\omega t\right)$ So,$A=220$  and $\omega =120\pi$ and hence we can find the period

$$\begin{gathered} T=\frac{2\mathrm{\pi }}{\omega } \\ =\frac{2\mathrm{\pi }}{12\mathrm{\pi }} \\ =\frac{1}{6} \end{gathered}$$

Step 3.(b) Graph V over two periods, beginning at t = 0

Step 4.(c) If a resistance of R = 10 o h m s is present, what is the current I ?

Substitute $220\sin \left(120\mathrm{\pi t}\right)$ for V and 10 for R in the equation $V=IR$

$$\begin{gathered} 220\sin \left(120\mathrm{\pi t}\right)=\mathrm{I}.10 \\ 10.\mathrm{I}=220\sin \left(120\mathrm{\pi t}\right) \end{gathered}$$

Divide both sides of the equation by 10 .

$$\begin{gathered} \frac{10.I}{10}=\frac{220\sin \left(120\mathrm{\pi t}\right)}{10} \\ I=22\sin \left(120\mathrm{\pi t}\right) \end{gathered}$$

Step 5.(d) What is the amplitude and period of the current I ?

Comparing $I\left(t\right)=22\sin \left(120\mathrm{\pi t}\right)$ to the standard form $y=A\sin \left(\omega t\right)$. we get $A=22$ ,$\omega =120\pi$ , and  $$\begin{gathered} T=\frac{2\mathrm{\pi }}{120\mathrm{\pi }} \\ =\frac{1}{6} \end{gathered}$$

Therefore, the period and amplitude of the function $I\left(t\right)=22\sin \left(120\mathrm{\pi t}\right)$ are $\frac{1}{60}$ and 22 , respectively.

Step 6.Graph I over two periods, beginning at t = 0

By using amplitude and period Ploting graph as