Question

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

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

(a) What is the amplitude? What is the period?
(b) Graph V over two periods, beginning att=0.
(c) If a resistance of R=20 ohms is present, what is the currentI ?
[Hint: Use Ohm's Law, V=IR
(d) What is the amplitude and period of the current I.
(e) Graph Iover two periods, beginning att=0 .

Short answer

(a) Amplitude and period of the given function $120and\frac{1}{60}$respectively

(b) Graph beginning at t=0

(c) Current $I=6\sin \left(120\mathrm{\pi t}\right)$

(d) Amplitude and period of the current $6and\frac{1}{60}$respectively

(e) Graph of the given function

Step-by-step solution

Step 1.Given information

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

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

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

$$\begin{gathered} T=\frac{2\mathrm{\pi }}{\omega } \\ =\frac{2\mathrm{\pi }}{120\mathrm{\pi }} \\ =\frac{1}{60} \end{gathered}$$

Therefore,$T=\frac{1}{60}andA=120$respectively

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

The graph of the given voltage function over 2 periods will be-:

Step 4.(c) If a resistance of R=20 ohms is present, what is the current I ?

Using Ohm's law

$$\begin{gathered} I\left(t\right)=\frac{V\left(t\right)}{R} \\ =\frac{120\sin \left(120\mathrm{\pi t}\right)}{20} \\ =6\sin \left(120\mathrm{\pi t}\right) \end{gathered}$$

Therefore,$I\left(t\right)=6\sin \left(120\mathrm{\pi t}\right)$

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

For the given $I\left(t\right)=6\sin \left(120\mathrm{\pi t}\right)$we can find the amplitude and period by comparing with $y=A\sin \left(\omega t\right)$.So,$A=6and\omega =120\pi$and we can find the period

$$\begin{gathered} T=\frac{2\mathrm{\pi }}{\omega } \\ =\frac{2\mathrm{\pi }}{120\pi } \\ =\frac{1}{60} \end{gathered}$$

Therefore,$T=\frac{1}{60}andA=6$

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

The given required graph.