Answer
(a) $220\ V$. $\frac{1}{60}\ sec$,
(b) See graph.
(c) $I(t)=22sin(120\pi t)$
(d) $22\ A$. $\frac{1}{60}\ sec$
(e) See graph.
Work Step by Step
Given $V(t)=220sin(120\pi t)$, we have
(a) amplitude $|A|=220\ V$. period $p=\frac{2\pi}{120\pi}=\frac{1}{60}\ sec$,
(b) See graph.
(c) Use $I=\frac{V}{R}$, we have $I(t)=22sin(120\pi t)$
(d) For the current, we have amplitude$=22\ A$. period $p=\frac{2\pi}{120\pi}=\frac{1}{60}\ sec$
(e) See graph.
