Chapter 6.6, Problem 27AYU

Alternating Current (ac) Circuits The current $I$, in amperes, flowing through an ac (alternating current) circuit at time $t$, in seconds, is

$I\left(t\right)\text{ = 120sin}\left(30\pi t-\frac{\pi }{3}\right)t\ge \text{ 0}$

What is the period? What is the amplitude? What is the phase shift? Graph this function over two periods.

To determine

To find: The period, amplitude, phase shift and graph of the function.

Answer to Problem 27AYU

Amplitude $=120 amp$, Period $=\frac{1}{15}s$, Phase shift $=\frac{1}{90}s$.

Explanation of Solution

Given:

$I\left(t\right)=120\sin \left(30\pi t-\frac{\pi }{3}\right)$, where Current $I$, in amperes and time $t$ in seconds.

Calculation:

Compare $y=120\sin \left(30\pi t-\frac{\pi }{3}\right)$ to $y=A\sin \left(\omega x-\varphi \right)$, note that $A=120$, $\omega =30\pi$, and $\varphi =\frac{\pi }{3}$. The graph is a sine curve with amplitude $\left|A\right|=120 amp$, period $T=\frac{2\pi }{\omega }=\frac{2\pi }{30\pi }=\frac{1}{15}s$ and phase shift $=\frac{\varphi }{\omega }=\frac{\frac{\pi }{3}}{30\pi }=\frac{1}{90}s$.