Chapter 5.9, Problem 64ES

Electricity is supplied to homes in the form of alternating current, which means that the voltage has a sinusoidal waveform described by an equation of the form

$V=V_p\sin \left(2\pi ft\right)$

(see the accompanying figure). In this equation, $V_p$ is called the peak voltage or amplitude of the current, $f$ is called its frequency, and $1/f$ is called its period. The voltages $V$ and $V_p$ are measured in volts $\left(\text{V}\right)$ , the time $t$ is measured in seconds (s), and the frequency is measured in hertz (Hz). ( $1\text{ Hz}=1$ cycle per second; a cycle is the electrical term for one period of the waveform.) Most alternating-current voltmeters read what is called the rms ox root -mean-square value of $V$ . By definition, this is the square root of the average value of $V^2$ over one period.

(a) Show that

$V_{\text{rms}}=\frac{V_p}{\sqrt{2}}$

(b) In the United States, electrical outlets supply alternating current with an rms voltage of $120\text{ V}$ at a frequency of $60\text{ Hz}$ . What is the peak voltage at such an outlet?