# Pumping the Power

What is an AC voltage? How much power does an AC power source
supply?

## Problem

An electric power source produces an AC voltage $v = V_0\sin(\omega t)$, where $V_0$ is the peak voltage. A filament light bulb resistor R is placed across the source. Can you find a formula for the power (which changes over time) produced in the bulb? What is the average power produced in terms of $V_0$ and $R$?

If you would like to keep using the power formulae you learned for DC electricity, what would be a good representative voltage for this AC source, in terms of $V_0$?

Can you draw a graph of the light produced by the bulb over time if the bulb has been on for a long time?

## Getting Started

You will need to know about complex numbers to attempt this
problem.

The necessary background physics and mathematics is contained in the article Electromagnetism

The necessary background physics and mathematics is contained in the article Electromagnetism

## Student Solutions

time-dependent current $i = V_0\sin(\omega t)/R$

time-dependent power $P = iv = V_0^2\sin^2(\omega t)/R$

$\sin^2(\omega t) = 1/2 - 1/2\times \cos(2\omega t)$

$$

\begin{align}

P_{average} &= 1/T\times \int^T_0{P}dt\quad\left(\mbox{where T is the period, }2\pi/\omega\right)\\

&= \frac{V_0^2}{RT}\left[\frac{t}{2} - \frac{1}{4\omega}\sin(2\omega t)\right]^T_0 = \frac{V_0^2}{2R}

\end{align}

$$

The DC power formula for voltage is $P = V^2/R$, so the good representative (RMS) voltage is $V_0/\sqrt{2}\;.$

If you draw out the power graph for the bulb, the frequency of the power is twice the frequency of the voltage, and the graph sits just above the $P=0$ line. The graph of the light output will follow this frequency, but will be raised up on the $y$-axis by some fixed amount, as of course the filament stays hot between cycles! There will also be a shift to the right because the filament will keep heating beyond the peak power for a short time.