Question

A resistor of resistance 100 $\Omega$ is connected to an AC source $\epsilon$ = (12 V) sin $\left(250\pi s^{-1}\right)$t. Find the energy dissipated as heat during t = 0 to t = 1.0 ms.

Answer 1

Given H = $\frac{V^2}{R}$ T;

$E_0=12V;\omega =250\pi$,

R= 100 $\Omega$

H= ${\int }_0^R$ dH

= $\int \frac{E_0^{2}si{n}^2\omega t}{R}$ dt

= $\frac{144}{100}\int si{n}^2\omega$ t

= $\frac{144}{100}\int \left(\frac{1-cos2\omega t}{2}\right)$ dt

= $\frac{1.44}{2}[{\int }_0^{{10}^{-3}}+{\int }_0^{{10}^{-3}}cos2\omega tdt]$

= $0.72[{10}^{-3}-{\left\{\frac{si{n}^2\omega t}{2\omega }\right\}}_0^{10-3}]$

= $0.72[\frac{1}{1000}-\frac{2}{1000\pi }]$

= $\left(\frac{\pi -2}{1000\pi }\right)\times 0.72$

= 0.0002614 = $2.61\times {10}^{-4}$ J