Question

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

Answer 1

Given,

Peak voltage, $v_o$= 12 v

Angular velocity, $\omega =250\pi s^{-1}$

R= 100$\Omega$

Energy dissipated heat, H=$I^{2}RT$

= $\frac{v^2}{R^2}$RT

= $\frac{v_{rms}^2}{R}T$

Energy dissipated heat(H) during t=0, t= 1ms

H=${\int }_o^{t}dH$

$H={\int }_0^t\frac{v^2{sin}^2\omega t}{R}dt$

(Since $v_{rms}=v_{o}Sin\omega t$)

=$\frac{144}{100}{\int }_0^{{10}^{-3}}{Sin}^2\omega tdt$

$1.44{\int }_0^{{10}^{-3}}\left(\frac{1-Cos2\omega t}{2}\right)dt$

=$\frac{1.44}{2}[{\int }_0^{{10}^{-3}}dt+{\int }_0^{{10}^{-3}}Cos2\omega tdt]$

=0.72[${10}^{-3}-\{\frac{Sin2\omega t}{2\omega }{\}}_0^{{10}^{-3}}$]

= 0.72 [$\frac{1}{1000}-\frac{1}{500\pi }$]

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

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

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