Question

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

Answer 1

Given :

the peak voltage of AC source $E_0=12V$

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

The resistance of the resistor, $R=100\Omega$

The energy dissipated as heat (H)is given by,

$H=\frac{E_{rms}}{R}T$

here, $E_{rms}$= RMS value of voltage

$R$= Resistance of the resistor

$T=temperature$

Energy dissipated as heat during $t=0$ to $t=1.0$ ms,

$H={\int }_0^{{10}^{-3}}dH$

$=\int \frac{E_0^{2}si{n}^{2}wt}{R}dt\left(since{E}_{Rms=E_{0}sinwt}\right)$

$=\frac{144}{100}{\int }_0^{{10}^{-3}}\left(si{n}^{2}wt\right)dt$

$1.44{\int }_0^{{10}^{-3}}\left(\frac{1-cos2wt}{2}\right)dt$

$=\frac{1.44}{2}[{\int }^{{10}_0^{-3}}dt+{\int }_0^{{10}^{-3}}cos2wtdt]$

$=0.72[{10}^{-3}-{\left(\frac{sin2wt}{2w}\right)}_0^{{10}^{-3}}]$

$=0.72[\frac{1}{100}-\frac{1}{500\pi }]$

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

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

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

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