Question

A resistor of resistance 100 ohm is connected to an AC source E = (12V)sin (250πs^-1)t. Find the energy dissipated as heat during t=0 to t=1ms

Answer 1

Heat energy dissipated $H=\int_0^{t}\space\frac{E^2}{R}dt$

where $t=10^{-3}$, $E=E_{rms}=E_0sin\omega t$ ,

Therefore

$\int_0^{t}\space \frac{(12sin\omega t)^{2}}{100}dt$

$\frac{144}{100}\int_0^{t}\space\frac{(1-cos2\times250\pi t)}{2}dt$

$\frac{144}{200}[1-\frac{sin500\pi t}{500\pi}]_0^t$

At $t=10^{-3}$

Energy dissipated =$\frac{144}{200}[10^{-3}-\frac{1}{500\pi}]$

Heat Energy= $2.61\times10^{-4}J$