The current through a resistor R from an a.c. source is given as $\mathrm{I}=\mathrm{I}_{\mathrm{m}}

Question

The current through a resistor R from an a.c. source is given as $\mathrm{I}=\mathrm{I}_{\mathrm{m}} \sin \omega \mathrm{t}$. The heat liberated in the resistor ' $R$ ' during one half cycle is\

Professor PC Thomas Classes · Answer

Heat $\int_{0}^{\mathrm{T} / 2} \mathrm{I}_{\mathrm{m}}^{2} \sin ^{2} \omega \mathrm{tR} \mathrm{dt}$\ $H=I_{m}^{2} R \int_{0}^{T / 2}\left(\frac{1-\cos 2 \omega t}{2}ight) d t$\ $\mathrm{H}=\frac{\mathrm{I}_{\mathrm{m}}^{2} \mathrm{R}}{2} 	imes \frac{\mathrm{T}}{2} \mathrm{H}=\frac{\mathrm{I}_{\mathrm{m}}^{2} \mathrm{R}}{4}\left[\frac{2 \pi}{\mathrm{w}}ight]$\