Question

A resistor of resistance 100 Ω is connected to an AC source ε = (12 V) sin (250 π s⁻¹)t. Find the energy dissipated as heat during t = 0 to t = 1.0 ms.

Answer 1

Given:
Peak voltage of AC source, E₀ = 12 V
Angular frequency, ω = 250 $\mathrm{\pi }$ s⁻¹
Resistance of resistor, R = 100 Ω
Energy dissipated as heat $\left(H\right)$ is given by,
$H=\frac{{E_{\mathrm{rms}}}^2}{R}T$
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,
$$\begin{gathered} H={\int }_{\mathit{0}}^{t}dH \\ =\int \frac{E_0^2{\sin }^2\mathrm{\omega }t}{R}dt\left(\because E_{rms}=E_0\sin \omega t\right) \\ =\frac{144}{100}{\int }_0^{{10}^{-3}}{\sin }^2\omega t\mathit{}dt \\ =1.44{\int }_0^{{10}^{-3}}\left(\frac{1-\cos 2\omega t}{2}\right)dt \\ =\frac{1.44}{2}\left[{\int }_0^{{10}^{-3}}dt+{\int }_0^{{10}^{-3}}\cos 2\omega tdt\right] \\ =0.72\left[{10}^{-3}-{\left\{\frac{\sin 2\omega t}{2\omega }\right\}}_0^{{10}^{-3}}\right] \\ =0.72\left[\frac{1}{1000}-\frac{1}{500\mathrm{\pi }}\right] \\ =0.72\left[\frac{1}{1000}-\frac{2}{1000\mathrm{\pi }}\right] \\ =\left(\frac{\mathrm{\pi }-2}{1000\mathrm{\pi }}\right)\times 0.72 \\ =0.0002614=2.61\times {10}^{-4}\mathrm{J} \end{gathered}$$