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.

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Solution

## Given:Peak voltage of AC source, E0 = 12 VAngular frequency, ω = 250 $\mathrm{\pi }$ s−1Resistance of resistor, R = 100 ΩEnergy dissipated as heat $\left(H\right)$ is given by,$H=\frac{{{E}_{\mathrm{rms}}}^{2}}{R}T$Here, Erms = RMS value of voltage R = Resistance of the resistor T = TemperatureEnergy dissipated as heat during t = 0 to t = 1.0 ms,$H={\int }_{\mathit{0}}^{t}dH\phantom{\rule{0ex}{0ex}}=\int \frac{{E}_{0}^{2}{\mathrm{sin}}^{2}\mathrm{\omega }t}{R}dt\left(\because {E}_{rms}={E}_{0}\mathrm{sin}\omega t\right)\phantom{\rule{0ex}{0ex}}=\frac{144}{100}{\int }_{0}^{{10}^{-3}}{\mathrm{sin}}^{2}\omega t\mathit{}dt\phantom{\rule{0ex}{0ex}}=1.44{\int }_{0}^{{10}^{-3}}\left(\frac{1-\mathrm{cos}2\omega t}{2}\right)dt\phantom{\rule{0ex}{0ex}}=\frac{1.44}{2}\left[{\int }_{0}^{{10}^{-3}}dt+{\int }_{0}^{{10}^{-3}}\mathrm{cos}2\omega tdt\right]\phantom{\rule{0ex}{0ex}}=0.72\left[{10}^{-3}-{\left\{\frac{\mathrm{sin}2\omega t}{2\omega }\right\}}_{0}^{{10}^{-3}}\right]\phantom{\rule{0ex}{0ex}}=0.72\left[\frac{1}{1000}-\frac{1}{500\mathrm{\pi }}\right]\phantom{\rule{0ex}{0ex}}=0.72\left[\frac{1}{1000}-\frac{2}{1000\mathrm{\pi }}\right]\phantom{\rule{0ex}{0ex}}=\left(\frac{\mathrm{\pi }-2}{1000\mathrm{\pi }}\right)×0.72\phantom{\rule{0ex}{0ex}}=0.0002614=2.61×{10}^{-4}\mathrm{J}$

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