NEET-XII-Physics
39: Alternating Current
- #11A 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.Ans : 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, Erms = 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 }_{\mathit{0}}^{t}dH``
`` =\int \frac{{E}_{0}^{2}{\,\mathrm{\,sin\,}}^{2}\,\mathrm{\,\omega \,}t}{R}dt\left(\because {E}_{rms}={E}_{0}\,\mathrm{\,sin\,}\omega t\right)``
`` =\frac{144}{100}{\int }_{0}^{{10}^{-3}}{\,\mathrm{\,sin\,}}^{2}\omega tdt``
`` =1.44{\int }_{0}^{{10}^{-3}}\left(\frac{1-\,\mathrm{\,cos\,}2\omega t}{2}\right)dt``
`` =\frac{1.44}{2}\left[{\int }_{0}^{{10}^{-3}}dt+{\int }_{0}^{{10}^{-3}}\,\mathrm{\,cos\,}2\omega tdt\right]``
`` =0.72\left[{10}^{-3}-{\left\{\frac{\,\mathrm{\,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\,}``
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