Question

# A resistor of resistance $100ohm$ is connected to an AC source, $\left(12V\right)\mathrm{sin}\left(250\pi s-1\right)t$. Find the energy dissipated as heat during$t=0$to $t=1$ $ms.$

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Solution

## Given dataThe AC voltage source is $E=\left(12V\right)\mathrm{sin}\left(250\pi s-1\right)t$.The resistance of the resistor is $R=100ohm.$Time duration from $t=0$to $t=1$ $ms.$Joule's law of heating If a conductor of resistance carries a current I for time t, then the heat developed in the conductor is proportional to the square of the current.Joule's law can be defined by the form, $H={I}^{2}Rt$, where, where, H is the heat developed in the conductor.Calculation of energy dissipationThe given voltage is $E={E}_{o}\mathrm{sin}\omega t$, where, ${E}_{o}=12volt$, $\omega =250\pi {s}^{-1}$.So, energy dissipated in the body is $H={I}^{2}Rt=\frac{{V}^{2}}{R}×t\left(\mathrm{sin}ce,V=IR\right)\phantom{\rule{0ex}{0ex}}orH=\frac{1}{R}{\int }_{0}^{{10}^{-3}}{{E}_{o}}^{2}{\mathrm{sin}}^{2}\omega tdt.\phantom{\rule{0ex}{0ex}}orH=\frac{1}{100}{\int }_{0}^{{10}^{-3}}{\left(12\right)}^{2}{\mathrm{sin}}^{2}\omega tdt.\phantom{\rule{0ex}{0ex}}orH=\frac{144}{100}{\int }_{0}^{{10}^{-3}}{\mathrm{sin}}^{2}\omega tdt\phantom{\rule{0ex}{0ex}}orH=1.44{\int }_{0}^{{10}^{-3}}\left[\frac{1-\mathrm{cos}2\omega t}{2}\right]dt\phantom{\rule{0ex}{0ex}}orH=\frac{1.44}{2}{\int }_{0}^{{10}^{-3}}dt-1.44{\int }_{0}^{{10}^{-3}}\frac{\mathrm{cos}2\omega t}{2}dt\phantom{\rule{0ex}{0ex}}orH=\frac{1.44}{2}\left[t\right]{}_{0}^{{10}^{-3}}-\frac{1.44}{2}{\left[\frac{\mathrm{sin}2\omega t}{2\omega }\right]}_{0}^{{10}^{-3}}\phantom{\rule{0ex}{0ex}}orH=0.72\left[{10}^{-3}-\frac{1}{2×250\mathrm{\pi }}\right]\phantom{\rule{0ex}{0ex}}orH=0.72\left[\frac{1}{1000}-\frac{1}{500\mathrm{\pi }}\right]\phantom{\rule{0ex}{0ex}}orH=0.72\left[\frac{\mathrm{\pi }-2}{1000}\right]=2.61×{10}^{-4}\phantom{\rule{0ex}{0ex}}orH=2.61×{10}^{-4}joule.$Therefore, the energy dissipated as heat is $2.61×{10}^{-4}joule.$

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