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(A) Given: Resistance $R = 100 \Omega$,$AC$ source $\varepsilon = 10 \sin (250 \pi t)$.Comparing with $\varepsilon = \varepsilon_0 \sin (\omega t)$,we

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$\frac{0.57}{\pi} 	ext{ mJ}$

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(A) Given: Resistance $R = 100 \Omega$,$AC$ source $\varepsilon = 10 \sin (250 \pi t)$.Comparing with $\varepsilon = \varepsilon_0 \sin (\omega t)$,we get $\varepsilon_0 = 10 	ext{ V}$ and $\omega = 250 \pi 	ext{ rad/s}$.The energy dissipated as heat $H$ is given by $H = \int_0^{t} \frac{\varepsilon^2}{R} dt$.$H = \frac{1}{R} \int_0^{10^{-3}} (10 \sin (250 \pi t))^2 dt = \frac{100}{100} \int_0^{10^{-3}} \sin^2 (250 \pi t) dt$.Using $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we get $H = \int_0^{10^{-3}} \frac{1 - \cos (500 \pi t)}{2} dt$.$H = \frac{1}{2} \left[ t - \frac{\sin (500 \pi t)}{500 \pi} ight]_0^{10^{-3}}$.$H = \frac{1}{2} \left[ 10^{-3} - \frac{\sin (500 \pi 	imes 10^{-3})}{500 \pi} ight] = \frac{1}{2} \left[ 10^{-3} - \frac{\sin (0.5 \pi)}{500 \pi} ight]$.Since $\sin (0.5 \pi) = 1$,$H = \frac{1}{2} \left[ \frac{1}{1000} - \frac{1}{500 \pi} ight] = \frac{1}{2} \left[ \frac{\pi - 2}{1000 \pi} ight] = \frac{\pi - 2}{2000 \pi} 	ext{ J}$.Approximating $\pi \approx 3.14$,$\pi - 2 \approx 1.14$. So $H \approx \frac{1.14}{2000 \pi} 	ext{ J} = \frac{0.57}{\pi} 	imes 10^{-3} 	ext{ J} = \frac{0.57}{\pi} 	ext{ mJ}$.

$A$ resistor of resistance $100 \Omega$ is connected to an $AC$ source $\varepsilon = 10 \sin (250 \pi t)$. The energy dissipated as heat during $t = 0$ to $t = 1 \text{ ms}$ is approximately.

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