A resistance R = 12 , Omega is connected acro | vedclass

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(B) From the graph,the $emf$ $E$ varies linearly with time $t$ as $E(t) = mt$. At $t = 4 \, s$,$E = 24 \, V$. Therefore,the slope $m = \frac{24}{4} =

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$64$

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(B) From the graph,the $emf$ $E$ varies linearly with time $t$ as $E(t) = mt$. At $t = 4 \, s$,$E = 24 \, V$. Therefore,the slope $m = \frac{24}{4} = 6 \, V/s$. Thus,$E(t) = 6t$.The heat developed $H$ in the resistance $R$ is given by the integral of power $P = \frac{E^2}{R}$ over the time interval from $0$ to $4 \, s$:$H = \int_{0}^{4} \frac{E^2}{R} \, dt = \int_{0}^{4} \frac{(6t)^2}{12} \, dt$$H = \int_{0}^{4} \frac{36t^2}{12} \, dt = \int_{0}^{4} 3t^2 \, dt$$H = [t^3]_{0}^{4} = 4^3 - 0^3 = 64 \, J$.

$A$ resistance $R = 12 \, \Omega$ is connected across a source of $emf$ as shown in the figure. Its $emf$ changes with time as shown in the graph. What is the heat developed in the resistance in the first four seconds (in $, J$)?

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