The charge flowing through a resistor R varie | vedclass

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Question

(A) The current $i$ is given by the rate of change of charge: $i = \frac{dQ}{dt} = \frac{d}{dt}(3t - 6t^2) = 3 - 12t$.The current becomes zero when $3

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$\frac{3R}{4}$

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(A) The current $i$ is given by the rate of change of charge: $i = \frac{dQ}{dt} = \frac{d}{dt}(3t - 6t^2) = 3 - 12t$.The current becomes zero when $3 - 12t = 0$,which gives $t = \frac{3}{12} = \frac{1}{4} 	ext{ s}$.The heat produced $H$ is given by the integral $H = \int_{0}^{t} i^2 R \, dt$.Substituting the values: $H = \int_{0}^{1/4} (3 - 12t)^2 R \, dt$.Let $u = 3 - 12t$,then $du = -12 \, dt$,or $dt = -\frac{du}{12}$.When $t = 0, u = 3$. When $t = 1/4, u = 0$.$H = R \int_{3}^{0} u^2 \left(-\frac{du}{12}ight) = \frac{R}{12} \int_{0}^{3} u^2 \, du$.$H = \frac{R}{12} \left[ \frac{u^3}{3} ight]_{0}^{3} = \frac{R}{12} 	imes \frac{27}{3} = \frac{27R}{36} = \frac{3R}{4}$.

The charge flowing through a resistor $R$ varies with time $t$ as $Q = 3t - 6t^2$. The heat produced in $R$ until the current in it becomes zero is:

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