The charge flowing through a resistance R var | vedclass

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Question

(A) Given the charge $Q = at - bt^2$.The current $I$ is the rate of flow of charge: $I = \frac{dQ}{dt} = a - 2bt$.The current becomes zero at time $t_

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$\frac{a^3R}{6b}$

Vedclass · Answer

(A) Given the charge $Q = at - bt^2$.The current $I$ is the rate of flow of charge: $I = \frac{dQ}{dt} = a - 2bt$.The current becomes zero at time $t_0$ when $a - 2bt_0 = 0$,which gives $t_0 = \frac{a}{2b}$.The heat produced $H$ in the resistance $R$ is given by $H = \int_0^{t_0} I^2 R \, dt$.Substituting $I = a - 2bt$:$H = \int_0^{a/2b} (a - 2bt)^2 R \, dt = R \int_0^{a/2b} (a^2 - 4abt + 4b^2t^2) \, dt$.Integrating with respect to $t$:$H = R \left[ a^2t - 2abt^2 + \frac{4b^2t^3}{3} \right]_0^{a/2b}$.Substituting the upper limit $t = \frac{a}{2b}$:$H = R \left[ a^2(\frac{a}{2b}) - 2ab(\frac{a^2}{4b^2}) + \frac{4b^2}{3}(\frac{a^3}{8b^3}) \right]$$H = R \left[ \frac{a^3}{2b} - \frac{a^3}{2b} + \frac{a^3}{6b} \right] = \frac{a^3R}{6b}$.

The charge flowing through a resistance $R$ varies with time according to $Q = at - bt^2$. The total heat produced in $R$ is: (assume that the direction of current is not reversed)

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