The charge q flows through a resistance R varies with time t as q = at - bt². The total heat produced in fi is

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

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

$\frac{a^{3} R}{2 b}$

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

answer is A.

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Detailed Solution

$i = \frac{dq}{dt} = a - 2 bt$
For $t = t_{0}, i = 0$
$∴ 0 = a - 2 {bt}_{0} or t_{0} = \frac{a}{2 b}$
So, $H = \int_{0}^{t_{0}} i^{2} Rdt = \int_{0}^{t_{0}} (a - 2 bt)^{2} Rdt$
$= \frac{{Ra}^{3}}{6 b}$

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