The charge flowing through a resistance varies with time as where and are positive constants. The total heat produced in is [2016]

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Solution

Q.
The charge flowing through a resistance $R$ varies with time $t$ as $Q = a t - b t^{2},$ where $a$ and $b$ are positive constants. The total heat produced in $R$ is [2016]

1 $\frac{a^{3} R}{2 b}$

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

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

4 $\frac{a^{3} R}{3 b}$

Ans.
(3)

Given, $Q = a t - b t^{2}$

$∴$ $I = \frac{d Q}{d t} = a - 2 b t$

At $t = 0, Q = 0 \Rightarrow I = 0$

Also, $I = 0$ at $t = \frac{a}{2 b}$

$∴$ Total heat produced in resistance $R,$

$H = \int_{0}^{a / 2 b} I^{2} R d t = R \int_{0}^{a / 2 b} {(a - 2 b t)}^{2} d t$

$= R \int_{0}^{a / 2 b} (a^{2} + 4 b^{2} t^{2} - 4 a b t) d t = R {[a^{2} t + 4 b^{2} \frac{t^{3}}{3} - 4 a b \frac{t^{2}}{2}]}_{0}^{a / 2 b}$

$= R [a^{2} \times \frac{a}{2 b} + \frac{4 b^{2}}{3} \times \frac{a^{3}}{8 b^{3}} - \frac{4 a b}{2} \times \frac{a^{2}}{4 b^{2}}]$

$= \frac{a^{3} R}{b} [\frac{1}{2} + \frac{1}{6} - \frac{1}{2}] = \frac{a^{3} R}{6 b}$

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