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The charge flowing through a resistance $R$ varies with time $t$ as $Q=2at=bt^2$ where $a$ and $b$ are positive arbituary constant.

The total heat produced in $R$ is
Option: 1
$\frac{4}{3} \frac{b^3 R}{a}$

Option: 2
$\frac{4}{3} \frac{a^3 R}{6}$

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

Option: 4
$\frac{5}{6} a^3 R .$

Answers (1)

Current through $R,$ $I=\frac{d Q}{d t}$

$\begin{aligned} & =\frac{d\left(2 a t-b t^2\right)}{d t} \\ & =(2 a-2 b t) \end{aligned}$

For, $I=0 \Rightarrow 2 a-2 b t=0$

$I\Rightarrow \quad t=\frac{2 a}{2 b}=\frac{a}{b} .$

So heat produced is $R$

Heat produced $=\int_0^{a / b}(2 a-2 b t)^2 R d t$

$\begin{aligned} & \left.=R \int_0^{a / b} (4 a^2+4 b^2 t^2-8 a b t)\right) d t \\ & =R\left[4 a^2 t+\frac{4 b^2 t^3}{3}-\frac{8 a b t^2}{2}\right]_0^{a / b} \end{aligned}$

$\begin{aligned} & =R\left[4 a^2 \cdot \frac{a}{b}+\frac{4 b^2 \cdot a^3}{3 b^3}-\frac{8 a b}{2} \cdot \frac{a^2}{b^2}\right] \\ & =R\left[\frac{4 a^3}{6}+\frac{4 a^3}{3}-\frac{4 a^3}{b}\right] \\ & =4 / 3 (a^3 R / b) \end{aligned}$

Posted by

Ramraj Saini

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