Question

The charge flowing through a resistance $R$ varies with time $t$ as $Q=at-b{t}^2$, where $a$ and $b$ are positive constants. The total heat produced in $R$, from $t=0$ till the current becomes zero, is:

• A. $\frac{a^{3}R}{6b}$
• B. $\frac{a^{3}R}{3b}$
• C. $\frac{a^{3}R}{2b}$
• D. $\frac{a^{3}R}{b}$

Answer 1

The correct option is A $\frac{a^{3}R}{6b}$

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

$\therefore I=\frac{dQ}{dt}=a-2bt$

Let at time $t_0$, current becomes zero.

$I=a-2b{t}_0=0$

$\therefore t_0=\frac{a}{2b}$

Now, total heat produced in the resistance is

$H={\int }_0^{t_0}{I}^{2}Rdt$

Here, $I^2=(a^2+4b^2{t}^2-4abt)$

$\therefore H={\int }_0^{a/2b}{I}^{2}Rdt$

$=R\left\{{\int }_0^{a/2b}{a}^{2}dt+{\int }_0^{a/2b}4b^2{t}^{2}dt-{\int }_0^{a/2b}4abtdt\right\}$

$=R[a^2\cdot \left(\frac{a}{2b}\right)+\frac{4b^2}{3}\cdot \left(\frac{a^3}{8b^3}\right)-\frac{4ab}{2}\cdot \left(\frac{a^2}{4b^2}\right)]$

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

$\therefore H=\frac{a^{3}R}{6b}$

Therefore, option $\left(A\right)$ is the correct answer.