Question

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

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

So, current, $i=\frac{dQ}{dt}=a-2bt$

At, $t=0,i=a$

And when, $i=0,t=\frac{a}{2b}$

Now, total heat produced in $R$ from $t=0$ to $t=\frac{a}{2b}$ is,

$H={\int }_0^t{i}^{2}Rdt$

$\Rightarrow H={\int }_0^{a/2b}(a-2bt{)}^{2}Rdt$

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

$\Rightarrow H=R[a^2.(\frac{a}{2b}-0)+\frac{4b^2}{3}.(\frac{a^3}{8b^3}-0)-\frac{4ab}{2}.(\frac{a^2}{4b^2}-0)]$

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

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