Question

The charge $Q$ flowing through a resistance $R$ varies with time $t$ as $Q=at-b{t}^2$. The total heat produced in $R$ is (up to the instant when instantaneous current is zero)

• 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

Answer: (A)

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

By definition current is rate of flow of charge, hence
$I=\frac{dQ}{dt}$

$I=\frac{d}{dt}(at-b{t}^2)$

$I=a-2bt$.............................................................$\left(1\right)$

For, $I=0,$
$0=a-2bt$
$2bt=a$
$\Rightarrow t=\frac{a}{2b}$..............................................................$\left(2\right)$

The total heat produced in resistor $R$ is:
$H={\int }_0^t{I}^{2}Rdt$

$H={\int }_0^t(a-2bt{)}^{2}Rdt$ ...............................from$\left(1\right)$

$H={\int }_0^t(a^2-4abt+4b^2{t}^2)Rdt$

$H=R{[a^{2}t-\frac{4ab{t}^2}{2}+\frac{4b^2{t}^3}{3}]}_0^t$

$H=R\left[\frac{6a^{2}t-12ab{t}^2+8b^2{t}^3}{6}\right]$

$H=R\left[\frac{6a^2\left(\frac{a}{2b}\right)-12ab{\left(\frac{a}{2b}\right)}^2+8b^2{\left(\frac{a}{2b}\right)}^3}{6}\right]$ .................from (2)

$H=R\left[\frac{3\left(\frac{a^3}{b}\right)-3\left(\frac{a^3}{b}\right)+\left(\frac{a^3}{b}\right)}{6}\right]$

$H=R\left[\frac{2\left(\frac{a^3}{b}\right)}{6}\right]$

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