Question

The charge through a conductor of resistance $R$ varies with time as $q=at-b{t}^2$ where a and b are constant. Total heat produced in R is

• A. $R{a}^3/3b$
• B. $Ra/6b$
• C. $R{a}^2/9b$
• D. $R{a}^3/6b$

Answer 1

Answer: (A)

$R{a}^3/3b$

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

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

$I=a-2bt$ ............(1)

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

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

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

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

$H=R{[a^{2}t-\frac{4ab{t}^2}{2}+\frac{8b^2{t}^3}{3}]}_0^t$

$H=R\left[\frac{6a^{2}t-12ab{t}^2+16b^2{t}^3}{6}\right]$

$H=R\left[\frac{6a^2\left(\frac{a}{2b}\right)-12ab(\frac{a}{2b}{)}^2+16b^2(\frac{a}{2b}{)}^3}{6}\right]$ .................from (2)

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

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