Question

The relation between current in a conductor and time is shown in the figure.

$\left(iii\right)$ If the resistance of the conductor is $R$ then the total heat dissipated across resistance $R$ is

​​​​​​​$[$1 Mark$]$

• A. $\frac{{\text{i}}_0^{2}R{\text{t}}_0}{2}$
• B. $\frac{{\text{i}}_0^{2}R{\text{t}}_0}{4}$
• C. $\frac{{\text{i}}_0^{2}R{\text{t}}_0}{3}$
• D. ${\text{i}}_0^{2}R{\text{t}}_0$

Answer 1

The correct option is C $\frac{{\text{i}}_0^{2}R{\text{t}}_0}{3}$

Let $dH$ be the small amount of heat dissipated in time $dt$
i.e $dH={\text{i}}^{2}Rdt$
Total heat dissipated
${\int }_0^{H}dH={\int }_0^{t_0}{\text{i}}^{2}Rdt$
The expression for $\text{i}$ in terms of $\text{t}$ is given by
$\text{i}={\text{i}}_0(1-\frac{\text{t}}{{\text{t}}_0})$

$H={\int }_0^{t_0}{\text{i}}_0^2{(1-\frac{\text{t}}{{\text{t}}_0})}^{2}Rdt$
$H={\text{i}}_0^{2}R{\int }_0^{t_0}{(1-\frac{\text{t}}{{\text{t}}_0})}^{2}dt$
$\therefore H={\text{i}}_0^{2}R{\left[\frac{{(1-\frac{\text{t}}{{\text{t}}_0})}^3}{3\left(\frac{-1}{{\text{t}}_0}\right)}\right]}_0^{t_0}$
$H=\frac{{\text{i}}_0^{2}R{\text{t}}_0}{3}$
So, option $\left(c\right)$ is correct.