Question

The amount of heat generated in a coil of resistance $R$ due to a charge $q$ passing through it if the current in the coil, decreases down to $0$ uniformly during a time internal $t_o$ is -

• A. $\frac{2}{3}\frac{q^{2}R}{t_o}$
• B. $\frac{4}{3}\frac{q^{2}R}{t_o}$
• C. $\frac{1}{3}\frac{q^{2}R}{t_o}$
• D. $\frac{4}{3}\frac{q{R}^2}{t_o}$

Answer 1

The correct option is B $\frac{4}{3}\frac{q^{2}R}{t_o}$

The corresponding $i-t$ graph will be a straight line with $i$ decreasing from a peak value $($say $i_o)$ to zero in time $t_o$.

$\therefore i=i_o-\left(\frac{i_o}{t_o}\right)t(y=-mx+c)...\left(i\right)$

Here, $i_o$ is unknown, which can be obtained by using the fact that area under $i-t$ graph gives the amount of charge.

Hence, $q=\frac{1}{2}\left(t_o\right)\left(i_o\right)$

$\therefore i_o=\frac{2q}{t_o}$

Substituting in Eq. $\left(i\right)$, we get,

$i=\frac{2q}{t_o}(1-\frac{t}{t_o})$

$\Rightarrow i=(\frac{2q}{t_o}-\frac{2qt}{t_o^2})$

Now, at time $t$, heat produced in a short interval $dt$ is,

$dH=i^{2}Rdt$

$\Rightarrow dH={(\frac{2q}{t_o}-\frac{2qt}{t_o^2})}^{2}Rdt$

$\therefore$ Total heat produced,

${\int }_0^{H}dH={\int }_0^{t_o}{(\frac{2q}{t_o}-\frac{2qt}{t_o^2})}^{2}Rdt$

$\Rightarrow H=\frac{4}{3}\frac{q^{2}R}{t_o}$

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