Question

A current passing through a resistance $R$ decreases uniformly to zero in a time interval $T$ and a total charge $q$ passes through the resistance. Find the total heat produced in resistance in this process.

• A. $\frac{4}{3}\frac{q^{2}R}{T}$
• B. $\frac{9}{4}\frac{q^{2}R}{T}$
• C. $\frac{8}{5}\frac{q^{2}R}{T}$
• D. $\frac{7}{3}\frac{q^{2}R}{T}$

Answer 1

The correct option is A $\frac{4}{3}\frac{q^{2}R}{T}$


The figure shows variation of current with time as described in the question. Here we consider at $t=0,i=i_0$ and after time $t=T$, current in the circuit becomes $i=0$.

From the equation of straight line, $y=mx+c$

Here, $m=-\frac{i_0}{T},c=i_0\text{and}y=i$ putting in the equation we get,

$i=i_0-\frac{i_0}{T}t$

As total charge $q$ passes through the circuit,

$q={\int }_0^{T}idt=\text{Area under the curve}$

$q=\frac{1}{2}{i}_{0}T$

$\therefore i_0=\frac{2q}{T}$

Thus, current in circuit as a function of time is given as,

$i=i_0-\frac{i_0}{T}t=\frac{2q}{T}[1-\frac{t}{T}]$

Now, total heat produced from $0$ to $T$ is given as,

$H={\int }_0^T{i}^{2}Rdt={\int }_0^T\frac{4q^{2}R}{T^2}{(1-\frac{t}{T})}^{2}dt$

$H=\frac{4q^{2}R}{T^2}{\int }_0^T(1+\frac{t^2}{T^2}-\frac{2t}{T})dt$

$H=\frac{4q^{2}R}{T^2}{[t+\frac{t^3}{3T^2}-\frac{t^2}{T}]}_0^T$

$H=\frac{4q^{2}R}{T^2}[T+\frac{T}{3}-T]$

$\therefore H=\frac{4}{3}\frac{q^{2}R}{T}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{A}\right)$ is the correct answer.