Question

In the circuit shown in the figure, $X$ is joined to $Y$ for a long time. Then $X$ and $Y$ are disconnected and $X$ is joined to $Z$. The total heat produced in $R_2$ if current in the circuit is reduced to zero is

• A. $\frac{L{E}^2}{2R_1^2}$
• B. $\frac{L{E}^2}{2R_2^2}$
• C. $\frac{2L{E}^2}{R_1^2}$
• D. $\frac{2L{E}^2}{R_2^2}$

Answer 1

The correct option is A $\frac{L{E}^2}{2R_1^2}$


The current in $L$, in steady state when $X$ remain connected with $Y$, is

$i_0=\frac{E}{R_1}$

Now, maximum energy stored in the inductor is,

$U=\frac{1}{2}L{i}_0^2=\frac{1}{2}L{\left(\frac{E}{R_1}\right)}^2=\frac{L{E}^2}{2R_1}$

When $X$ is connected to $Z$ and current reduces to zero, all the stored energy in the inductor will be dissipated by resistor $R_2$.

So, total heat produced, $H=U=\frac{L{E}^2}{2R_1}$

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