Question

Switch S is closed for a long time at t = 0. It is opened, then

• A. total heat produced in resistor R after opening the switch is $\frac{1}{2}\frac{L{V}^2}{R^2}$
• B. total heat produced in resistor $R_1$ after opening the switch is $\frac{1}{2}\frac{L{V}^2}{R^2}\left(\frac{R_1}{R_1+R_2}\right)$
• C. total heat produced in resistor $R_1$ after opening the switch is $\frac{1}{2}\frac{R_{2}L{V}^2}{(R_1+R_2)R^2}$
• D. none of these

Answer 1

The correct option is C total heat produced in resistor $R_1$ after opening the switch is $\frac{1}{2}\frac{R_{2}L{V}^2}{(R_1+R_2)R^2}$

Given that;-

Switch S is closed for a long time

When switch S is closed

$I=\frac{E}{R}$

And this current will not flow through $R_1$ and $R_2$ as L is acting like a simple wire

Energy stored in inductor L

$U_L=\frac{1}{2}L{I}^2=\frac{1}{2}{\left(\frac{\in }{R}\right)}^2$

At $t=0$, when switch is opened R will not be part of circuit. Here, no heat will produce in R.

The stored energy in inductor will result in heat in $R_1$ and $R_2$.

The inductor energy will dissipate in resistance $R_1$ and $R_2$ in the ratio $\frac{1}{R_1}$ and $\frac{1}{R_2}$

Heat produce in $R_1$

$H_{R_1}=\left(\frac{R_2}{R_1+R_2}\right)U_L$

$=\left(\frac{R_2}{R_1+R_2}\right)\times \frac{1}{2}\times L{\left(\frac{\in }{R}\right)}^2$

$H_{R_1}=\frac{1}{2}L\frac{{\in }^2}{R^2}\left(\frac{R_2}{R_1+R_2}\right)$

hence, option C is correct.