Question

In the circuits as shown in the figure, $S_1$ and $S_2$ are switches. $S_2$ remains closed for a long time and $S_1$ is opened. Now $S_1$ is also closed. Just after $S_1$ is closed, find the potential difference ($V$) across $R$ and $\frac{di}{dt}$ in $L$.

• A. $\frac{\epsilon }{3},\frac{\epsilon }{3L}$
• B. $\frac{\epsilon }{3},\frac{2\epsilon }{3L}$
• C. $\frac{2\epsilon }{3},\frac{2\epsilon }{3L}$
• D. $\frac{2\epsilon }{3},\frac{\epsilon }{3L}$

Answer 1

The correct option is B $\frac{\epsilon }{3},\frac{2\epsilon }{3L}$

Before closing $S_1$, current in the inductor is $i=\frac{ϵ}{2R}$.
Just after closing $S_1$, current in inductor will remain same, and
other currents are as shown
For left loop: $\epsilon =i_{1}R+L\left(\frac{di}{dt}\right)$ .....(i)
For right loop: $\epsilon =i_2\left(2R\right)+L\left(\frac{di}{dt}\right)$ .....(ii)
and $i_1+i_2=i$ .....(iii)
Solve to get $\frac{di}{dt}=\frac{2\epsilon }{3L}$
p.d. across $R$: $V_R=i_{1}R=\epsilon -L\left(\frac{di}{dt}\right)=\epsilon -L\left(\frac{2E}{3L}\right)=\frac{\epsilon }{3}$