Question

A source of constant voltage $V$ is connected to a resistance $R$ and two ideal inductors $L_1$ and $L_2$ through a switch $S$ as shown. There is no mutual inductance between the two inductors. The switch $S$ is initially open. At $t=0$, the switch is closed and current begins to flow. Which of the following options is/are correct?

• A. At $t=0$, the current through the resistance $R$ is $\frac{V}{R}$
• B. After a long time, the current through $L_2$ will be $\frac{V}{R}\frac{L_1}{L_1+L_2}$
• C. After a long time, the current through $L_1$ will be $\frac{V}{R}\frac{L_2}{L_1+L_2}$
• D. The ratio of the currents through $L_1$ and $L_2$ is fixed at all times $(t>0)$

Answer 1

Answer: (B), (C), (D)

The ratio of the currents through $L_1$ and $L_2$ is fixed at all times $(t>0)$

Let the currents through inductors be $i_1$ and $i_2$. Then,

$V_L=\frac{L_{1}d{i}_1}{dt}=L_2\frac{d{i}_2}{dt}$

i.e. ${\int }_0^{t}d\left(L_1{i}_1\right)={\int }_0^{t}d\left(L_2{i}_2\right)\Rightarrow L_1{i}_1=L_2{i}_2$

i.e. $\frac{i_1}{i_2}=\frac{L_2}{L_1}$ is fixed at all times.

Hence 'D' is correct.

Also, since inductor acts as an open circuit at $t=0,$ current through all branches is zero at $t=0$.
After a long time current through battery is

$i_0=\frac{V}{R}$ and $L_1{i}_1=L_2{i}_2$

and $i_1+i_2=i_0$

So, $i_1=\frac{i_0{L}_2}{L_1+L_2}$ and $i_2=\frac{i_0{L}_1}{L_1+L_2}$

Hence option B, C are correct.