Question

In the given circuit diagram, the key $K$ is switched on at $t=0$. The ratio of the current $i$ through the cell at $t=0$ and at $t=\infty$ will be

• A. $3:1$
• B. $1:3$
• C. $1:2$
• D. $2:1$

Answer 1

The correct option is A $3:1$

At time $t=0$, the capacitor acts as a short circuit and hence all the current will pass through the capacitor. While at time $t=\infty$, the capacitor will act as a open circuit and the inductor acts as a closed circuit. The two equivalent circuits and thus the current through the circuits are as given below.


At $t=0,i_{t=0}=\frac{\epsilon }{R}$
At $t=\infty ,i_{t=\infty }=\frac{\epsilon }{R_{eff}}=\frac{\epsilon }{\frac{6R\left(3R\right)}{6R+3R}+R}=\frac{\epsilon }{3R}$
hence ratio is $\frac{i_{t=0}}{i_{t=\infty }}=\frac{3}{1}$