Question

In the circuit shown below, the key $K$ is closed at $t=0$. The current through the battery is

Answer 1

Inductor

1. The inductor serves as an open switch at $t=0$ since there is no current flowing through it.
2. This occurs because, when the circuit’s current builds, the inductor, due to Lenz’s law, tends to counter the current buildup.
3. As a result, even in our given circuit, the inductor behaves as an open switch at $t=0$, with no contribution from the entire branch containing the inductor.

Calculate the Current

We know that, at $t=0$, inductor behave like an infinite resistance.

So, at $t=0$, $i=\frac{V}{R_2}$

Now, at $t=\infty$, the inductor behaves like a conductive wire.

So, $i=\frac{V(R_1+R_2)}{R_1{R}_2}$

The current in the circuit stabilizes after a certain amount of time has passed, and the inductor serves as a closed switch with a saturation current $I_0$ running through it.

As a result, the current flowing in the circuit in question will be a contribution of the parallel combination of the two resistances.

Hence, the current is $i=\frac{V}{R_2}$ at $t=0$ and $i=\frac{V(R_1+R_2)}{R_1{R}_2}$ at $t=\infty$.