Question

In the figure, the steady-state current through the inductor will be:

Answer 1

Step 1: Given data

Given voltage $=5V$

Two resistance $=40\Omega ,1\Omega$

Capacitance $=4\mu F$

Inductance $=3mH$

Step 2: Draw the steady-state circuit

Given that the circuit is in a steady state.

Therefore, the capacitor is completely charged.

So, the circuit will break at the capacitor.

In the steady-state, the inductor will simply behave as a conductor or a short-circuited path.

Since the $1\Omega$ resistance path is open, we only consider the $4\Omega$ resistance.

Step 3: Calculate the current

Given voltage $=5V$

We know that according to Ohm's law

$V=IR$

$\therefore I=\frac{V}{R}$

Upon substituting the values we get,

$I=\frac{5}{4}$

$=1.25A$

Hence, the current through the inductor is $1.25A$.