Question

Three identical resistors with resistance \(R =12\, \Omega\) and two identical inductors with self inductance \(L\) \(=5\, mH\) are connected to an ideal battery with emf of \(12\, V\)as shown in figure. The current through the battery long after the switch has been closed will be ______A.

Hint:

Inductors behave as short circuits after a long time in DC circuits. Use resistor combi nations to calculate the equivalent resistance.

Answer 1

Correct Answer: 3

Step 1: Behavior of inductors after a long time After a long time, the inductors behave as short circuits (resistance-less paths) because the current through them becomes steady, and there is no changing magnetic flux.

Step 2: Simplified circuit analysis With the inductors acting as short circuits, the circuit reduces to three resistors connected in parallel across the battery.

Step 3: Equivalent resistance of the parallel resistors The equivalent resistance (\( R_\text{eq} \)) of three identical resistors in parallel is given by:

\[ \frac{1}{R_\text{eq}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} \] \[ \frac{1}{R_\text{eq}} = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{3}{12} \] \[ R_\text{eq} = \frac{12}{3} = 4 \, \Omega \]

Step 4: Total current through the battery Using Ohm's law, the total current through the battery is:

\[ I = \frac{\text{EMF}}{R_\text{eq}} = \frac{12}{4} = 3 \, \text{A} \]

Answer 2

The correct answer is 3.
After long time an inductor behaves as a resistance-less path. 
So current through cell 
I=R/312​=3A{∵R=12Ω}