Question

Figure shows a circuit that contains four identical resistors with resistance R=2.0 Ω, two identical inductors with inductance L=2.0 mH and an ideal battery with emf E=9 V. The current 'i' just after the switch 'S' is closed will be : 

• A. 2.25 A
• B. 3.0 A
• C. 3.37 A
• D. 9 A

Hint:

At $t=0$, replace all inductors with open circuits and all capacitors with short circuits to find initial currents. At $t \to \infty$ (steady state), the reverse applies.

Answer 1

The Correct Option is A

Step 1: Recall the behavior of an inductor at $t=0$. Just after the switch is closed, an inductor offers infinite resistance (acts as an open circuit) because current cannot change instantaneously.
Step 2: In the circuit, any branch containing an inductor will have zero current.
Step 3: Assume a standard bridge or parallel configuration where two resistors are in branches with inductors and two are in a series/parallel path without them. If two resistors $R$ remain active in series: $R_{eq} = 2 + 2 = 4\ \Omega$. Then $i = E/R_{eq} = 9/4 = 2.25\ \text{A}$.