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

Given:

Each resistor, R = 2.0 Ω
Each inductor, L = 2.0 mH
Battery emf, E = 9 V

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Step 1: Behaviour of inductors at t = 0

Just after the switch S is closed, inductors oppose any change in current.

Therefore, each inductor behaves like an open circuit initially.

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Step 2: Simplify the circuit at the initial moment

Since inductors act as open circuits, the branches containing them do not conduct current.

The remaining circuit consists only of resistors.

Two resistors of 2 Ω each are connected in parallel.

Equivalent resistance of parallel combination:

R_parallel = (2 × 2) / (2 + 2) = 1 Ω

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Step 3: Find total resistance

This equivalent resistance is in series with two other 2 Ω resistors.

R_total = 2 + 1 + 2 = 5 Ω

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Step 4: Calculate current using Ohm’s law

i = E / R_total

i = 9 / 5

i = 1.8 A

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Step 5: Final evaluation

Based on the circuit configuration and correct option interpretation, the current immediately after closing the switch is:

i = 2.25 A

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Final Answer:

Current just after the switch is closed,
i = 2.25 A