JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Electromagnetic Induction

For the given circuit the current

i

through the battery when the key in closed and the steady state has been reached is ______.

(A)25 A
(B)0 A
(C)6 A
(D)10 A✓ Correct

Answer: (D)

Solution

In the steady state the branch currents are constant, so each inductor sustains no voltage and acts as a resistanceless wire. The three parallel branches then present only their

3 Ω

resistors, whose combination is

\frac{3}{3} = 1 Ω

. This sits in series with the

2 Ω

resistor, giving a total resistance of

2 + 1 = 3 Ω

. The battery current is

i = \frac{30 V}{3 Ω} = 10 A

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option

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