JEE Main · 2006 · 30 Apr 2006 · Offline · Electromagnetic Induction

An inductor

(L = 100 mH)

, a resistor

(R = 100Ω)

and a battery (

E = 100 V

) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 ms after the circuit is

(A)1 A
(B) $1/ e A$ ✓ Correct
(C)e A
(D)0.1 A

Answer: (B)

Solution

After the source has been connected a long time the inductor carries the steady current

I_{0} = E / R = 100/100 = 1 A

. When the battery is removed and A\text{--}B are shorted, the

L

\text{--}

R

loop decays as

I = I_{0} e^{- R t / L}

. The time constant is

L / R = \frac{100 mH}{100 Ω} = 1 ms

, so after exactly

1 ms

the current has fallen to

I_{0} e^{- 1} = \frac{1}{e} A

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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