An inductor ( $\mathrm{L}=0.03 \mathrm{H}$ ) and a resistor ( $\mathrm{R}=0.15 \mathrm{k} \Omega$ ) are connected in series to a battery of 15 V EMF in a circuit shown. The key $\mathrm{K}_{1}$ has been kept closed for a long time. Then at $\mathrm{t}=0, \mathrm{~K}_{1}$ is opened and key $\mathrm{K}_{2}$ is closed simultaneously. At $\mathrm{t}=1 \mathrm{~ms}$, the current in the circuit will be: ( $\mathrm{e}^{5} \cong 150$ )

JEE Main · 2015 · 04 Apr 2015 · Offline · Electromagnetic Induction

An inductor (

L = 0.03 H

) and a resistor (

R = 0.15 k Ω

) are connected in series to a battery of 15 V EMF in a circuit shown. The key

K_{1}

has been kept closed for a long time. Then at

t = 0, K_{1}

is opened and key

K_{2}

is closed simultaneously. At

t = 1 ms

, the current in the circuit will be: (

e^{5} ≅ 150

)

Answer: (C)

With

K_{1}

long closed the inductor has reached steady state and carries

I_{0} = \frac{E}{R} = \frac{15}{150} = 0.1

A, i.e.

100

mA. The moment

K_{1}

opens and

K_{2}

closes, the inductor drives this current through

R

while it decays as

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

. At

t = 1

ms the exponent is

\frac{R t}{L} = \frac{150 \times 1 0 ^{- 3}}{0.03} = 5

, and with

e^{5} \approx 150

this leaves

I = \frac{0.1}{150} \approx 0.67

mA.

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option