An inductor (L=0.03H) and a resistor ( R = 0.15\,k\Omega ) are connected in series to a battery of 15VEMF in a circuit shown below. 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=1ms, the current in the circuit will be : ( e^5 \cong 150 )

Alternating Current PYQ — JEE Main 2015 · Shift 0 · Q46 with Solution

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Correct answer: B

Initial steady current before switching

Since key $K_1$ has been closed for a long time, the circuit reaches steady state with the battery, resistor, and inductor in series.

At steady state for DC, the inductor behaves like a short circuit.

So the initial current is

I_0 = \frac{V}{R} = \frac{15}{0.15\times 10^3}

I_0 = \frac{15}{150} = 0.1\,\text{A} = 100\,\text{mA}

Current decay after switching

At $t=0$ , $K_1$ is opened and $K_2$ is closed. Now the battery is removed and the $R$ - $L$ circuit forms a closed loop, so current decays exponentially:

I(t)=I_0 e^{-t/\tau}

where the time constant is

\tau = \frac{L}{R}

Given:

L=0.03\,\text{H}, \qquad R=0.15\,\text{k}\Omega =150\,\Omega

So,

\tau = \frac{0.03}{150}=2\times 10^{-4}\,\text{s}=0.2\,\text{ms}

Current at $t=1\,\text{ms}$

Given:

t=1\,\text{ms}=10^{-3}\,\text{s}

Then

\frac{t}{\tau} = \frac{10^{-3}}{2\times 10^{-4}}=5

Hence,

I(1\,\text{ms}) = 100\,\text{mA}\cdot e^{-5}

Using $e^5\approx 150$ ,

e^{-5}=\frac{1}{150}

So,

I(1\,\text{ms})=\frac{100}{150}\,\text{mA}=0.67\,\text{mA}

Option check

A: $6.7\,\text{mA}$ ❌
B: $0.67\,\text{mA}$ ✅
C: $100\,\text{mA}$ ❌
D: $67\,\text{mA}$ ❌

Therefore, the correct answer is

\boxed{0.67\,\text{mA}}

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