JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Electromagnetic Induction

An inductor of 10 mH is connected to a 20 V battery through a resistor of

10 k Ω

and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after

1 μ s

\frac{x}{100} mA

. Then x is equal to ____.

(Take e^{- 1} = 0.37)

Answer: 74

Solution

After a long time the inductor offers no opposition and the steady current is

I_{0} = \frac{V}{R} = \frac{20}{10 \times 1 0 ^{3}} = 2 mA

Once the source is cut out, the current decays through the same resistance with time constant

τ = \frac{L}{R} = \frac{10 \times 1 0 ^{- 3}}{10 \times 1 0 ^{3}} = 1 0^{- 6} s = 1 μ s

so the elapsed time is exactly one time constant and

I = I_{0} e^{- 1} = 2 \times 0.37 = 0.74 mA

Written as

\frac{x}{100}

mA this gives

x = 74

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Integer answer