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

An inductor of 10 mH is connected to a 20 V battery through a resistor of 10 kΩ 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 μs is x100 mA. Then x is equal to ______ . (Take e-1=0.37)

Solution

Imax=VR=20 V10 KΩ=2 mA

For LR-decay circuit, 

I=Imaxe-Rt/L

I=2 mA e-10×103×1×10-610×10-3

I=2 mA e-1

I=2×0.37 mA

I=74100 mA

x=74

Asked in: JEE Main 2021 (25 Jul Shift 1)

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