QJEEM21Ph0936: JEE Main 2021 Physics - Class 12 JEEM21Ph0936 Solution: An inductor of 10 mH is connected to a 20V battery through a…

Question

Verified Solution • 2025–26 Session moderate

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

\Omega

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

\mu

s is

{x \over {100}}

mA. Then x is equal to ___________. (Take e^{$-$ 1} = 0.37)

Accepted Answer

{I_{\max }} = {V \over R} = {{20V} \over {10K\Omega }} = 2

mA

For LR - decay circuit

I = {I_{\max }}{e^{ - Rt/L}}

I = 2mA{e^{{{ - 10 \times {{10}^3} \times 1 \times {{10}^{ - 6}}} \over {10 \times {{10}^{ - 3}}}}}}

I = 2mA e^{$-$ 1}

I = 2

\times

0.37 mA

I = {{74} \over {100}}

mA

x = 74

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