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  4. 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 is (x/100) m A. Then x is equal to. (Take e-1=0.37 )

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Q. An inductor of $10\, mH$ is connected to a $20\, V$ 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 $\frac{x}{100} m A$. Then $x$ is equal to_______.
(Take $e^{-1}=0.37$ )

JEE MainJEE Main 2021Alternating Current

Solution:

$I_{\max }=\frac{V}{R}=\frac{20 V }{10 K \Omega}=2\, mA$
For $LR -$ decay circuit
$I=I_{\max } e^{-R / L}$
$I=2\, m\,Ae \frac{-10 \times 10^{3} \times 1 \times 10^{-6}}{10 \times 10^{-3}}$
$I=2 mA e^{-1}$
$I=2 \times 0.37 mA$
$I=\frac{74}{100} mA$
$x=74$