Alternating Current

JEE Main level, numerical answer, medium

109 of 190

Question 109 of 190

An inductor of 10mH10mH is connected to a 20V20V battery through a resistor of 10kΩ10k\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 1μs1\mu s is x100mA\dfrac{x}{100}mA. Then xx is equal to ______ . (Take e−1=0.37{e}^{-1}=0.37)

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Answer: 74

Imax⁡=VR=20V10KΩ=2mA{I}_{\max }=\dfrac{V}{R}=\dfrac{20V}{10K\Omega }=2mA
For LRLR-decay circuit,
I=Imax⁡e−Rt/LI={I}_{\max }{e}^{-Rt/L}
I=2mAe−10×103×1×10−610×10−3I=2mA{e}^{\dfrac{-10\times {10}^{3}\times 1\times {10}^{-6}}{10\times {10}^{-3}}}
I=2mAe−1I=2mA{e}^{-1}
I=2×0.37mAI=2\times 0.37mA
I=74100mAI=\dfrac{74}{100}mA
x=74x=74

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