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)
Answer (integer)
74
Solution
${I_{\max }} = {V \over R} = {{20V} \over {10K\Omega }} = 2$ mA<br><br>For LR - decay circuit<br><br>$I = {I_{\max }}{e^{ - Rt/L}}$<br><br>$$I = 2mA{e^{{{ - 10 \times {{10}^3} \times 1 \times {{10}^{ - 6}}} \over {10 \times {{10}^{ - 3}}}}}}$$<br><br>I = 2mA e<sup>$-$1</sup><br><br>I = 2 $\times$ 0.37 mA<br><br>$I = {{74} \over {100}}$ mA<br><br>x = 74
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
This question is part of PrepWiser's free JEE Main question bank. 115 more solved questions on Alternating Current are available — start with the harder ones if your accuracy is >70%.