JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2021
- Hard
- Numerical
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 \(\frac{{x}}{100}\, {mA}\). Then \({x}\) is equal to ...... . (Take \(\left.{e}^{-1}=0.37\right)\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(74\)
Step-by-step Solution
Detailed explanation
\(I_{\max }=\frac{V}{R}=\frac{20 V}{10 K \Omega}=2\; mA\)
For LR - decay circuit
\(I=I_{\max } e ^{- R / L}\; mA\)
\(I=2 e ^{\frac{-10 \times 10^{3} \times 1 \times 10^{-6}}{10 \times 10^{-3}}}\; mA\)
\(I=2 e ^{-1}\; mA\)
\(I=2 \times 0.37\; mA\)
\(I=\frac{74}{100}\; mA\)
\(x=74\)
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