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JEE Mains · Physics · STD 12 -6. Electromagnetic induction

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 \(\left.{e}^{-1}=0.37\right)\)

  1. A \(71\)
  2. B \(73\)
  3. C \(74\)
  4. D \(80\)
Verified Solution

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 \(I=I_{\max } e ^{- R / L}\; mA\) \(I=2 e ^{\frac{-10 \times 10^{3} \times 1 \times 10^{-6}}{10 \times 10^{-3}}}\; mA\) \(I=2 e ^{-1}\; mA\) \(I=2 \times 0.37\; mA\) \(I=\frac{74}{100}\; mA\)…
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