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 \(\left.{e}^{-1}=0.37\right)\)
- A \(71\)
- B \(73\)
- C \(74\)
- D \(80\)
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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