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JEE Mains · Physics · STD 12 -Electromagnetic induction
  • JEE Mains 2015
  • Hard
  • MCQ

An inductor \((L = 0.03\;H)\) and a resistor \((R = 0.15 k\Omega )\) are connected in series to a battery of \(15\;V\) \(EMF\) in a circuit shown below. The key \(K_1\) has been kept closed for a long time. Then at \(t = 0\), \(K_1\) is opened and key \(K_2\) is closed simultaneously. At \(t = 1 \;ms\), the current in the circuit will be ........... \(mA\). (\({e^5} \cong 150)\)

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Answer & Solution

Correct Answer

(C) \(0.67\)

Step-by-step Solution

Detailed explanation

According to given conditions:

\(\mathrm{i}_{0}=\frac{\mathrm{V}}{\mathrm{R}}\)

\(=\frac{15}{0.15 \times 10^{3}}\)

\(=0.1 \,A\)

\(i=i_{0} e^{-\frac{R t}{L}}\)

\(=0.1 \times e^{-\frac{0.15 \times 10^{3} \times 10^{-3}}{0.03}}\)

\(=0.1 \times \mathrm{e}^{-5}=\frac{0.1}{150}=0.67\, \mathrm{mA}\)

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