JEE Mains · Physics · STD 12 -6. Electromagnetic induction
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 ........... \(mA\). (\({e^5} \cong 150)\)
- A \(67\)
- B \(6.7\)
- C \(0.67\)
- D \(100\)
Answer & Solution
Correct Answer
(C) \(0.67\)
Step-by-step Solution
Detailed explanation
According to given conditions: \(\mathrm{i}_{0}=\frac{\mathrm{V}}{\mathrm{R}}\) \(=\frac{15}{0.15 \times 10^{3}}\) \(=0.1 \,A\) \(i=i_{0} e^{-\frac{R t}{L}}\) \(=0.1 \times e^{-\frac{0.15 \times 10^{3} \times 10^{-3}}{0.03}}\)…
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