A Coil Of Inductance 300mH And Resistance 2Ω Is Connected To A Source Of Voltage 2v. The Current Reaches Half Of Its Steady State Value In____seconds.(A) 0.1.(B) 0.3 .(C) 0.05.(D)0.105 .

The correct answer is (A) 0.1 seconds.

\(I = I_{0} [1 – e^{-Rt/L}] \) \(\frac{I_{0}}{2} = I_{0} (1 – e^{-Rt/L} ) \) \(e^{-Rt/L} = ½ \) \(\frac{Rt}{L} = ln 2 \)

Therefore,\( t = \frac{L}{R} ln 2 \) \(\Rightarrow \frac{300 * 10^{-3}}{2} * 0.693 \) \(\Rightarrow 150 * 0.693 * 10^{-3} \)

Therefore, t = 0.10396 seconds = 0.1seconds.

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