# 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.

Explore more such questions and answers at BYJU’S.