**A coil of inductance 300 mH and resistance 2 is connected to a source of voltage 2 V. The current reaches half of its steady state value in**

(a) 0.15 s

(b) 0.3 s

(c) 0.05 s

(d) 0.1 s

Correct answer is option d) 0.1 s

Explaination:

During growth of charge in an inductance, I = I_{0} (1 – e ^{–Rt/L})

or \(\frac{I_0}{2} = I_0(1 – e^{ –\frac{Rt}{L}})\)

\( e^{ –\frac{Rt}{L}}=\frac{1}{2}=2^{-1}\)

or \( \frac{Rt}{L} =ln2\)

⇒t = (\(\frac{L}{R}\))In 2

t =[\(\frac{(300 \times 10^{-3})}{2}\)] x (0.693)

or t = 0.1 sec