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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

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Correct answer is option d) 0.1 s

Explaination:

During growth of charge in an inductance, I = I0 (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

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