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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

in Physics | 4 views

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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