Question

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

Answer 1

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