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Question

Physics Question on Alternating current

A coil of inductance 300 mH and resistance 2 Ω\Omega 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

0.1 s

Explanation

Solution

The charging of inductance is given by
I=I0[1eRt/L]I = I_0 [1 - e^{-Rt/L} ]
I02=I0[1eRT/L]\therefore \: \frac{I_0}{2} = I_0 [ 1 - e^{-RT/L} ]
or eRt/L=12e^{-Rt/L} = \frac{1}{2} or RtL=\frac{Rt}{L} = 1n 2
or t=0.693×300×1032t = \frac{0.693 \times 300 \times 10^{-3}}{2}
= 0.1 s