A coil of inductance $$300 mH$$ and resistance $$2\,\Omega $$ is connected to a source of voltage $$2 V.$$ The current reaches half of its steady state value in
A.
$$0.01\,s$$
B.
$$0.05\,s$$
C.
$$0.3\,s$$
D.
$$0.15\,s$$
Answer :
$$0.01\,s$$
Solution : KEY CONCEPT : The charging of inductance given by, $$i = {i_0}\left( {1 - {e^{ - \frac{{Rt}}{L}}}} \right)$$
$$\frac{{{i_0}}}{2} = {i_0}\left( {1 - {e^{ - \frac{{Rt}}{L}}}} \right) \Rightarrow {e^{\frac{{Rt}}{L}}} = \frac{1}{2}$$
Taking log on both the sides,
$$\eqalign{
& - \frac{{Rt}}{L} = \log 1 - \log 2 \cr
& \Rightarrow t = \frac{L}{R}\log 2 = \frac{{300 \times {{10}^{ - 3}}}}{2} \times 0.69 \cr
& \Rightarrow t = 0.1\,{\text{sec}} \cr} $$
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