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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Alternating Current

A coil of inductance 300 mH and resistance $2\Omega$ is connected to a source of voltage 2V. The current reaches half of its steady state value in

$\begin {array} {1 1} (A)\;0.05s & \quad (B)\;0.1s \\ (C)\;0.15s & \quad (D)\;0.2s \end {array}$

 

1 Answer

$I=I_0 \bigg( 1-e^{-\large\frac{R}{L}t} \bigg)$
$ 0.693 =\large\frac{R}{L}t$
$ t = \large\frac{3 \times 0.693}{2} $ $= 0.2 \: sec$
Ans : (D)
answered Feb 15, 2014 by thanvigandhi_1
edited Mar 24, 2014 by sreemathi.v
 

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