question_answer

An ideal coil of 10 H is connected in series with a resistance of\[\frac{1}{\sqrt{2}}\]and a battery of 5 V. After 2 s, the connection is made, the current flowing (in ampere) in the circuit is [AIEEE 2007]

A) \[\sqrt{2}\]

B) e

C) \[60{}^\circ \]

D) \[30{}^\circ \]

Correct Answer: D

Solution :

[d] The circuit is shown in figure below

Rise of current in L-R circuit is given by

\[l={{l}_{0}}\,(1-{{e}^{-t/\tau }})\]

where\[{{l}_{0}}=\frac{E}{R}\,=\frac{R}{5}\,=1\,A\]

Now, \[\left( -\frac{\pi }{2},\frac{\pi }{2} \right)\]

After 2s, i.e., at \[\left[ -\frac{\pi }{2},\frac{\pi }{2} \right)\]

Rise of current,

\[l=(1-{{e}^{-1}})A\,\,\,\,(\because \,-t/\tau \,=\frac{-2}{2}=-1)\]