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  4. A capacitor of capacitance 1 μ F is charged to a potential of 1 V. It is connected in parallel to an inductor of inductance 10- 3 H. The maximum current that will flow in the circuit has the value

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Q. A capacitor of capacitance $1 \, \mu \, F$ is charged to a potential of 1 V. It is connected in parallel to an inductor of inductance $10^{- 3}$ H. The maximum current that will flow in the circuit has the value

Alternating Current

Solution:

Change on the capacitor,
$ q_0 = CV = 1 \times 10^{ - 6} \times 1 = 10^{ - 6 } $ C
here, $ q = q_0 \, \sin \, \omega t $
or $ I_0 = \omega q_0 = $ maximum current
Now, $ \omega = \frac{ 1}{ \sqrt{ LC}} = \frac{ 1}{ \sqrt{ 10^{ - 9}}} = ( 10^9 )^{ 1/2} $
$\therefore i_0 = ( 10^9)^{1/2} \times (1 \times (10^{ - 6 }) $
$= \sqrt{1000} $ mA