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  4. A 16 μ F capacitor is charged to a 20 V potential. The battery is then disconnected and pure 40 mH coil is connected across the capacitor, so that L-C oscillations are setup. The maximum current in the coil is

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Q. A $16 \, \mu \, F $capacitor is charged to a 20 V potential. The battery is then disconnected and pure 40 mH coil is connected across the capacitor, so that L-C oscillations are setup. The maximum current in the coil is

KEAMKEAM 2006Alternating Current

Solution:

Energy of charged capacitor = energy generated across inductor coil
E = $ \frac{ 1}{ 2} CV^2 = \frac{ 1}{ 2} Li^2 $
Given, $ C = 16 \mu F = 16 \times 10^{ - 6 } \, F, \, V = 20 $ volt
L = 40 m H = 40 $\times 10^{- 3 } $ H
$\therefore \frac{ 1}{ 2} \times 16 \times 10^{ - 6 } \times ( 20)^2= \frac{ 1}{ 2} \times 40 \times 10^{ - 3} \times i^2$
$ \Rightarrow i^2 = \frac{ 16 }{ 100} $
$ \Rightarrow i = 0.4$ A