An Lc circuit contains a 40 mH inductor and a `25 mu F` capacitor. The resistance of the circuit is negligible.The time is measured from the instant the circuit is closed. The energy stored in the circuit is completely magnetic at time (in milliseconds)

Options

A. `0,3.14,6.28`
B. `0,1.57,4.71`
C. `1.57,4.71,7.85`
D. `1.57,3.14,4.71`

Answer

C

Explanation

Here, `L = 40 mH = 40xx10^(-3)H`
`C = 25 mu F = 25xx10^(-6)F, upsilon = (1)/(2pi sqrt(LC))`
Substituting the given values, we get
`upsilon = (1)/(2pi sqrt(40xx10^(-3)xx25xx10^(-6)))=(10^(3))/(2pi)Hz`
`therefore T = (1)/(upsilon)=(2pi)/(10^(3))s=2pixx10^(-3)s=2pi ms`
Energy stored is completely electrical at times
`t=0, (T)/(2), T,(3)/(2)T`, .......
Energy stored is completely magnetic at times
`t=(T)/(4),(3)/(4)T,(5)/(4)T`, .........
Hence, `t=(pi)/(2)ms, (3pi)/(2)ms, (5pi)/(2)ms=1.57ms, 4.71 ms, 7.85 ms`