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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)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` |
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Answer» Correct Answer - C 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` |
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