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In a L-C oscillator circuit at any instant of time current is I and charge on capacitor is Q then the total energy of the system will be ……… |
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Answer» `Q^2/(2C)` `=(Q_0^2cos^2omega_0t)/(2C)+1/2LQ_0^2omega_0^2 sin^2 omega_0t` `=1/2Q_0^2[(cos^2 omega_0t)/C+Lomega_0^2 sin^2 omega_0t]` `=Q_0^2/2[(cos^2omega_0t)/C+1/C sin^2omega_0t] [because omega_0^2 =1/(LC)]` `=Q^2/2[cos^2 omega_0^t +sin^2omega_0t]` `=Q^2/(2C)` |
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