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Show that the sum of electrostatic energy and magnetic energy in an LC oscillator equals (q_(0)^(2))/(2C). |
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Answer» Solution :Total energy = `=U_(E)+U_(m)` `=(1)/(2)(q_(0)^(2))/(C)cos^(2)omegat+(1)/(2)Lq_(0)^(2)omega^(2)sin^(2)omegat` `=(1)/(2)(q_(0)^(2))/(C)cos^(2)omegat+(1)/(2)q_(0)^(2)L((1)/(SQRT(LC)))^(2)sin^(2)omegat` `=(1)/(2)(q_(0)^(2))/(C)cos^(2)omegat+(1)/(2)(q_(0)^(2))/(C)sin^(2)omegat` `=(1)/(2)(q_(0)^(2))/(C)[cos^(2)omegat+sin^(2)omegat]` `=(1)/(2)(q_(0)^(2))/(C)` `q_(0)` and C are time independent constants. Note : Maximum energy of a CAPACITOR = maximum energy of an inductor = `(1)/(2)(q_(0)^(2))/(C)=(1)/(2)Li_(m)^(2)` At any given INSTANT, the total energy, in an LC oscillator is the sum of energy in each of them. |
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