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Show that in the free oscillations of an LC circuit, the sum of energies stored in the capacitor and the inductor is constant in time. |
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Answer» Solution :SUPPOSE at time `t=0`, a capacitor with charge `q_(0)` is connected to an inductor ( without any magnetic field inside it) and the circuit is MADE closed. Now, capacitor starts getting discharged through an inductor. If charge on capacitor is q at time t then, `q = _(0 ) cos ( OMEGA t ) `...(1) ( Where` omega =` angular frequency of LC oscillations `= ( 1)/( sqrt( LC ))` )..(2) Energy in the electric field between the plates of a capacitor at time t will be. `U_(E ) = ( 1)/(2)(q^(2))/( C )` `:. U_(E ) = ( 1)/(2) (q_(0)^(2)cos^(2) ( omegat))/( C ) `....(3) Now, current through an inductor at time t is, `i = ( dq)/( dt)` `= ( d)/( dt) { q_(0) cos ( omega t )}` `= q_(0) { - sin ( omega t )} omegat` `:. i = - q_(0) omega sin ( omega t )`....(4) Energy in the magnetic field inside an indcutor at time t will be, `U _(B) = ( 1)/(2) Li^(2)` `:. U_(B ) = ( 1)/(2) Lq_(0)^(2) omega^(2) ( omega t )` [ From equation (4) ] `:. U_(B) = (1)/(2) Lq_(0)^(2) xx (1)/( LC)sin^(2) ( omegat )` `:. U_(B )= (1)/(2) ( q_(0)^(2))/( C ) sin^(2) ( omega t ) `...(5) TOTAL energy stored in capacitor and in an inductor at time t will be , `U_(E ) + U_(B ) = (1)/(2) ( q_(0)^(2))/( C ) { cos ^(2) ( omega t ) |+ sin^(2) ( omega t) }` `:. U_(E )+ U_(B ) = ( 1)/(2) (q_(0)^(2))/( C )` ...(6) `:. U_(E ) + U_(B )`= constant `( :. q_(0)` and C are constants ) |
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