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6 RLC circuits: Transcient analysis Con elow. The capacitor is initially hold a charge of 0.25 nano-Coul. (a) Find the initia

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Given IM Q = 0.25 h coul. cv = 0 Voor = 0.2SX10 0.02x15 = 0.0125 Volta time constant : RC - 1x106 x 0.0 2x10G 0.02 sec + 20+ 0 switch connect to A. 2012 BOU Vc for long time capacitor aut as open circuit Vx10 VX10 Vc Ct -00 to 0) = to +20 30 30xiowhen switch connect B Capautor discharge through Resor Vcco=o. 2ooks lout - 22ook lou VC & UR 20. Vrt IRO Vc + cdVcR=0 at deQI VC = 12.50 sotmu 12.5 Mv - t-o msee 20 nsce 02 VR A loml t 2 sec

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