Example:A 300-V dc power supply is used to charge a 25-µF capacitor. After the capacitor is fully charged, it is disconnected fromthe power supply and connected across a 10-mH inductor. The resistance in the circuit is negligible. (a) Find the frequencyand period of oscillation of the circuit. (b) Find the capacitor charge and the circuit current 1.2 ms after the inductor andcapacitor are connected. Then find for the magnetic and electric energies (c) at and (d) at t = 1.2 ms.Given:C = 25 x 106 FL = 10 x 10³ Ht = 1.2 x 10³ s(c) Solving for magnetic (UB) and electric (UE) energiesat time t=0.Solution:Q²(7.5 x 10-3 C)²(a) Solving for angular frequency (w)and period (T)UB=Li²=0 UE 2C 2(25 × 106 F)1@=LC(10 x 10-³H) (25 x 10 °F)(d) Solving for magnetic (UB) and electric (UE)energies at time t = 1.2 ms.f =320 Hz= 2.03 x 10³ rad/s(1)2.03 x 10³ radSUB=Li²(10 x 10-3 H) (-10 A)² = 0.5 J2π2πq² (-5.5 x 10-³ C)²1T=UE= 0.6 J3.1 x 10 ³sf 320 Hz2C2(25 X 10 F)(b) Solving for capacitor charge (q) and circuit current (i)Q = CVm= (25 x 10-6F) (300 V) = 7.5 x 10-³Csince there is no initial given charge so > = 0. Therefore:q=Qcos (wt + 4) =(7.5 10-³ C)cos [(2.03 x 10³rad/s) (3.1 x 10-³s) +0] = -5.5 x 10-³ Ci = -wQsin(wt + 4) = -(2.03 x 10³ rad/s) (7.5 10-³C)sin [(2.03 x 10³ rad/s) (3.1 x 10-³s) +0] = -10 AActivity:If a 500-V dc power supply is used to charge a 25-µF capacitor. After the capacitor is fully charged, it is disconnected fromthe power supply and connected across a 20-mH inductor. The resistance in the circuit is negligible. (a) Find the frequencyand period of oscillation of the circuit. (b) Find the capacitor charge and the circuit current 1.6 ms after the inductor andcapacitor are connected. Then find for the magnetic and electric energies (c) at and (d) at t = 1.6 ms.-

Given:initial voltage across capacitor, V = 500 Vcapacitance value, C = 25×10-6 Finductance of the…

Answered: If a 500-V dc power supply is used to…

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