A 300 V power supply is used to charge a 25 F capacitor. After the capacitor is

Question

A 300 V power supply is used to charge a 25 F capacitor. After the capacitor is fully charged, it is disconnected from the power supply and connected across a 10 mH inductor. The resistance of the circuit is negligible.

a) Find the capacitor charge and the circuit current 1.2 ms after the inductor and capacitor are connected.

b) Find the magnetic energy and electric energy at t=0.

c) Find the magnetic energy and electric energy at t=1.2 ms.

d) If at some instant of time the voltage across the capactitor is 10 V, what is the voltage across the inductor?

e) What is the total energy of the circuit?

f) What are the maximum charge stored in the capacitor and the maximum current flowing through the circuit?

Explanation / Answer

A 300 V power supply is used to charge a 25 µF capacitor. After the capacitor is fully charged, it is disconnected from the power supply and connected across a 10 mH inductor. The resistance of the circuit is negligible. a) Find the capacitor charge and the circuit current 1.2 ms after the inductor and capacitor are connected. Q=U*C=7.5e-3(C). W=1/sqrt(LC)=2000(rad/s). we have that q=Q*cos(wt) so q=5.5e-3(C). so i=sqrt(Q^2w^2-q^2w^2)=10.2(A) b) Find the magnetic energy and electric energy at t=0. at t=0. U_E=Q^2/2C=1.125J. U_L=I^2*L/2=0. c) Find the magnetic energy and electric energy at t=1.2 ms. U_e=Q^2/2C=0.6(J). U_l=10.2^2*10e-3/2=0.52(J) d) If at some instant of time the voltage across the capactitor is 10 V, what is the voltage across the inductor? it is 10 V too. e) What is the total energy of the circuit? it is 1.125J. because energy is conserved. f) What are the maximum charge stored in the capacitor and the maximum current flowing through the circuit? maximum charge Q=7.5e-3C. and I=Q*w=7.5e-3*2000=15(A)

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