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Question:

An LC circuit of a 4-mH inductor and a 200μF capacitor.?

An LC circuit of a 4-mH inductor and a 200μF capacitor. If the maximum energy stored in the circuit is 10?4 J, what are the maximum charge on the capacitor and the maximum current in the circuit? What are the minimum values?Can someone please explain this in detail. thanks!

Answer:

The energy of the capacitor is stored as an electric field and is maximum when its fully charged. Once fully charged the capacitor then discharge sending current through the inductor where the energy of the inductor increases as it is stored as in the magnetic field around it. So, there is an exchange of energy between to two components: L and C back and forth. Normally, resistance will will cause the amount exchanged per cycle to be reduced over time as energy is lost as resistance heating, however, the frequency remains the same. Energy of a capacitor,Ec 0.5 CV^2; However, Q CV therefore, VQ/C and V^2 Q^2/C^2 Ec 0.5 C x Q^2/C^2 Ec 0.5 Q^2/C Q^2 (2 x Ec x C) Q root(2 x Ec x C) Q 200uC E 0.5 Li^2 i^2 (2 xE)/L i^2 (2 x 10^-4)/4 x 10^-3 i^2 0.05 I 0.224A
In a RLC circuit w0 a million/sqrt(LC), fw0/(2*pi) (you could convert it from radians to hertz to get 352) w0 is the resonance frequency, it extremely is mutually as the resistance is lowest via making use of certainty the capacitor and inductor cancel one yet yet another out. So at ww0, the resistance is in ordinary terms R IV/R one hundred 55/10 15.5A i'm doubtful at what element you're on your reading, yet convert the two the inductor and capacitor into their impedance type Z. capacitor a million/(jwC), inductor jwL you could desire to apply calculus to discover the minimum of the series resistance (it is w0). and you will study that the impedance of the capacitor and inductor cancel out at this frequency.

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