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  • Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?

Q7.8  Suppose the initial charge on the capacitor in Exercise 7.7 is  6mC . What is the total energy stored in the circuit initially? What is the total energy at later time?

Answers (1)

best_answer

Given

Capacitance  C=30\mu F=30*10^{-6}

Inductance L = 27mH = 27*10^{-3}H

Charge on the capacitor  Q=6mC=6*10^{-3}C

Now,

The total energy stored in Capacitor :

E=\frac{Q^2}{2C}=\frac{(6*10^{-3})^2}{2*30*10^{-6}}=\frac{6}{10}=0.6J

Total energy later will be same because energy is being shared with capacitor and inductor and none of them loses the energy, they just store it and transfer it.

Posted by

Pankaj Sanodiya

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