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)

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.

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