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Question -

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



Answer -

Capacitance of thecapacitor, C = 30 μF = 30×10−6 F

Inductance of the inductor, L =27 mH = 27 × 10−3 H

Charge on the capacitor, Q =6 mC = 6 × 10−3 C

Totalenergy stored in the capacitor can be calculated by the relation,

Total energy at a later time willremain the same because energy is shared between the capacitor and theinductor.

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