7.8 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?

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Payal Gupta | Contributor-Level 10

8 months ago

7.8 Capacitance of the capacitor, C = 30 $μ F$ = 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

Total energy stored in the capacitor can be calculated as:

E = $\frac{1}{2} \frac{Q^{2}}{C}$ = $\frac{1}{2} \frac{({6 * 10^{- 3})}^{2}}{30 * 10^{- 6}}$ = 0.6 J

Total energy at a later time will remain same because energy is shared between the capacitor and the

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Physics Ncert Solutions Class 12th 2023

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