Ncert Exercises & Solutions

Question 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 a later time?

Given Capacitance value of the capacitor, 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

Total energy stored in the capacitor can be calculated as :

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

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

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