Question

C1= 5.6μF is charged by 220 V. Another capacitor C2 = 2.8μF is connected in parallel. Find a loss in energy.

Answer 1

C1= 5.6μF is charged by 220 V. Another capacitor C2 = 2.8μF is connected in parallel.

To find : The loss in energy.

solution : initial energy = energy due to capacitor 1 +

= 1/2 × C1V²

= 1/2 × 5.6 × 10^-6 × (220)²

= 2.8 × 10^-6 × 48400

= 0.13552 J ........(1)

now capacitor 2 is connected in parallel combination.

so, initial charge = final charge [from law of conservation of charge ]

⇒C1V = (C1 + C2)v'

⇒v' = 5.6 × 220/8.4 = 146.67 V

final energy = energy due to combination of capacitors in parallel

= 1/2 [C1 + C2] V'²

= 1/2 × [5.6 + 2.8 ] × 10^-6 × (146.67)²

= 4.2 × 10^-6 × (146.67)²

= 0.0903507 J .......(2)

loss in energy = initial energy - final energy

= 0.13552 - 0.0903507

= -0.0451693 ≈ 0.045 J

Therefore the loss in energy is 0.045 J