Two condensers of capacity 2uF and 8uF are each charged to a potential of 1000V and connected with their unlike plates together The loss in energy.
Answers
Given : Two condensers of capacity 2μF and 8μF are each charged to a potential of 1000 volts and connected with their unlike plates together.
To find : The loss in energy.
solution : charge on first condenser, Q₁ = C₁V
= 2 × 10^-6 × 1000 = 2 × 10¯³ Coulombs
charge on 2nd condenser, Q₂ = C₂V
= 8 × 10^-6 × 1000 = 8 × 10¯³ Coulombs
both condensers are connected with their unlike plates together. means, negative plate of first condenser is connected with positive plate of 2nd plate.
from conservation of charge,
(C₁ + C₂)v' = Q₂ - Q₁
⇒(2 + 8) × 10^-6 v' = (8 - 2) × 10^-3
⇒10^-5 v' = 6 × 10^-3
⇒v' = 600 volts.
the loss in energy = final energy - initial energy
= 1/2 × (C₁ + C₂)v'² - 1/2 C₁V² - 1/2 C₂V²
= 1/2 × (2 + 8) × 10^-6 × (600)² - 1/2 × 2 × 10^-6 × 10^6 - 1/2 × 8 × 10^-6 × 10^6
= 5 × 10^-6 × 36 × 10⁴ - 1 - 4
= 180 × 10^-2 - 5
= 1.80 - 5
= -3.20
Therefore loss in energy is 3.2 J