A body of capacity 4 F is charged to 80 ~V and another body of capacity 6 F is charged to 30 ~V . When they are connected the energy lost by 4 F capacitor is

Step-by-step solution

C ₂=4 F , V₁=80 volt, C₂=6 F , V₂=30 volt Energy loss, aligned U & = 1 2 C₁ C₂ C₂+C₂ (V₁-V₂ )^2 & = 1 2 4 10⁻⁶ 6 10⁻⁴ 4 10⁻⁶+6 10⁻⁴ (80-30)^2 & = 12 10⁻⁶ 10 2500=1.2 25 10⁻⁴ & =30 10⁻⁴=3 10⁻³ ~J aligned Initial energy of 4 F capacitor. aligned E_ n & = 1 2 C₁ V₁^2 & = 1 2 4 10⁻⁴ (80)^2 & =2 10⁻⁴ 6400 & =12.8 10⁻³ ~J aligned Energy lost by C₁=12.8 10⁻¹-3 10⁻¹ aligned & =9.8 10⁻³ ~J & =9.8 ~mJ aligned