Question

a 4uF capacitor is charged by a 200 v supply .the supply is then disconnected and the charged capacitor is connected to another uncharged 2uF capacitor .how much electroststic energy of the first capacitor is lost in the process of attaining the steady situation

Answer 1

Here, the loss in energy is given by the following relation

ΔU = C₁C₂.(V₁ - V₂)² / 2(C₁ + C₂)

here,

C₁ = 4 uF = 4 x 10^-6 F

C₂ = 2 uF = 2 x 10^-6 F

V₁ = 200 volts

V₂ = 0 volts

so, we get

ΔU = [(4 x 10^-6 x 2 x 10^-6) x (200 - 0)²] / [2x(4 x 10^-6 + 2 x 10^-6)]

= (8/3) x 10^-2

thus,

ΔU = 0.026 J