Question

600pF capacitor is charged by 200 V supply. Is then disconnected from the supply and is connected to another uncharged 300pF capacitor. Calculate how much electrostatic energy is loss and source of energy loss?

Comments

C1= 600 pF, C2=300 pF V = 200 V Energy lost = Initial Energy - Final Energy '^' = Power Before connecting to new capacitor Initial Energy (I.E) = (1/2)(C1)(V^2) I.E = (1/2)(600×10^-12)(200)^2 I.E = 12×10^-6 F I.E = 12 micro F After connecting to new capacitor Equivalence Capacitance = Cs 1/Cs = (1/C1) + (1/C2) 1/Cs = (1/600×10^-12) + (1/300×10^-12) Cs = 200×10^-12 F Cs = 200 pF Final Energy (F.E) = (1/2)(Cs)(V)^2 F.E = (1/2)(200×10^-12)(200)^2 F.E = 4×10^-6 F F.E = 4 pF Energy Lost (E.L ) = 12 pF - 4 pF E.L = 8 pF Therefore, energy lost is 8 pF