Question

A 400pf capacitor charge by a 100v dc supply is disconnected from the supply and connected to another uncharged 300pf capacitor. calculate the loss of energy

Answer 1

initial energy=1/2CV^2=200×10^4×10^-12
final energy=1/2(C1+C2)V^2=1/2(400+300)×V
where,V=C1V1+C2V2/C+C2
loss in energy is initial energy minus final energy.
use the above to solve my brother

Answer 2

Dear Student,

◆ Answer -

∆U = 5×10^-5 J

◆ Explanation -

Energy stored in the capacitor is given by -

U = CV²/2

Initially,

U1 = C1.V²/2

U1 = 400×10^-12 × 100² / 2

U1 = 2×10^-6 J

Finally,

U2 = C2.V²/2

U2 = 300×10^-12 × 100² / 2

U2 = 1.5×10^-6 J

So loss of energy is -

∆U = U1 - U2

∆U = 2×10^-6 - 1.5×10^-6

∆U = 5×10^-5 J

Hence, loss of energy is 5×10^-5 J.

Thanks dear...