WHere is has the other half of the energy gone ??

this is quite a common question asked in many interviews.There are two capacitors C.Intially one capacitor is charged to V volts while the other is at zero volts.At t=0,a switch connects both of them in parallel.Due to charge sharing the voltage at the node connecting both of the transistors become V/2 volts.Now the question is the energy before the switch was closed was 1/2 CV^2 and the total energy after the switch was closed is 1/4 CV^2.WHere is has the other half of the energy gone ?????

WHere is has the other half of the energy gone ??
Ohmic heat and radiation.
Reply:The "ideal case" as you define it is inconsistent with the final state assumed. It will oscillate forever at frequency that increases in proportion to the inverse square root of the circuit inductance. Report It

Reply:I assume that the formula for STORED energy is correct for each capacitor, hence at start, total E =E1(of C1) + E2(ofC2)





total E = 1/2 CV^2 + 0





After charge sharing,


total E =1/2 C(V/2)^2 + 1/2 C(V/2)^2


ie tot stored E =1/4 CV^2 as you said.......


However, consider that to GET the first capacitor charged, we needed a CURRENT, which required work (energy expended) to be done. With charge sharing, we find HALF the amount of charge moves ( ie ~half as much current) to the second capacitor, requiring HALF as much as the initial amount of work (energy expenditure) to be done. Thus you still have half the initial energy stored in the two capacitors and half the energy "lost 'from the partial (half) discharge of the first.





[Does it sound reasonable or do I need to go back to my text books, its been a long time...]