stupid basic dilemma
Can anyone answer this basic dilemma please?
I have a 1 F capacitor with 1 V across it.
I have a second capacitor with no charge, V = 0, same capacitance
(E = 0.5*CV^2)
Initial energy = 0.5*C1*V1^2 + 0.5*C2*V2^2 = 0.5*1*1^2 + 0 = 1 J
I put the two in parallel.
Original charge = q = C1V1 + C2V2 = 1 * 1 + 0 = 1 C
New capacitance = Ct= 1 + 1 = 2 F
The charge eventually distributes itself across the total capacitance.
Total voltage across both capacitors = Vt = q/C = 1/2 = 0.5 V
so far so good ...
but Final energy = 0.5*Ct*Vt^2 = 0.5*2*0.5^2 = 0.5 J
where has the other 0.5 J gone? There is no resistance to dissipate energy ....
If you put them in series it is even stranger, you double your energy ...
What's going on here?
I have a 1 F capacitor with 1 V across it.
I have a second capacitor with no charge, V = 0, same capacitance
(E = 0.5*CV^2)
Initial energy = 0.5*C1*V1^2 + 0.5*C2*V2^2 = 0.5*1*1^2 + 0 = 1 J
I put the two in parallel.
Original charge = q = C1V1 + C2V2 = 1 * 1 + 0 = 1 C
New capacitance = Ct= 1 + 1 = 2 F
The charge eventually distributes itself across the total capacitance.
Total voltage across both capacitors = Vt = q/C = 1/2 = 0.5 V
so far so good ...
but Final energy = 0.5*Ct*Vt^2 = 0.5*2*0.5^2 = 0.5 J
where has the other 0.5 J gone? There is no resistance to dissipate energy ....
If you put them in series it is even stranger, you double your energy ...
What's going on here?
Replies
-
reachrkataAha !! Nothing is lost my friend !!!
Initial energy = 0.5*C1*V1^2 + 0.5*C2*V2^2 = 0.5*1*1^2 + 0 = 1 J
0.5*1*1^2 = 0.5 and not 1 !!!😁
Karthik -
Tsardoz
Well I feel pretty stupid!reachrkataAha !! Nothing is lost my friend !!!
0.5*1*1^2 = 0.5 and not 1 !!!😁
Karthik
Initial energy = 0.5*C1*V1^2 + 0.5*C2*V2^2 = 0.5*1*1^2 + 0 = 0.5 J
AND
Final energy = 0.5*Ct*Vt^2 = 0.5*2*0.5^2 = 0.25 J
so there were 2 mistakes but the energy has still gone down.
I found this on internet ...
The two-capacitor problem reconsidered
Mayer, R.P.; Jeffries, J.R.; Paulik, G.F.
Education, IEEE Transactions on
Volume 36, Issue 3, Aug 1993 Page(s):307 - 309
Digital Object Identifier 10.1109/13.231509
Summary:The two-capacitor problem involving connecting a charged capacitor to an uncharged capacitor and accounting for the difference in energy between the initial and equilibrium states, is discussed. Heat due to electrical resistance in the connecting wires is usually cited for the energy loss. In this study, the wires are assumed to be perfectly conducting and without electrical resistance. The circuit then behaves as a loop antenna and radiates energy in the form of electromagnetic radiation (EMR). All loss of energy in the system can be accounted for through EMR considerations. Examples illustrate the rate of decay of the current in the circuit
also this link
#-Link-Snipped-#
My final statement
If you put them in series it is even stranger, you double your energy ...
was wrong. The two capacitor problem only happens when they are connected in parallel.
In series, if you have a charged capacitor and put an uncharged one in series, the circuit is open loop and no charge transfer occurs. Final conditions equal initial conditions. -
Saandeep SreerambatlaSo I think as you said this is due to EMI.
I have some questions here , if the second capacitor is charged initially to the same voltage and placed will the same thing happens?
I guess no , because as sudden voltage appears across the Capacitor which is initially uncharged this happens.
If we put a inductor in between what happens?
Will the voltage dissipation doesn't happen? -
reachrkataI think then it would act like a Tank circuit which can be used as an oscillator.
-Karthik
You are reading an archived discussion.
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