CrazyEngineers
CrazyEngineers

  • stupid basic dilemma

    Tsardoz

    Tsardoz

    @tsardoz-e1LOGo
    Updated: Jan 7, 2010
    Views: 1.0K
    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?
    0
    Replies
Howdy guest!
Dear guest, you must be logged-in to participate on CrazyEngineers. We would love to have you as a member of our community. Consider creating an account or login.
Replies

  • reachrkata

    Member • Jan 9, 2010

    Aha !! 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
    Are you sure? This action cannot be undone.
    Cancel

  • Tsardoz

    Member • Jan 9, 2010

    reachrkata
    Aha !! Nothing is lost my friend !!!

    0.5*1*1^2 = 0.5 and not 1 !!!😁

    Karthik
    Well I feel pretty stupid!

    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.
    Are you sure? This action cannot be undone.
    Cancel

  • Saandeep Sreerambatla

    Member • Jan 12, 2010

    So 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?
    Are you sure? This action cannot be undone.
    Cancel

  • reachrkata

    Member • Jan 13, 2010

    I think then it would act like a Tank circuit which can be used as an oscillator.

    -Karthik
    Are you sure? This action cannot be undone.
    Cancel
Home Channels Search Login Register