Capactance homework problem

Discussion in 'Homework Help' started by mentaaal, Dec 11, 2007.

  1. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Hey guys i dont really have a clue on how to attempt to answer this question:
    Ex 3: A 50 F capacitor is charged from a 200 V supply and after being disconnected is immediately connected in parallel with a 30 F capacitor. The 30 F capacitor is initially uncharged. Calculate:
    (a)the p.d. across the combination;
    (b) the electrostatic energy stored before and after the capacitors are connected in parallel. Account for any difference in energy stored.
    [125 V, 1 J, 0.625 J]

    Could anyone help me? The answers to the above question are in the [ ]

    Thanks, Greg
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    The key to solving this problem is recognizing that charge is conserved. The charge (coulombs) on the two paralleled capacitors after they are connected is the same as the charge on the 50F capacitor just before the 30F is connected.

    The last part of (b) is the most interesting. Since we have the law of conservation of energy, and we also have conservation of charge, where can the energy ( 1J - .625J) have gone? Even though (total) energy must be conserved, there is no reason why energy cannot be changed from one type to another.
     
  3. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    Hi, thanks for the reply yeah i understand now how to solve the first part. i solved it using c = q/v but like you said the last part is the most interesting and i dont understand it at all. As you say that energy has to be conserved as well, where did the difference in energies go to? I didnt take into account things like resistance so where did it go? I think this problem is similar to that of momentum. In an elastic collision kinetic energy is conserved where in inelastic collisions, kinetic energy is not. I dont understand how this occurs without taking into account these energy loss mechanisms like heat etc.

    Any explanation on this subject will be GREATLY appreciated!
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You said "I dont understand how this occurs without taking into account these energy loss mechanisms like heat etc."

    The fact is, it can't be understood without taking into account energy loss mechanisms. Of course, heat is the form of energy where many things send their energy losses.

    For your problem, if you actually were to do the experiment you would find that when you connected the 30F capacitor there would be a big spark. So you have several energy loss mechanisms. The spark would product light, sound and heat. There would also be some loss by radiation at radio frequencies.
     
  5. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    I know what you mean but it seems to me that when we do the calculations like i have done above, just using c=q/v and the energy formula this would be an ideal case and the energy should be conserved. Once i take into acount like what you said about sparks etc then we would see the real world losses mentioned. I hope i am not sounding too retarded by repeating myself again.
     
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    This is the *apparent* paradox that many beginning students encounter when working out this problem. Energy *is* conserved, just not as electrostatic energy in the capacitor.

    Charge *must* be preserved on the capacitors because charge only comes in one form: it can only be charge.

    But, energy comes in many forms, and that is how it escapes the physical capacitors, by transforming into another form.

    So, when we say energy is conserved, we mean that it isn't destroyed. In this case, it is transformed.

    But, charge can't be destroyed and it can't be transformed into something else. It is the charge which cannot escape the capacitors, and that is how we solve the problem.

    If you were to assume that the energy had to be the same after the capacitors were connected, you would get a different solution for the voltage on the caps. This solution would not preserve the charge, and is contrary to the experimental result obtained when the procedure is performed in the lab.

    To reiterate, there is no mechanism for charge to escape when you connect the caps together, but there are various mechanisms for energy to escape. And, when energy escapes the two physical capacitors, conservation of energy is not violated.
     
  7. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
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    thanks alot for the reply, i think that clarifies it a bit more for me i must just think on it. I guess i assumed that both charge and energy were conserved but i see that that is not possible.

    thanks again!
     
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