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Homework Help: Circuits and recharging battery

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data

    I decided to repost the question and state my problems as I believe I was getting too confusing in the other thread.

    a)Suppose that a battery of EMF E and internal resistance r is being recharged: another EMF sends a current I through the battery in the reverse direction. At what rate is the electrical energy converted in chemical energy in the recharging batter?
    b) what is the power supplied by the recharging circuit to the battery?

    3. The attempt at a solution

    for a)
    So, bassically a EMF source is being recharged by another EMF source. However, it does not state whether the EMF source that is doing the charging has an internal resistance. Do we assume there is no internal resistance?

    This is what my answer is:

    P= EI - I^2r

    However, if there is an internal resistance on the battery that is doing the charging, then would it not be:

    P= EI - I^2r - I^2R

    I also don't know how to go about solving b)
  2. jcsd
  3. Nov 20, 2008 #2


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    The statement of the problem is apparently not concerned with the real world aspects of the charging battery. I think it is appropriate to treat it as an ideal EMF source. (0 internal resistance.)
  4. Nov 20, 2008 #3
    Okay, would the power supplied by the recharging battery simply be:

    P=EI ??
  5. Nov 20, 2008 #4


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    No the battery that's doing the charging is the one without any internal resistance for your purpose I think.

    I think your internal resistance power consumption should be considered for what you are asked.
  6. Nov 20, 2008 #5
    But if that is the case, then would the answer for a simply be the answer for b? If not, sorry. I'm just not following.
  7. Nov 20, 2008 #6
    Actually, it does not matter what the charging source is. Assume an ideal current source I. That makes things easier. That way you don't have to worry about how energy is dissipated in the charging (source) circuit, what its resistance is, etc. Note that the question only asks you at what rate energy is being suppled to the battery being charged, it says nothing about dissipation in the charging circuit.

    To obtain the solution, it helps to draw a circuit diagram.
  8. Nov 20, 2008 #7
    I did draw a circuit, and I'm sure the picture is correct. However, I don't know if what I am doing is right.

    For a)

  9. Nov 20, 2008 #8
    Or wait. I think I got it now. Correct me if I am wrong, but it tells you the EMF source is E, but it does not say what the other EMF source is. Hence, we can define the EMF source that is doing the charging is having a potential difference of V.

    So for a)
    it would be P = VI - rI^2

    b) P = VI
  10. Nov 20, 2008 #9
    Yes, how would you relate V to the EMF of the battery E (which is the known quantity)?
  11. Nov 20, 2008 #10
    Well, these EMFs would be in series so would I add them? So Instead, it would be for b)
    P=(V + E)I .. or were you refering to a)?
  12. Nov 20, 2008 #11
    Sorry, it seems that I'm only confusing you more. Your answers in the previous post were correct, (a) VI - I^2 r, (b) VI.

    The problem is that we don't know what V is. There are a couple of ways to find that. Have you studied Kirchoff's Laws yet? Otherwise you need to use the energy conservation relations differently.
  13. Nov 20, 2008 #12
    Is that where the sum of all the potential differences in a closed loop=0?

    So would I have to put V in terms of E?

    V + E = 0
    V = -E

    so for a) P = -EI + rI^2

    wow am I ever confused. Sorrry
  14. Nov 20, 2008 #13
    The statement of the law is correct, but you have to worry about the proper signs. And, there is a potential difference across the resistor.

    Let's try the energy conservation approach. The charging EMF supplies some power, you've worked out that it is given by VI. What happens to this energy?

    The battery being charged has an EMF E and is being charged by a current I. How much energy does it consume per unit time?
  15. Nov 20, 2008 #14
    The energy that is produced by VI travels through the circuit and charges the EMF source E. It does so at a rate of VI-I^2r does it not?
  16. Nov 20, 2008 #15
    wait, would V = Ir + E
  17. Nov 20, 2008 #16
    Ah yes, V = Ir + E from Kirchoff's law.

    You can plug this into your equations and get the answer.

    The answer the problem was looking for is much simpler, using only energy conservation. The EMF source V driving out a current I is losing power VI. Similarly the EMF source E is driving a current -I (note the negative sign), so it is gaining power EI. This means EI Joules per second are converted to chemical energy. The source V provides this energy plus the heat dissipated in the resistor.
  18. Nov 20, 2008 #17
    So really, my answers were backwards?

    Is this right, or are they backwards again?

    a) EI J/s
    b)P = VI = I^2r + EI
  19. Nov 20, 2008 #18
    These answers are correct, but please make sure you understand why they are correct :smile:
  20. Nov 20, 2008 #19
    I do, finally! Thanks a lot.
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