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Circuits with two sources of Emf

  1. May 5, 2015 #1
    1. The problem statement, all variables and given/known data
    This isn't a homework problem per say but I am struggling with what we are doing in class at the moment which is to do with circuits with two sources of Emf. More specifically, how one source can be used to re-charge a battery. I have looked at the following website which i will refer to below (https://www.boundless.com/physics/t...battery-emfs-in-series-and-parallel-536-5597/)

    2. Relevant equations
    Kirchhoff's laws.

    3. The attempt at a solution
    I get that when one the positive side of one source of Emf is wired to the positive side of the other source then the combined Emf is the difference between the two. What I don't get is why this charges a battery. Imagine the source Emf (to charge the battery) is 12V and it is charging a battery which is running low at 4V (compared to its normal voltage of 10V). A diagram of this set up that I have seen has nothing else in the circuit just these two Emf sources. How does Kirchoff's voltage loop rule apply? If I gain 12V through the source voltage then drop 4V I am not at zero volts? What am I missing?

    Many thanks to any help offered!
     
  2. jcsd
  3. May 5, 2015 #2

    BvU

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    You are missing 8 Volts. Corny, isn't it ? But there is some truth in it, nevertheless. Point is that you can't do this with ideal sources: connecting these two would be the equivalent of short-circuiting an 8 V battery. In which case your Kirchoff rule has the same difficulty.

    In reality there is always something that limits the current from going to infinity: the internal resistance of the charger, the resistance of the wires to connect the two, some other current limiting device (or a simple resistor). The r1 and r2 in the second picture in your link.
     
  4. May 5, 2015 #3
    I have found and uploaded a question similar to what I described. I really don't understand how the battery is being re-charged - i.e. what is causing this. Doesn't there need to be a voltage drop across the battery being charged in order to charge it? Is this 7.6V or is it the difference between the two i.e. 6.4V?
     

    Attached Files:

  5. May 5, 2015 #4

    BvU

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    Must be my old eyes. Can't read a word.
     
  6. May 5, 2015 #5
    I have just changed the question to a word doc as the jpg was too small.
     
  7. May 5, 2015 #6
    Sorry! Yes it was very small
     
  8. May 5, 2015 #7

    BvU

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    Good. So we are in the right place: a homework forum with a problem statement (plus all variables and given/known data), relevant equations (well, one, to start with) and an attempt at solution.

    The diagram in the word document can't be the diagram you have seen, because there is a lot more to see than just the two emf sources.

    And I'm quite convinced that you can fill in the blanks (or rather: write the answers on the dots).
    (b) (i) 1 is easy. Oblige me.
    (b) (i) 2 well, since we've brought along good old Kirchoff, let's assume his law holds: 0 (zero) Volts would seem a good answer to me. Do you agree ?

    The fact that we are looking at a (b) part, more specifically at a (b) (i) part, makes me suspect there is more to follow !

    (And I'm curious to see the diagram you describe in post #1 !! Who knows we will need another equation - one connecting voltage and current)

    By the way, connecting a flat car battery to a charger (or a full battery in another car to jump-start (*)) generally causes a hefty spray of sparks to fly !


    (*) make sure the other car has the engine running when you have to do this in practice

    --
     
  9. May 5, 2015 #8
    Thanks. I can easily get the answers. b (i) is the sum of the internal resistance which is 0.80 Ohms and b (ii) is the driving emf minus the battery emf which is 6.4V. I can do the questions easily but I just really don't understand how the circuit works to charge the battery. Am I correct in saying that if you do the loop rule there is a voltage gain of 14V and a voltage drop of 7.6V across the battery. Does the rest get dropped across the internal resistance? What is it that is charging the battery? Is it the fact that the battery charger drives current backwards through the battery?
     
  10. May 5, 2015 #9

    BvU

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    I would maintain (b) (ii) zero volt: there is a voltage drop of 6.4 V over the two resistors. So yes, the 'rest gets dropped' over the internal resistors (assuming the cabling has negligible resistance).

    It (and now we bring in uncle Ohm's law) causes a current of 6.4 V/0.8 ##\Omega## = 8 A (or am I now giving away (b) (i) 3 or (b) (ii) 1 ?). That is litterally pumped into the flat battery and does some obscure electrochemical work to charge it untill it's full. Then, when disconnected from the charger and re-installed in the car, it can deliver the nominal 12 V again.
     
  11. May 5, 2015 #10
    Ah ok that makes sense thanks. So how would you find the power delivered to the battery being charged - would it be P = V I = 7.6V x 8A = 60.8W. I am confused whether the voltage drop across the battery is 7.6V as this seems to also be the emf of the battery which is confusing?
     
  12. May 5, 2015 #11

    BvU

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    Yeah, the voltage drop over the flat battery is 7.6 V. At least:

    upload_2015-5-6_0-34-50.png

    By the time it is fully charged, it should be 12 V again (at least if that is the nominal voltage of the thing).

    Google "how does a car battery work" to find out more.

    [edit] I did and Wikipedia says flat means 11.7 V -- perhaps your exercise battery is an 8 V battery.
    Anyway: you need a hgher voltage than the battery gives off to charge it -- otherwise there simply won't be any current flowing.
     
  13. May 5, 2015 #12
    Thanks for all your help! Much more informed now.
     
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