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Two different EMFs in parallel

  1. Feb 22, 2008 #1
    Good day, mates.

    I was wondering if anyone could me out with this circuit question involving two different EMFs. I can't seem to figure out how to do it and I don't quite know where to begin.

    [​IMG]

    What must the emf be in order for the current through the 7.00 Ohm resistor to be 1.78 A? Each emf source has negligible internal resistance.

    I've exhausted all but one try for this question and I was wondering if anyone can shed some light on it.

    Cheers.
     
  2. jcsd
  3. Feb 22, 2008 #2

    Kirchoff's Laws are pretty good for this sort of problem...

    Alternatively, since you know the value for the current through the 7R resistor, you can work out the voltage across it.

    The solution follows from that...
     
  4. Feb 22, 2008 #3
    Well yeah, the voltage across it is 12.46 V but I'm not sure where to go from there.
     
  5. Feb 22, 2008 #4
    the voltage in the other two branches also have to be 12.46. So,

    24 + 3*i1 = 12.46

    E + 2*i2 = 12.46

    i1+i2 = 1.78

    That's 3 equations and 3 unknowns
     
  6. Feb 24, 2008 #5
    No.

    That's 3 equations and 2 unknowns...

    Like I said, it's Kirchoff's laws...
     
  7. Feb 25, 2008 #6
    Just curious . . I know how to solve this problem mathematically, but isn't putting two voltage sources in parallel a bad thing in real life?

    Also, it appears the EMFs are not really in parallel according to the figure you provided.
     
  8. Feb 26, 2008 #7
    It can be... though the example above is somewhat similar to connecting a battery charger to a car.

    It's not a real life problem as such, but if you did that with real batteries, one would probably explode... or be otherwise damaged...
     
  9. Feb 26, 2008 #8

    CEL

    User Avatar

    Putting two voltage sources in parallel goes against Kirchoff's voltage law. In your circuit the 24V and the E source are not in parallel, since there are the 3 ohm and 2 ohm resistors in the circuit.
    If both sources are providing current to the load, there is no problem. If one of the sources provides current both to the load and to the other source, there could be a problem if the source receiving the current is not rechargeable.
     
  10. Feb 26, 2008 #9
    i know this but what about real life when there is no such thing as an ideal voltage source. since there is impedance between all components of a voltage source, can they really be considered in parallel? And there are limits on current and power on the components as well .. so i've never actually put two in parallel. Do they explode or do they overheat or melt? What if the voltage difference is almost 0 (say a 5V and a 5.01V supply are put in parallel) to where KVL is almost satisfied, will a power supply still funciton then??
     
  11. Feb 26, 2008 #10

    CEL

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    Real voltage sources have internal resistances. In the circuit of the example you can see the two ideal voltage sources in series with the 2 and 3 ohm resistors as models of real voltage sources with the open circuit voltage represented by the ideal sources and the internal resistances as the resistors in series.
    We do put real voltage sources in parallel often. If your car battery is discharged, you can put a good battery in parallel using jumpers and you are able to start the car.
    The charging of the car battery is another example. The alternator plus rectifier is a voltage source that is in parallel with the battery and charges it.
     
  12. Mar 19, 2008 #11
    Okay, I said that

    [tex]24-E -2I_2-3I_1=0[/tex]

    [tex]24-7I_3-3I_1=0[/tex]

    [tex]E-7I_3-2I_2=0[/tex]

    [tex]I_1 + I_2 = 1.78[/tex]

    And I proceed to solve for I1 using the 2nd equation, but when I plug that into equation 4 to see what I2 equals, and then use equation 3 to find E, I get a wrong answer.
     
    Last edited: Mar 19, 2008
  13. Mar 19, 2008 #12

    CEL

    User Avatar

    You are confusing the directions of the currents in your equations. Subtracting equation 3 from equation 2 you get:
    [tex]24-E +2I_2-3I_1=0[/tex]
    where the term [tex]2I_2[/tex] has a different sign from that in equation 1.
    Redraw your circuit with the directions of the currents clearly specified and use the Kirchoff's laws.
     
  14. Mar 19, 2008 #13
    Oh, thanks. It works out now.
     
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