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Circuits with two voltage sources

  1. Feb 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Well this is the problem:

    Untitled_zpsb38107eb.png

    Whats the current in I1 and I2?
    2. Relevant equations
    I'm guessing you use Kirthoff's Laws.


    3. The attempt at a solution
    Well, what I did was I broke it up to two different circuits but then it would have two different currents flowing into the 2 ohms resistor
     
  2. jcsd
  3. Feb 17, 2013 #2

    gneill

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    Hello Chaso, Welcome to Physics Forums.

    Yes, Kirchhoff's laws are applicable. In addition, a handy technique for this type of circuit is superposition, which you may have covered in your coursework. You can consider the effects of one source at a time, suppressing all the others, then calculate their effects individually and sum together the results.

    To "suppress" a voltage source you set its value to zero (replace the source with a wire). Then apply the usual circuit laws (Kirchhoff, Ohm, etc.) to solve for the parameters of interest (currents, potentials). Repeat for each source. Sum the results.
     
  4. Feb 17, 2013 #3
    So what I would do is just consider the current for the first half and than the other half and then I add the currents together for the 2 ohms resistor?
     
  5. Feb 17, 2013 #4

    gneill

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    That is essentially it. Remember, when you suppress a voltage source you effectively replace it with a wire. When you suppress the 2 V source, this puts its 10 Ω resistor in parallel with the 2 Ω resistor. Similarly, when you suppress the 3 V source the 4 Ω resistor will parallel the 2 Ω resistor.
     
  6. Feb 17, 2013 #5
    Ohhh. So I don't divide the circuits then I just take the current by replacing the voltages with a wire and do it for both voltage source and add up the current then?
     
  7. Feb 17, 2013 #6

    gneill

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    Yes.
     
  8. Feb 17, 2013 #7
    Hmm is there a method to check if the sums are correct? Like Kirthoff's Sum of Current law for a point. Do the sums add up to zero?
     
  9. Feb 17, 2013 #8

    gneill

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    You can take your total current through the center resistor and use Ohm's law to determine the potential across it. Then with that potential in place you can find the potential drops across the other resistors (KVL around the individual loops), and hence the currents in those resistors. All the currents should satisfy KCL at the top center node.
     
  10. Feb 17, 2013 #9
    Well, I'm guessing thats what I did. I added up the I-total. that was 12/17.
    and for the resistors:
    2 ohm-10/17
    4 ohm-19/34
    10 ohm-9/34

    and it doesnt seem to add up :/

    But I got the I1 and I2 to be 10/17 and 18/17
     
  11. Feb 17, 2013 #10

    gneill

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    I'm not seeing the same numbers that you are, so one of us (or both) has a problem :smile:

    Can you show more of your calculations of how you arrived at your results? Let's say the 2V supply is suppressed and only the 3V supply is operating. Can you show your work for finding the current through the 2Ω resistor?
     
  12. Feb 17, 2013 #11
    Well here's are all my numbers :).
    HP0003_zps5ce8fbfe.jpg
     
    Last edited: Feb 17, 2013
  13. Feb 17, 2013 #12
    And I just added up the numbers to get the totals.
     
  14. Feb 17, 2013 #13

    gneill

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    Okay, so the currents through the 2Ω resistor due to the 3V source is (15/34)A, and due to the 2V source is (2/17)A, giving a total of Ir = (19/34)A . Note that both currents flow in the same direction, down through the 2Ω resistor, so they add. Adding the voltages produced across the resistor in a similar fashion yields Vr = (19/17)V across the 2Ω resistor.

    With potential Vr across the resistor, for the full circuit the current I1 must be (3V - Vr)/4Ω. The current I2 is (Vr - 2V)/10Ω according to the current direction specified in the diagram. Do these expressions yield the results you've fond?
     
  15. Feb 17, 2013 #14
    Hmm I'm kind of lost with the 2nd paragraph. I don't know what Vr is and how you are finding it and its connection with I1 and I2.
     
  16. Feb 17, 2013 #15

    gneill

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    Vr is the net voltage across the 2Ω resistor; the sum (superposition) of the two voltages due to the two currents you found by suppressing the sources in turn. It's also equal to the voltage across that resistor due to the net current flowing through it.

    Knowing the net voltage across the 2Ω resistor gives you the potential at the top center node. The individual currents I1 and I2 can then be calculated by considering the potential drops across the two other resistors.
     
  17. Feb 17, 2013 #16
    Hmm okay so so Vr is universal to all the resistors then? That's why we find Vr with the 2Ω resistor and we can apply it to the other resistors? And to find the I in the 4Ω Resistor we subtract 3 by the Vr because 3 is flowing into it while Vr is flowing out and that's why its vice-versa for the 10Ω resistor?
     
  18. Feb 17, 2013 #17

    gneill

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    Vr is the potential across the 2Ω resistor in the complete circuit (with both voltage sources operating). So the top of the 2Ω resistor is at Vr above the ground potential. You can find the currents in the branches by doing a "KVL walk" from ground to the node where Vr sits. For example, for the left branch you have +3V - I1*(4Ω) = Vr. Solve for I1.
     
  19. Feb 17, 2013 #18
    So what I did was I1 = (3V - 19/17V)/4Ω = 8/17A

    I2 = (19/17V - 2V)/10Ω = -3/34A

    I'm assuming Vr is universal from 2Ω to 4Ω and 10Ω. If not, how would I find Vr for the other resistors? using the same method just add them up?
     
  20. Feb 17, 2013 #19

    gneill

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    Those currents look good.
    Vr is the potential across the 2Ω resistor only. To find the others, use KVL since you know the potential at ground is 0 and that at the top node is Vr. The potential across the given resistor in the path must be such that it makes sum of potentials along the path equal this total. Another method is to use Ohm's law. If you know the current through the 4Ω resistor is I1, then the potential drop across it must be I1*4Ω.
     
  21. Feb 17, 2013 #20
    Hm okay that makes sense but why are we using Vr for the 2Ω to solve for the current of the other 2 resistors?
     
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