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Potential difference on complex circuits

  1. Aug 5, 2012 #1

    Hello dear Physics Forums users,

    I have no idea about how to find the potential difference between two points in a complex circuit. I m not even sure if I really know how to find it in a basic circuit. Please help me and enlighten me with your wisdom and experience. Thank you.
    Last edited: Aug 5, 2012
  2. jcsd
  3. Aug 5, 2012 #2

    Philip Wood

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    We need some diagnostic tests. Try the ones on the attached thumb-nail. Work out what the voltmeter reads in each case. Give us your answers, or tell us where – if anywhere – you get stuck.

    Attached Files:

  4. Aug 5, 2012 #3
    Thanks for spending time on that. I think I have solved them, here are my steps:


    12/24=0.5 A

    0.5*6=3 V

    b) 0.3=12/(6+R)



    c)2-0.5=1.5 A

    1.5*5=7.5 V, which is equal the one on the parallel line.

    d) Well eh, with some doubt:



    12/8=1.5 A


    Current on left line=0.5*2

    Current on right line= 0.5*1

    The voltage difference= 18*0.5-3*1=6
  5. Aug 6, 2012 #4

    Philip Wood

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    Well, you've certainly got the basic ideas! Here are some comments/alternative methods...

    (b) Alternatively, p.d. across 6Ω = 6 x 0.3 = 1.8 V,
    So pd across lower resistor = 12 - 1.8 = 10.2 V.

    (c) And top voltmeter reads 12 - 7.5 = 4.5 V

    (d) No need to evaluate resistors in parallel.
    Left and right branches each have 12 V across them
    So pd across 3Ω = [itex]\frac{3}{12}\times12[/itex] = 3 V,
    And p.d. across 18Ω =[itex]\frac{18}{24}\times12[/itex] = 9 V,
    So lhs of voltmeter is 3 V downhill from the top, and rhs of voltmeter is 9 V downhill from the top, so rhs of voltmeter is 6 V downhill from lhs of voltmeter.

    Can you now supply a problem of the sort which gives you trouble?
  6. Aug 6, 2012 #5


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    Cetullah - Perhaps worth a look at this trick you can use to avoid a two stage approach..


    Allows you to work out voltages without first explicitly calculating the current.
  7. Aug 6, 2012 #6

    Philip Wood

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    Yes indeed.
    This is one of the tricks I used in (d) above.
  8. Aug 6, 2012 #7
    That is one of the kind of problems I get stuck with, in the attachment. :/

    Sorry for the super paint skills by the way :D

    Attached Files:

  9. Aug 6, 2012 #8

    Philip Wood

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    Hope you're familiar with Kirchhoff's laws...

    See thumb-nail.

    Get back if you're not happy.

    Attached Files:

  10. Aug 6, 2012 #9
    Thank you for the effort Philip. I m familiar with the Kirchhoff's laws, however the part that I ve been requesting aid for was the "simple" Step 7 :3 Can you explain a little bit, about how did you find the voltage difference, please?
  11. Aug 6, 2012 #10

    Philip Wood

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    The 40 V battery has no internal resistance, so the p.d. across its terminals (i.e. between A and B) is equal to its emf: 40 V.
  12. Aug 6, 2012 #11
    Oh, now things started to settle :) So, first we find which path every current follows, then if there are any resistances along the path, we just find the potential spent on resistance, then substract it from the positive potentials. I m not sure if I have managed to tell the thing in my mind, but correct me if I m wrong. :)
  13. Aug 7, 2012 #12

    Philip Wood

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    Yes, though I'm not sure we do "find the path every current follows".

    Good luck!
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