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Homework Help: Determine the difference in potential between A and B

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Determine the difference in potential between A and B.
    5ecB5.png

    2. Relevant equations

    Kirchhoff's rules for current (loop, junction), V = IR

    3. The attempt at a solution

    I have found the currents through the resistors:
    12 - 3.9a - 1.2b - 9.8a = 0 (loop rule)
    12 - 3.9a - 6.7c - 9 - 9.8a = 0 (loop rule)
    a = b + c (junction rule)
    c = -1.0295 & b = 1.75196 & a = 0.722456

    6.7 ohm resistor has 1.02 A counterclockwise
    3.9 ohm and 9.8 ohm resistors have 0.72 A clockwise
    1.2 ohm resistor has 1.7 A clockwise.

    What do I do from here? I know that if A and B were in series, the difference in potential between them would be a drop which equals the potential differences across each component between them in series, but this is in parallel, and so I don't know what to do.
     
  2. jcsd
  3. Nov 22, 2011 #2

    gneill

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    Staff: Mentor

    Kirchhoff's voltage law works for any components along a continuous path. If you can trace a path from B to A and can add up all the voltage changes along the way, you're done!
     
  4. Nov 22, 2011 #3
    There is a 1.2 ohm resistor on one path between A and B, and the current through it is 1.7 A, so the potential difference across the resistor is 2.04 V. Is this the difference in potential between A and B? The correct answer is 2.2 V.
     
  5. Nov 22, 2011 #4

    gneill

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    Staff: Mentor

    Your method is correct.

    If you keep a few more decimal places in your intermediate values you should find that the voltage is a bit higher than what you got (although not quite 2.2 V).

    You could also try the same thing for other paths between A and B.
     
  6. Nov 22, 2011 #5
    1.2 ohms * 1.75196 A = 2.102352 V

    Still coming up short. :smile:

    Using the other available path,
    -6.7*1.0295 + 9 = 2.102352 V

    Hmm... I suppose it's close enough.
     
  7. Nov 22, 2011 #6

    gneill

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    Staff: Mentor

    Not only is it close enough, 2.1V is the correct answer! Sometimes books can be, shall we say, not entirely correct.
     
  8. Nov 22, 2011 #7
    Got it, thanks for your help. :smile:
     
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