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Find the potential difference between a and b

  1. Jun 27, 2009 #1
    1. The problem statement, all variables and given/known data
    A network below consists of with three bat-
    teries, each having an internal resistance, and
    five resistors.

    http://images.upload2world.com/get-6-2009-upload2world_com_vypln.jpg [Broken]

    Find the magnitude of the potential differ-
    ence between points a and b. Answer in units
    of V.

    2. Relevant equations

    i solved it using the loop equation



    3. The attempt at a solution


    VA-VB= 4*i1+35-i1-7*i1-6*i2-i2-28
    VA-VB=7*i3+18+i3+5*i3-6*i2-i2-28
    i1=i2+i3


    but im stuck her i dont knowhat to do next
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 27, 2009 #2
    following what you've done, you need one extra equation. If you take the first two and substract them, you get 0 on one side, and a linear combination of i1 i2 and i3 on the other, so with the other equation you have in total two linear equations on three variables (i1, i2, i3), so you have infinitely many solutions. you're gonna need one extra equation before solving for the currents (see http://en.wikipedia.org/wiki/System_of_linear_equations#General_behavior).
    for that,
    write down the "loop" equation for the big loop in the circuit (the big loop being, well, the biggest loop you can make in that circuit, it does not go through a and b so the total voltage difference will be 0.)
    Once you solve for the currents, you can get VA - VB from one of the equations you already wrote.
     
  4. Jun 27, 2009 #3
    ahaaa

    ok thanx for your help i got it now
     
  5. Jun 27, 2009 #4

    vk6kro

    User Avatar
    Science Advisor

    There seems to be a simpler way of doing this problem.

    The centre battery is open circuited isn't it?

    So take the current in the outer loop only. 35 volts minus 18 volts and a total of 25 ohms. So the total current is.......

    Then work out the voltage at point A relative to the right vertical connector.

    The voltage at B is available because the centre battery isn't supplying any current. So the resistors in series with it are irrelevant.

    The difference between A and B is available by subtraction.
     
    Last edited: Jun 27, 2009
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