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Circuit Problem

  1. Jun 18, 2008 #1
    1. The problem statement, all variables and given/known data


    The circuit in the above figure contains two batterires, each with an emf and an internal resistance, and two resistors. Find a) the current in the circuit (magniturde and direction) b) the terminal voltage Vab of the 16V battery c) the potential difference Aac of paoint a with respect to point c

    2. Relevant equations

    I = V/R

    V = [tex]\epsilon[/tex] - Ir

    Vac = Vab-Vbc

    3. The attempt at a solution

    I1 = V/R = 16V/1.6ohms = 10A (first battery)

    I2 = V/R = 8V/1.4ohms = 5.7A (2nd battery)

    Vbc = [tex]\epsilon[/tex] - IR = (16+8V) - (10A + 5.7A)(9ohms) = -117.3V

    Vab = [tex]\epsilon[/tex] - IR = (16V) - (10A)(1.6ohms) = 0V

    Vac = Vab-Vbc = 0 - (-117.3V) = 117.3V

    I'm not quite sure if I did the problem correctly, if anyone could check my method I would greatly appreciate it.
    Last edited: Jun 18, 2008
  2. jcsd
  3. Jun 18, 2008 #2
    I don't know if I follow your method, but here's the method I would use:

    Sum of voltages around the circuit loop is zero, as Kirchoff's Voltage Law dictates. So starting at node 'a'...

    1.6I-16V+9V+...(you finish)=0
  4. Jun 18, 2008 #3
    um, I understance where you get the intitial resistance and voltage but where does 9V come into play?

    would it be:

    I(1.6ohms) + 16V - I(9ohms) - 8V - I(1.4ohms) + I(5ohms) = 0 and then I could solve for I?
    Last edited: Jun 18, 2008
  5. Jun 18, 2008 #4
    using the setup I made using the Loop Rule I got I = 2A through the circuit, does this seem correct?
  6. Jun 18, 2008 #5
    okay well using my attemtped model, I did the terminal voltage for the 16V battery

    Vab = E - Ir = 16 - (2*1.6) = 12.8V

    Vac was done taking what looked like the easier path first going through the 5ohm resistor and then through the 8V emf and the 1.4ohm resistor with the 2A current and did:

    (-2A*5ohm) + 8V + (2A*1.4ohm) = Vac = .8V

  7. Jun 18, 2008 #6


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    Why are you making some of the voltage drops across resistors positive and some negative? You should be assuming the current is flowing in the same direction through the whole circuit.
  8. Jun 18, 2008 #7
    I assumed I could pick direction and going below the a node I would be going in the negative direction going through the 5ohm resistor.
  9. Jun 18, 2008 #8


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    Here's what I would do. And I'm not very good at this because I confuse plus and minus a lot but let's assume current is flowing counterclockwise and put a voltage drop across each resistor. Start at a. -5*I-1.4*I-8-9*I+16-1.6*I=0. Am I doing ok, or did I get a sign wrong?
  10. Jun 19, 2008 #9
    Ah, yes, I meant to say 9I. So going from point 'a' all the way around the loop, and assuming a clockwise current, gives.

    1.6I-16V+9I+8V+1.4I+5I=0 (the same as Dick's if you multiply by through by -1)
    I=4/9 and clockwise

    Does it make sense to you?
    Last edited: Jun 19, 2008
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