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Find unknown EMF magnitude in circuit

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data
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    A DC circuit which includes Emfx of unkown magnitude and orientation, at the location indicated is shown below. Note the 'ground' (V=0) in the circuit.
    The potential between 'b' and 'a', Vb-Va, is measured to be 2.00 V.
    a) The magnitude of Emfx is:

    b) Calculate the current through the 8Ω resistor located on the far right. Use + sign for current flowing toward top of page.

    2. Relevant equations

    Kirchoff's Current law (Sum Iin=Iout)
    Kirchoff's Voltage law (Sum V=0)

    3. The attempt at a solution

    First, to simplify this circuit, I added up all the resistors which were in series, so instead of having 6 resistors, I now have 3.
    So my R1=5ohms, R2= 15Ohm, and R3= 11Ohm.
    for a, I tried solving for the E magnitude by doing V(gE)= VE-Vg (Where g Vg is ground voltage which equals to 0). I dont know how the Kirchoff's law could help me with this.
    I dont know where to go from here, I am absolutely confused because my teacher does not want to give me a little push either :(.
    as for b, I guess i could get it once i get the answer to a.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 12, 2011 #2

    gneill

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    Staff: Mentor

    Just before you simplify the circuit by combining the resistances, can you think of any interesting information you can tease out of the circuit from the given information? (hint: why do you suppose Vba was given?)
     
  4. Oct 12, 2011 #3
    I may sound stupid, but I honestly have no idea how the Vab could help me with this. I calculated the current flowing from b to a, which gives me 1A...does this help me in any way?
    This assignment is due in 8 hours!
     
  5. Oct 12, 2011 #4

    gneill

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    Knowing the current that's flowing into node b from that branch allows you to write the node b KCL equation without having to know what Emfx is. That will allow you to calculate the voltage at that node independently of Emfx. Knowing that current also allows you to calculate the voltage across the 3 Ohm resistor...
     
  6. Oct 12, 2011 #5
    Okay this is what im doing:

    I3= I1+I 2


    E+6-11(I3)-2= 0 ----> E+4-11(I3)=0

    8+6-11(I3)-15I2=0
    14-11(I3)-15I2=0
    I3= 1.27-1.36I2


    How do I find E?? I am so close! I know it
     
  7. Oct 12, 2011 #6

    gneill

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    Sorry, but I can't follow your math. I don't know what your equation is supposed to represent, and you haven't defined the variables. A diagram showing them in place would help.
     
  8. Oct 12, 2011 #7

    NascentOxygen

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    You have overlooked the 3 ohm resistor, otherwise your working is correct.

    Solve your 3 current equations to determine I3, then substitute into the above equation (when corrected), to find E.
     
  9. Oct 27, 2012 #8
    I'm trying to figure out the same problem right now. And the solution provided isn't correct, and I think it has something to do with that "note the 'ground'" statement.

    Any help? What does that ground thing mean?
     
  10. Oct 27, 2012 #9

    gneill

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    The ground represents the assumed reference point for potentials in the circuit.

    (Imagine that there's a voltmeter with its negative lead attached there, and its positive lead is applied to various places to read the potential at those places)
     
  11. Oct 28, 2012 #10
    So when I apply the loop rule, how does the "ground" change it? I still don't get it...
     
  12. Oct 28, 2012 #11

    gneill

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    It doesn't change the application of KVL around the loops. It might, however, influence your thinking if you decided to apply nodal analysis.
     
  13. Oct 28, 2012 #12
    It doesn't. I think it's only mentioned so that you know what the ground symbol looks like.

    A voltage is a difference in potential between *two* points. You cannot say point A is at 5 volts without implicitly saying it's 5 volts higher than that point over there that is at 0 volts.

    But this is irrelevant to this problem. I can see immediately looking at that circuit that KVL will give you two unknown currents and the unknown EMF will add a third unknown. KVL alone gives two equations, you have three unknowns so you need one more piece of information that is provided to you in the question.
     
  14. Oct 31, 2012 #13
    Okay, I got the correct answer for EMF, but now I can't get b).

    I treated I1 as the 1 A current, and I3 is therefore 1.14 A. It works out, because I plugged in those values to get EMF. So then I2, which is equal to I3-I1 gives me 0.14 A, but the answer given is 0.929 A.

    PS. My version of the problem has different values. Starting from EMF and going upwards around the whole circuit, I have values 2,9,9,8,3 and in the middle from top to bottom I have 5,6,6.
     
  15. Nov 1, 2012 #14

    gneill

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    Can you provide a diagram that indicates your current choices and directions? Also, it would help if you could provide details for your calculations (equations).
     
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