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Find the oc voltage and the voltage across the 5 ohm load

  1. Feb 24, 2013 #1
    1. The problem statement, all variables and given/known data
    https://dl.dropbox.com/u/3405118/24-02-2013%2018-46-54.png [Broken]

    Find (a) the open-circuit voltage VL , and (b) the voltage across the 5 ohm load, by the following methods.

    Method 1: Direct approach. First find the open circuit voltage by simple ‘potential divider’ approach
    (but with complex impedances Z replacing the simple resistances we used under d.c. conditions). Then
    find the equivalent series impedance of the two parallel branches on the right, and again use the ‘potential
    divider’ idea to find VL. The calculations are lengthy, and you will probably find it best to use polar form
    for the multiplications and divisions.

    Method 2: Thévenin. Find the Thévenin equivalent circuit for the transformer and 100 V supply. Then
    apply the load and use the potential divider approach.
    Method 3: Nodal Analysis. Call the node at the top of j35 node a and that at the top of the 5  load
    node b, with the reference (zero) node at the bottom. Write down the nodal equations in terms of the
    complex impedances, and solve for Vb.

    no drawings required

    3. The attempt at a solution

    I did method one and got the answer to be : Voc :98.63 e^-j3.38 , VL: 98.45 e^-j6.810

    and tried to do the rest but I couldn't get the same answer can some one explain how do we the same answer for method 2 and 3
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 24, 2013 #2


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    Staff: Mentor

    Hi blue_tiger30, Welcome to Physics Forums.

    How do you know that the result you obtained by method 1 is correct? Did you get different values for both methods 2 and 3?

    Can you show more detail of your work for method 1 so that we can check it?
    Last edited by a moderator: May 6, 2017
  4. Feb 24, 2013 #3
    in the start I replaced the resistances that I got into a complex impedance z = ((5+j0.3)*j35))/(j35+4(5+j0.3)=4.818+j0.9804ohm
    Voc=z*I=v*(z/(z+j0.3))= 98.63 e^-3.38
    V across R = I*5=(Voc/(5+j0.3))*5= 98.45 e^-j6.810
    I think this is th right answer becuse I UNDERSTAND THIS METHOD
    but for the rest I did as follows :
    The Thevenin equivalent is
    a voltage source = 100(j35/(j35 +j0.3)= 100(j35/j35.3) = 99.15 volts, this is the open circuit voltage
    in series with R thevenin = j0.3 + (j0.3)(j35)/(j0.3 + j35) = j0.597Ω

    The voltage across RL is 99.15( 5/(5 + j0.597)) = 97.76 -j11.67 volts

    Nodal analysis
    Node A: (100 - Va)/(j0.3) = Va/(j35) + (Va - Vb)/(j0.3)
    100/j0.3 = 2Va/j0.3 + Va/j35 - Vb/j0.3
    100 = 2Va + Va(0.3/35) - Vb = Va( 2.0086) -Vb

    Node B: (Va - Vb)/(j0.3) = Vb/5
    Va/j0.3 = Vb/j0.3 + Vb/5
    Va = Vb + Vb(j0.3/5) = Vb( 1 +j0.06)

    100 = (Vb(1 +j0.06)(2.0086) -Vb = Vb(1.0086 + j0.121)
    Vb = 100/(1.0086 +j0.121) = (100.86 - j12.1)/1.016 = 99.3 -j11.9 volts

    which give me different answers and sice I m not that good at these methods I thinks that I did them in the wrong way
  5. Feb 24, 2013 #4


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    Staff: Mentor

    Okay, looks good.
    Yes, your Thevenin components look fine, and your result for the load voltage looks good.
    Your "Direct Approach" and Thevenin methods both look fine, but things have gone wrong with your Nodal Analysis.

    When doing Nodal Analysis you should first choose a reference node which is defined to have zero potential, then identify the independent nodes. It doesn't look like you've done this. The circuit has only two nodes of interest: a and b. The rest of the junctions lie along series connected branches. Which one will you take to be the reference node?
  6. Feb 24, 2013 #5
    so for method 2 shouldn't the answer be = to the answer I got in method one ? can u please tell me what went wrong there ?

    to be honest I dont really get nodal analysis I only know how to do loop analysis but thats not what he asked for in this hw so If you could correct my steps if possible

    many thanks for your help
  7. Feb 24, 2013 #6


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    Staff: Mentor

    The answers are the same; one is expressed in polar form, the other in rectangular form :wink:
    Your node equations look fine, but in the simultaneous solving of them something went wrong. So check your math in the last two lines.
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