Calculating Total Impedance

Discussion in 'Homework Help' started by nshepar2, Jul 6, 2012.

  1. nshepar2

    Thread Starter New Member

    Oct 13, 2011
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    I need to find the total impedance in the attached series-parallel problem, but I am not sure how to do it when I don't know the impedance of two of the elements. The information on the two elements provided is the power, power factor, and lead/lag.

    I have calculated the complex power, with the power triangle, for the right leg. 32+j24VA with a .8 lagging power factor. But how does this help me calculate the total impedance at ab?

    Thanks
     
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  2. WBahn

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    Mar 31, 2012
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    Something is missing from the description. How can the impedance connected to Node 'a' be absorbing 6W when it is only connected to the circuit at one point? Have they given you any other information, such as the total power in the entire circuit or the voltage or current at 'a'/'b'?

    As for how to determine the impedance if you know the power and either the current or the voltage, just express the power in terms of the impedance and either the current or the voltage, which ever you known, and solve for the impedance.
     
  3. nshepar2

    Thread Starter New Member

    Oct 13, 2011
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    For the premise of the homework at school, circuit diagrams like this are drawn with the assumption that there is either a current or voltage source connected across ab. It's not drawn for "simplicity". As for other information....nope. What you see is what you get. the only Vs is the 20<0˚ across the resistor and inductor.

    Since I don't have a source voltage or current, and I don't have the value of each element in ohms, I haven't the first clue how to start. This is not in the textbook nor did our professor cover a way of solving.

    If I had the values of the elements in ohms I know it's a matter of calculating as if it's a series-parallel circuit in DC.
     
  4. WBahn

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    Mar 31, 2012
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    That's a pretty unconventional assumption.

    Well, which is it? It makes a huge difference. And what is it's value?

    What you come up with for the components in the left hand impedance are completely dependent on that supply.

    I see two ways to proceed. Arbitrarily set the unknown (but assumed to be there) supply to some value. The obvious choice is to put a 0V voltage source there, which will place it in parallel with the other two branches and each branch has a defined 20V<0˚ across it. Easy to solve. The other obvious choice is to solve for the total current flowing from the supply and put a current supply with that current there.

    The more general way is to pick an unknown voltage supply and solve for the component values in terms of the supply voltage.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    If one assumes an unknown voltage source is connected from a to b the impedance may be found.
     
  6. nshepar2

    Thread Starter New Member

    Oct 13, 2011
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    Not sure what you mean about the assumption and which is it, but the other part I follow. To verify, you are saying treat this as if I'm looking for the Thevenin equivalent resistance with dependent sources in the circuit? Define a test voltage/current across ab and then use that to use the S=(Vrms)^2/Z* or Z(Irms)^2? Did not realize I could do that. Cool. Thanks.
     
  7. The Electrician

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    Oct 9, 2007
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    You know that the branch labeled "15W, .8 lagging" has 20 volts across it, and it can be represented as a series connection of a resistor and inductor just like the branch to its right.

    You can find values for the resistor and inductor such that 15 watts are dissipated by the resistor and the power factor is .8 lagging.

    Then you can calculate the current drawn by the two parallel branches when each one has 20 volts across it.

    That total current must pass through the branch labeled "6W, .6 leading" and this branch can be represented by a resistor in series with a capacitor.

    Calculate the value of a resistor such that when the current drawn by the two parallel branches passes through it, 6 watts will be dissipated. Calculate the value of a capacitor to give a .6 leading power factor.

    Now you have all the component values and it should be a piece of cake to calculate the impedance seen at a-b.
     
  8. WBahn

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    Yes and no.

    I realized just after I walked away from the computer that the problem is most definitely solvable as is. The key is actually in my last post. Notice that I said that you could assume a current source that happend to be set at the combined current in the two parallel branches to the right. Well, given that current, you can determine the elements needed in the top-left impedance. I had just gotten done pointing out that given the power and either the voltage or the current you can determine the impedance, but then overlooked that little fact because I was too focused on the missing supply.

    Now, when the diagram says 6W and 15W, does this truly mean watts, as in real power? Or does it mean apparent power (more properly denoted with VA)?
     
  9. t_n_k

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    As a point of reference, I have Z=4.752+j5.013Ω
     
  10. nshepar2

    Thread Starter New Member

    Oct 13, 2011
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    So after working it out, and getting different answers every time, I finally have an answer. I didn't get t_n_k's answer of Z=4.752+j5.013ohms. I calculated the Ztotal to be 25.766+j17.872. Would anybody be so kind as to verify this for me?

    I calculated the amps across the two parallel legs, far right=1.6-j1.2Arms and the center leg = .74-j.56Arms which I used along with the complex powers to find Z for each leg. Added the currents and used that to find the value across the 6W element at the top.

    To WBahn: for the notations of 6W and 15W, they are for real power.

    I appreciate the help and hope I finally got the answer correct.
     
  11. t_n_k

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    A useful check of your result is to calculate the unknown source voltage as well and then make sure the source power equals the individual load power summation.
     
  12. WBahn

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    It would really help if you could show your work. Short of that, tell us what you got for all of the intermediate steps. Not only the current through each of the two parallel branches, but the component values for the center leg, the total current for the element at the top, and the component values for the element at the top. List those in the order that you got them.
     
  13. nshepar2

    Thread Starter New Member

    Oct 13, 2011
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    First off, my bad on the value I typed in. I forgot to treat this as a series-parallel when I found that number. The actual value I found is 6.143+j3.157 Ohms.

    Call the 6W element A, the 15W element B, and the last one C, and the * denote conjugate. Using the given info in the diagram and the power triangle, I found the complex power on A to be 6-j8VA, B to be 15+j11.25VA and C to be 32+j24VA. I then took the 20<0 Vrms across B and C, along with the complex power values in the equation |Vrms|^2/S=Z* to find the impedance for B and C. B = 17.07+j12.8ohms and C = 8+j6ohms. (C is known but I calculated to verify my methodology is correct.)

    Next I used S=Vrms Irms* to find current. B = .75-j.56Arms and C = 1.6-j1.2Arms. Added these together to get the total current across A, which equals 2.35-j1.76Arms.
    Taking this current and |Vrms|^2/S=Z*, I found A to = .696-j.928ohms.

    Add the impedance values for a series-parallel circuit, A+B//C, to get 6.143+j3.157 Ohms.
    B//C = 5.447+j4.085 and add A to get 6.143+j3.157 Ohms.

    Circuit analysis is difficult for me and staring at this problem for 5-6 hours has melted my brain to a pile of mush.
     
  14. t_n_k

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    Yes I agree with the OP's result. I had a phase error in the series element.
     
  15. The Electrician

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    Oct 9, 2007
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    I got all the same numbers except for some small differences in the 4th decimal place. Probably because my calculator does complex arithmetic and carries 12 digits.
     
  16. nshepar2

    Thread Starter New Member

    Oct 13, 2011
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    Thank you all....You all rock....I hate complex numbers!!!
    I will now stick my head in the freezer to get some semblance of a brain back.
    Again, thanks.
     
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