Wye-Delta Transformation

Discussion in 'Homework Help' started by jdole, Sep 11, 2012.

  1. jdole

    Thread Starter New Member

    Sep 11, 2012
    7
    0
    http://i.imgur.com/1lO8E.png

    I was given the circuit on the top and told to find I

    I used Y-Δ/T-π to make the one on the bottom.

    I was wondering If I could add the voltage sources together.
     
  2. mlog

    Member

    Feb 11, 2012
    276
    36
    No, you cannot combine the voltage sources. However, you can transform a voltage source / series resistor combination into a current source / parallel resistor combination. Then you will have some parallel resistors to combine.

    For example, the 3 V / 9 Ω in series will give you a 1/3 A source in parallel with a 9 Ω. You can then combine the parallel 9 Ω and a 18 Ω resistors. Make sense?
     
  3. jdole

    Thread Starter New Member

    Sep 11, 2012
    7
    0
    Yeah, I just remembered you can only combine voltage sources in series.
     
  4. jdole

    Thread Starter New Member

    Sep 11, 2012
    7
    0
    I found the equivalent resistance and voltage and got 1/3. So I could've just stopped when combining the parallel resistors right after the wye-delta transformation. Unless I'm wrong.

    http://i.imgur.com/2nvH2.png
     
  5. mlog

    Member

    Feb 11, 2012
    276
    36
    I went about it with a slightly different approach. Refer to your last drawing.

    From your 1st wye-delta transormation, you got a symmetric 18 Ω delta. At the top of the delta, you then combined a parallel 9 Ω with an 18 Ω to get a new delta with a 6 Ω on top and 18 Ω's along both sides.

    Next is where I deviated with a voltage to current source transormation. I transformed each series combination of 3V voltage source and 9 Ω resistor to parallel combinations of a 1/3A current source and 9Ω resistor. You will find that each 18 Ω resistor is in parallel with a 9 Ω resistor.

    After you combine the pairs of 9Ω and 18Ω resistors, you will get a new balanced delta of 6 ohms per side. There are also the two 1/3A current sources, one each in parallel with the resistor on each side. The current sources are share a common node (at the bottom of the figure) and are of opposite polarity. (The source on the left points down and the source on the right points up.)

    Next you can transform the balanced 6Ω delta to a balanced 2 Ω wye. You'll see that the 1/3A current sources force a current around the outer loop. It's probably hard to visualize without a figure, but the results are the same as yours. Your loop current is also 1/3A, which comes from I=V/R=(27/7)/(81/7).
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    If you are up for it, the symmetry of the topology allows you to solve the problem without having to use any transformations.
     
  7. mlog

    Member

    Feb 11, 2012
    276
    36
    I'm guessing you split the 9-ohm resistor on top in half to get a pair of 4.5 ohms. Then you connect the center of them to the node connecting the 6 ohm resistors. Next you're able to parallel each 4.5 ohm and 6 ohm. Since the circuit is balanced and the voltage sources are equal and opposing, you know the center node is 0 volts. You're left with the 3 volt source on the right with a couple of series resistors feeding the zero volt point at the center node. Is that it?
     
Loading...