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Help With A Thevenin/norton Problem?

By antivirus software Posted in: Norton

Hi, I was doing this homework problem dealing with Thevenin and Nortons and was wondering if someone could look over and see if it’s good. If not, could you point out what I did wrong and were?
I’m supposed to find the Thevenin and Norton equivalent of the circuit
Ckt: http://i40.tinypic.com/2mmsbjm.jpg
My work:http://i41.tinypic.com/ojo5g9.jpg (sorry, a little sloppy)
To be honest here, I was a little lost. I tried to redraw the ckt into something more bearable for me and that’s what I came up with. I put the 2 resistors in series because they looked like they only shared one node before they went in parallel with another.
Then I tried to find the VOC and attempted to answer the problem.

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  1. Daiii Says

    You’re right that these bridge sort of circuits can be confusing, but your redrawn circuit is not correct. It’s easy to see why — the redrawn X and Y nodes are now directly connected to the voltage source, rather than through the resistors.
    It helps to analyze these sorts of circuits in two steps. First focus on just the X node and it’s two resistors. Since those two resistors are tied to an ideal voltage source that has zero Ohms source impedance, the presence of the Y resistors can be ignored for the moment. Then, we simply have the 30V source with two “voltage divider” resistors feeding X:
    Voltage at X: 30(12)/(6+12) = 20V
    Source impedance to X: 12||18 = (12)(18)/(12+18) = 7.2 Ohms
    Note that the two resistors are really in parallel — they are shorted at one end by the X node, and at the other by the zero-Ohm voltage source. This is often where folks get confused since the resistors sort of look like they’re in series, but they’re not.
    Analyze the Y node similarly:
    Voltage at Y: 30(5)(5+20) = 6V
    Source impedance to Y: 20||5 = (20)(5)/(20+5) = 4 Ohms
    Okay, we’re almost done. We now know the Thevenin equivalents for the X and Y nodes. Simply hook them together. The net voltage is the difference of the two voltages, and the net source impedance is the sum of the two. The Thevenin equivalent for the XY node pair is therefore:
    Voltage: 20-6 = 14V
    Source impedance: 7.2+4 = 11.2 Ohms

  2. Tim C Says

    When you redrew it you changed the location of the output terminals. That is why your answers are wrong.

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