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Homework Help: Thevenin/Norton Equivalents

  1. Jan 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Asn3.jpg

    I am being asked to draw the Thevenin/Norton equivalents for this circuit. The current sources are throwing me off... how do I deal with these?
     
  2. jcsd
  3. Jan 22, 2010 #2

    ehild

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    The current source supplies the current written on it no matter what is the voltage across its terminals.

    ehild
     
  4. Jan 22, 2010 #3
    I understand that, I should have been more specific. The problem is turning this into a Thevenin equivalent.
     
  5. Jan 23, 2010 #4
    I need help with this one too
     
  6. Jan 23, 2010 #5

    ehild

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    You have to show some effort. What is the Thevenin-equivalent of an active two-pole?
     
  7. Jan 23, 2010 #6
    I can't even say I know what an active two-pole is... any hints to get this started? I would use superposition if there were only one current source, but even still this wouldn't help me in finding the Thevenin/Norton equivalents..?
     
  8. Jan 23, 2010 #7

    ehild

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  9. Jan 23, 2010 #8
    ...no mention of an active two-pole there either. I know what Thevenin/Norton theorems are, it's the application that is causing confusion. Like I said before I would use superposition to find the currents in all the loops, if there was only one current source.
     
  10. Jan 23, 2010 #9

    vela

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    Start by replacing the 20K resistor and the 12-V source with its Norton equivalent.
     
  11. Jan 23, 2010 #10

    ehild

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    You have a box, with resistors and sources inside and an output or two terminals, it is equivalent either with a voltage source or a current source.
    The emf of the voltage source is equal to the open circuit voltage Uo of the "box", the current of the Norton equivalent source is equal to the short-circuit current flowing through the terminals. Th einternal resistance is Uo/Is.
    You need to obtain the open-circuit voltage: The voltage across the terminals when nothing is connected there. Assume that the upper terminal is at potential Uo with respect to the ground. The 9 mA current from the right current source flows through the 4 kohm resistor and produces 36 V drop across it. At the green node, the potential is Uo-36 V. Here, 3 mA flows downward and 6 mA flows through the voltage source where the potential rises by 12 V up to U-36+12. 6 mA flows through the 20 kohm resistor, causing potential drop of 120 V and here we arrive to zero potential: Uo-36+12-120=0 --> Uo=144 V, so the emf of the Thevenin equivalent is 144 V.

    We need the short-circuit current Is which flows through a resistance-free wire connected across the output terminals, both of which are at zero potential now. Assume downward current. Then we have 9-Is current flowing through the 4 kohm resistor, at a potential drop of (9-Is)*4. At the green node, 3 mA current flows through the current source, so 6-Is flows through the voltage source and through the 20 kohm resistor. The voltage source rises the potential by 12 V, the resistor lowers it, so the overall change of potential = -(9-Is)*4 +12-(6-Is)*20=0 -->Is=6 mA. The internal resistance of the Thevenin equivalent source is Ri= Uo/Is=144/6=24 kohm.

    It is a bit easier procedure to get the internal resistance by determining the resultant resistance between the output terminals, replacing voltage sources by their internal resistance in series, and current sources by their parallel internal resistance. The sources are ideal here, so the resultant output resistance is 24 kohm.

    As for the Norton equivalent, it is a current generator with current Is=6 mA and 24 kohm resistor connected in parallel with it.

    ehild
     
  12. Jan 23, 2010 #11
    O.O

    Thank you so much.
     
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