Thevenin equivalent of a circuit

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  • Thread starter aid
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  • #1
aid
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Homework Statement



I am attempting to solve problem number 4 from the following picture:

http://img641.imageshack.us/img641/7950/imageisx.jpg [Broken]

In the picture you can see the suggested line of "cut" for the first application of Thevenin's theorem.


The Attempt at a Solution



I have tried to find the Thevenin equivalent voltage (for the part on the left from the sketched line). I get:
[tex]V_{oc} = 12 \angle 0 - 2 * 6 \angle 0 = 0[/tex]
which would render the whole exercise surprisingly trivial.

What the heck is wrong with the above equation? I would be grateful for any help on this one.
 
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Answers and Replies

  • #2
176
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KVL in left part is not possible, if that is what you are doing ! KVL is not possible because of the presence of current source across which you dont know the voltage drop !
 
  • #3
NascentOxygen
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Hi Aid. Is that circle with + and | marked on it supposed to represent an independent voltage source? If so, it fixes the voltage across the network of passive components to its right.
 
  • #4
aid
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KVL in left part is not possible, if that is what you are doing ! KVL is not possible because of the presence of current source across which you dont know the voltage drop !
What method is possible to apply in this case, then? Is it really so that the Thevenin equivalent voltage of the marked part is equal to:

[tex]V_{oc} = 12 \angle0 V[/tex]?
 
  • #5
gneill
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What method is possible to apply in this case, then? Is it really so that the Thevenin equivalent voltage of the marked part is equal to:

[tex]V_{oc} = 12 \angle0 V[/tex]?
A fixed voltage source effectively isolates branches parallel to it. In this case the six amp current supply and its 2 Ω series resistor cannot affect the voltage across that voltage source in any way. So, if you are looking to find the voltage Vo which is to the right of the voltage source, you can discard anything to the left of that isolating supply from consideration. That is, ignore entirely the 6 Amp supply and its 2 Ω series resistance; the 12V voltage source will be the starting point for your march across the circuit towards Vo.

The question requires you to employ Thevenin's theorem to find the result, so I'd suggest applying it several times, accumulating chunks of the circuit into successive Thevenin models (source voltage and series impedance) as you go. Where you've drawn your first "cut" in pencil is a good staring point -- the Thevenin equivalent of the voltage source and capacitor alone is very straightforward.
 

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