Thevenin Equivalent Circuit

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  • Thread starter nobodyuknow
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  • #1
nobodyuknow
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Homework Statement


3VGbyPa.png


Above given is the circuit I am supposed to simplify where the 1/5Ω resistor is the load. I'm unsure how to convert this circuit into an equivalent Thevenin Circuit.

Homework Equations



V = I*R (Ohm's Law)

The Attempt at a Solution



Using Microcap and it's Dynamic DC Analysis Tool, I measured the current flow at the load. I managed to break it down to this:

http://img844.imageshack.us/img844/282/5x8t.png [Broken]

Where V5 was a 2/3V voltage source. (2A current source in series with the 1/3Ω resistor).

If this is correct, I would then say that the 2/3V voltage source and the 4A current source were in series and that the V4 voltage source and the 1/6Ω resistor were in series. Is this the correct approach?
 
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Answers and Replies

  • #2
gneill
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Remember that a Thevenin equivalent is a single voltage source with a single series resistance (for DC circuits).

What's the usual algorithm for finding the Thevenin equivalent analytically?
 
  • #3
nobodyuknow
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Would the first step be setting all those current sources to 0? Thus left with the remaining voltage source and the two resistors?

http://img820.imageshack.us/img820/2318/q2z4.png [Broken]
 
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  • #4
gneill
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No. You must suppress ALL sources to find the Thevenin resistance: Remove all current sources and short all voltage sources.

To find the Thevenin voltage you must find the open-circuit potential across the load (remove the load and find the potential across the open terminals).

Your course notes or text should cover this!
 

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