Thevenin equivalent circuit parameters

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Homework Statement


https://www.physicsforums.com/attachment.php?attachmentid=32837&stc=1&d=1299373667

Determine the Thevenin equivalent circuit parameters, VTh and RTh, referred to terminals A and B of the circuit shown below?

Homework Equations





The Attempt at a Solution



My only question is this step:
https://www.physicsforums.com/attachment.php?attachmentid=32838&stc=1&d=1299373667
Converted current source to voltage source (V = IR = 10 x 2 = 20 V), then combined the 10 ohm and 5 ohm resistor.

I'm pretty sure I've just ignored the polarity of the current source to get here - I've always been confused as to what to do with converting sources and polarity. Is this correct? If not, what is the proper way to convert the current source to a voltage source (assuming that is the correct method to solve the problem)?

Thanks!
 
Attachments
  • Circuit schematic with terminals A and B: a 10 ohm resistor on the left branch, a 2 A current source with arrow pointing down in the middle branch, and a 5 ohm resistor on the top wire and another 5 ohm resistor on the right branch.
    thevenin.PNG
    1.1 KB · Views: 695
  • Circuit schematic with a 20 V voltage source (+ on top) in the left branch, a 15 ohm resistor on the top wire, and a 5 ohm resistor in a vertical branch, with terminals A and B at the right.
    theveninstep2.PNG
    1 KB · Views: 710
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ya, that looks right. Just think that the 2 amps is making a loop through the 10 ohm resistor. If you made a circle on the left loop with an arrow pointing in the same direction as the current source. The current would go up through the 10 ohm resistor. So put your voltage source so the + side is pointing up.
 


1. The polarity of the voltage source in the second figure should be reversed.
2. You're not done yet. You've still got to resolve the 15Ω/5Ω voltage divider into a single resistance and voltage source.
 


Change your current source into a voltage source, which should give you a 20V voltage source. When you redraw the circuit, put the resistor in series with the voltage supply and that will give you the voltage divider that gneill was referring to.