Circuits Question

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Homework Statement



See attached

Homework Equations





The Attempt at a Solution



I found Vth
(V1 - V2)/2 + V1/10 + 2 = 0

(V2-6)/4 + (V2-4)/6 + (V2 - V1)/2 = 0

Solving for V1 I get -5/2 and V2 I get 1
V1 = Vth = -5/2

I believe this is correct. It's just if these two initial equations are wrong or correct. In the first equation I added the current entering the node close to a as V1. I wasn't sure how to handle the 5 ohm resistor.

I choose V2 as the node between the 2 ohm resistor, the 6 ohm resistor and the 6V power source.

I choose the ground reference node to be the whole right side.

I'm not sure how to procede and find the current across the short because drawing a wire to a and b results in a huge essential node.

Thanks for any help.
 

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Answers and Replies

  • #2
gneill
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Sometimes its easier to work step by step through the circuit, converting pieces to Thevenin or Norton models as you go, swallowing up the circuit until you're left with a single source and resistor, or a least a circuit that is trivial to analyze further.

For example, suppose you were to convert both of the voltage supplies and their resistors to Norton equivalents. You could then combine them both easily into a single Norton equivalent since the currents will be in parallel as will their resistors. Then convert it to a Thevenin model and swallow up the next two resistors. Eventually you should be able to get it down to two current sources and a couple of resistors. Open circuit voltage and short circuit current should be obtainable practically by inspection.
 
  • #3
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So I do something like this.

Where the resistor I drew has the same resistance as the resistor next to the 6V source (in this case 4 ohms) and the current source is just 6/4 A or 3/2 Amps?
 

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  • #4
gneill
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So I do something like this.

Where the resistor I drew has the same resistance as the resistor next to the 6V source (in this case 4 ohms) and the current source is just 6/4 A or 3/2 Amps?

Yup. Do the same for the next voltage source/resistor. You should then be able to combine them into a single current source and resistance...
 
  • #5
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This is what I get
what do I do from here?
 

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  • #6
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Is it a voltage source of 380/53 with a resistor of 190/53 in parallel?
 
  • #7
gneill
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It seems odd that you'd end up with a voltage supply in parallel with a resistance if you've been doing Thevenin/Norton reductions along the way. Can you outline the steps you've taken to get there?
 
  • #8
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Ya
see attached
 

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Last edited:
  • #9
gneill
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Okay. A Thevenin model is always a voltage source in series with a resistance. A Norton model is a current source in parallel with a resistance. If you convert a Norton model to a Thevenin model, the resistance is the same value, and vice versa.

I see from your attachment that the second voltage source, 4V with 6 Ohms in series, has been turned into a 1/2 Amp source with 8 Ohms in parallel. That's not right. The 2 Ohm resistor is NOT in series with the 4V supply. Only the 6 Ohm resistor is. So turn the 6 Ohm resistor and the 4V source into its Norton equivalent. Then combine the two parallel Norton models into one. This in turn can become a Thevenin model with a series resistance that can take then take the 2 Ohm resistor on board.
 

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