Finding equivalent voltage between Thevenin and Norton transformations

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Homework Statement
Use source transformation to reduce the circuit to a single voltage source Vs in series with a single resistor R. What are the values of Vs and R?
Relevant Equations
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Circuit schematic with a 5 A current source on the left in parallel with three branches: a 40 Ω resistor, a 30 Ω resistor in series with a 15 V source, and a 20 Ω resistor in series with a 20 V source, polarities marked +/−.

From this, I can transform the current source to:
Circuit schematic with a 200 V source in series with a 40 Ω resistor feeding a parallel network: a 30 Ω resistor in series with a 15 V source, and a 20 Ω resistor in series with a 20 V source.


I can then find the equivalent resistance to be (30//20)//40 = 9.23 ohms.
However, I am not sure how to find the equivalent voltage in this case?
What should I do to start?
 
That's a wrong transformation. Read this https://people.clarkson.edu/~jsvoboda/eta/dcWorkout/sourceXfrm.pdf a bunch of examples.
 
archaic said:
That's a wrong transformation. Read this https://people.clarkson.edu/~jsvoboda/eta/dcWorkout/sourceXfrm.pdf a bunch of examples.
I don't seem to get it, isn't the transformation correct for the left side though?
 
jisbon said:
I don't seem to get it, isn't the transformation correct for the left side though?
Circuit diagram showing a voltage source V in series with resistor R on the left, an equivalence sign, and a current source I in parallel with resistor R on the right.

You have an extra wire that is making a short circuit.
And, by the way, wouldn't you think that transforming the voltage sources instead would be a better strategy?
 
archaic said:
View attachment 267867
You have an extra wire that is making a short circuit.
And, by the way, wouldn't you think that transforming the voltage sources instead would be a better strategy?
Ah ok I get what you mean, so something like:

Hand-drawn circuit schematic: a 200 V source in series with a 40 Ω resistor feeds two parallel branches, one with a 30 Ω resistor in series with a 15 V source, the other with a 20 Ω resistor in series with a 20 V source.

How do I exactly find the 'total' voltage in this case tho?
 
jisbon said:
Ah ok I get what you mean, so something like:

View attachment 267868
How do I exactly find the 'total' voltage in this case tho?
You use the source transform to make the problem easier. non-ideal voltage sources in parallel are hard. Maybe you could transform the sources into something else?
 
jisbon said:
Ah ok I get what you mean, so something like:

View attachment 267868
How do I exactly find the 'total' voltage in this case tho?
That is correct, but follow my advice, don't do that transformation.
 
jisbon said:
I can then find the equivalent resistance to be (30//20)//40 = 9.23 ohms.
How do you know between which two points the equivalent resistance is asked? Ditto for "single voltage Vs".

I am not sure if I understood this problem correctly.
 
cnh1995 said:
How do you know between which two points the equivalent resistance is asked? Ditto for "single voltage Vs".

I am not sure if I understood this problem correctly.
The drawing does not make it explicit, but the fact that all of the top endpoints of the devices in the middle merge onto a single wire and that all of the bottom endpoints of the devices in the middle merge onto a single wire strongly suggests that those two wires are the two points between which equivalent resistance is desired.

[I had the same reaction at first]