Find Norton Equiv. for Homework: V=IR

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Homework Help Overview

The discussion revolves around finding the Norton equivalent circuit at specific terminals, utilizing the relationship defined by Ohm's law (V=IR). Participants are exploring the transition from a Thévenin equivalent to a Norton equivalent, addressing the necessary steps and considerations involved in this process.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the validity of previous calculations and the implications of resistor configurations (series vs. parallel). Questions arise regarding the conversion process between Thévenin and Norton equivalents, as well as the correct application of voltage division in different circuit configurations.

Discussion Status

There is an ongoing exploration of the relationships between circuit components and their configurations. Some participants have offered guidance on how to approach the conversion between Thévenin and Norton equivalents, while others are questioning the assumptions made in earlier calculations. The discussion is active, with multiple interpretations being examined.

Contextual Notes

Participants are navigating through potential misunderstandings regarding the arrangement of resistors and the application of circuit laws. The original problem constraints and the need to redraw the circuit for clarity have been acknowledged, but no definitive conclusions have been reached.

Marcin H
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Homework Statement


Find the Norton equivalent at the terminals.

Homework Equations


V=IR

The Attempt at a Solution


I did the thevenin equivalent, but I'm not sure where to go from there. How do I get the Norton equivalent out of this? [/B]
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Once you've gone to the trouble of determining the Thévenin equivalent, and are sure its correct, there is no need to repeat the process to determine the Norton. The impedance is the same for each, so simply short-circuit your Thévenin equivalent and that gives you the Norton's current.

Though before you calculate the Norton you'd best take another look at your Thévenin, as you have taken invalid shortcuts there. What became of the 20 and 80 ohm resistors?
 
NascentOxygen said:
Though before you calculate the Norton you'd best take another look at your Thévenin, as you have taken invalid shortcuts there. What became of the 20 and 80 ohm resistors?
I just learned this and it's still a bit confusing. Can't I just V=IR and, using the current and resistor, make the 2 sections into voltage sources? Or do I have to add those resistors in series in my thevenin circuit?
 
You can replace any current source + parallel resistor with a voltage source + series resistor.

Once you have done that, draw your new circuit.
 
This might help.
images?q=tbn:ANd9GcSvMaZwjm5cDFNoc9x_A2YDKfw06RtsS1ipPWgp669W0KVPEp_OGQ.png
 
NascentOxygen said:
You can replace any current source + parallel resistor with a voltage source + series resistor.

Once you have done that, draw your new circuit.
So would I just add a 100ohm resistor in series with my picture on the bottom left and then change the 400ohms to 500ohms on the picture on the right?
 
Marcin H said:
So would I just add a 100ohm resistor in series with my picture on the bottom left and then change the 400ohms to 500ohms on the picture on the right?
No. You should not convert 400 into 500.
You are supposed to find the resistance between a and b. 100 and 400 are not in series then. What is the procedure to get the Thevenin resistance?
 
cnh1995 said:
No. You should not convert 400 into 500.
You are supposed to find the resistance between a and b. 100 and 400 are not in series then. What is the procedure to get the Thevenin resistance?
Oh, they would be in parallel. So would I find my thevenin voltage by doing 80V(400ohms/500ohms)=64V?
 
Marcin H said:
Oh, they would be in parallel. So would I find my thevenin voltage by doing 80V(400ohms/500ohms)=64V?
Right!
 
  • #10
  • #11
R1 and R2 are in series. What is the quantity V1/(R1+R2)?
 
  • #12
cnh1995 said:
R1 and R2 are in series. What is the quantity V1/(R1+R2)?
The current? And then you multiply by the resistance R2 to find the voltage there and that gives you the voltage Vout?
 
  • #13
Marcin H said:
The current? And then you multiply by the resistance R2 to find the voltage there and that gives you the voltage Vout?
Right! It is derived from Ohm's law,V=IR.
 
  • #14
cnh1995 said:
Right! It is derived from Ohm's law,V=IR.
Then why can I use that here in this circuit if they are in parallel?
 
  • #15
Marcin H said:
Then why can I use that here in this circuit if they are in parallel?
For the original voltage source in your diagram, 100Ω and 400Ω are in series. For calculating Thevenin voltage, you make use of that original voltage source. To find the Thevenin resistance, you short the original voltage source and then proceed. 100Ω and 400Ω are in parallel when "viewed from port a-b".
Hence, 100Ω and 400Ω will be parallel for Thevenin voltage source but are in seeies fot the original voltage source.
 
  • #16
cnh1995 said:
For the original voltage source in your diagram, 100Ω and 400Ω are in series. For calculating Thevenin voltage, you make use of that original voltage source. To find the Thevenin resistance, you short the original voltage source and then proceed. 100Ω and 400Ω are in parallel when "viewed from port a-b".
Hence, 100Ω and 400Ω will be parallel for Thevenin voltage source but are in seeies fot the original voltage source.
Ok so if they are parallel for the thevenin voltage source, then why can I use that same equation for finding the thevenin voltage? if they are in parallel.
 
  • #17
Marcin H said:
Ok so if they are parallel for the thevenin voltage source, then why can I use that same equation for finding the thevenin voltage? if they are in parallel.
Thevenin voltage will be dependent on original voltage source and resistance
combinations "seen" by the original voltage source. You convert this circuit into Thevenin equivalent. It doesn't affect the load.
ANd9GcSvMaZwjm5cDFNoc9x_A2YDKfw06RtsS1ipPWgp669W0KVPEp_OGQ&hash=f5aefea7d75d010020a7658f31c53235.png
 
  • #18
Another method to find Thevenin resistance is as follows:
Short the a-b path and calculate total current (with original voltage source). This is your Norton current(IN). Find Vth using voltage divider.
Rth=Vth/IN.
=Voc/Isc
This method is commonly used when there are dependent sources in the circuit.
 
  • #19
cnh1995 said:
Thevenin voltage will be dependent on original voltage source and resistance
combinations "seen" by the original voltage source. You convert this circuit into Thevenin equivalent. It doesn't affect the load.
ANd9GcSvMaZwjm5cDFNoc9x_A2YDKfw06RtsS1ipPWgp669W0KVPEp_OGQ&hash=f5aefea7d75d010020a7658f31c53235.png
oh, ok. that makes more sense now. So how would I find the norton equivalent? Would I use the do In= Vth/Rth? Would Rth be found by adding the 100ohm and 400ohm in parallel?
 
  • #20
Marcin H said:
Would Rth be found by adding the 100ohm and 400ohm in parallel?
Yes. And it will be in parallel with the Norton current source.
 
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  • #21
cnh1995 said:
Yes. And it will be in parallel with the Norton current source.
Ok. I got .8A. The original problems says to find the Norton equivalent at terminals a-b for the circuit shown. Do I just have to find the value or do I have to redraw the circuit with a current source and norton(equivalent) resistance?
 
  • #22
Marcin H said:
Ok. I got .8A. The original problems says to find the Norton equivalent at terminals a-b for the circuit shown. Do I just have to find the value or do I have to redraw the circuit with a current source and norton(equivalent) resistance?
I guess they are expecting you to redraw the circuit.
 

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