Thevenin circuit

  • Engineering
  • Thread starter vikasagartha
  • Start date
  • #1

Homework Statement


Ive never done Thevenin equivalents before, and im not sure if I have removed the correct "load component". Could someone please take a look @ my solution and let me know if I have done this correctly? Thanks in advance! I really appreciate it!
Circuit Diagram:
https://www.dropbox.com/s/q5kubgapyu89blt/Screenshot from 2014-09-06 10:15:12.png?dl=0


Homework Equations


V=IR, KCL

The Attempt at a Solution


Written by hand, screenshot attached.
https://www.dropbox.com/s/3u4ku96uvksioul/2014-09-06-101902.jpg?dl=0
https://www.dropbox.com/s/dg1meg27ojsyvtc/2014-09-06-101846.jpg?dl=0
 

Answers and Replies

  • #2
gneill
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As drawn the circuit has no load between terminals A and B. A and B is where a load would be attached. The 5 k resistor between A and B is not a load resistor, it is part of the network to be analyzed so don't remove it!
 
  • #3
Clearly a conceptual misunderstanding. Makes much more sense now. Thanks for pointing it out:)
 
  • #4
Second try

New procedure:
1) Left 5k resistor in as A and B have no load in between

2) Used three current loops I1, I2, I3 and calculated three currents using a system of equations.
Current equations
Currents solved
* I1=173.91 micro Amps
* I2= 8.69 micro Amps
* I3= 382.6 micro Amps

3) Found voltage drops over resistors
https://www.dropbox.com/s/xbx05mykzjvjw2y/2014-09-07-061535.jpg?dl=0

4) Used the voltage drops to find the Vth = 826mV
vthevenin

5) Now im stuck. I have no idea how to find Rth for this circuit! Help!?

PS, sorry the webcam is garbage. My laptop was made ~200 yrs ago;)
 
  • #5
gneill
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Check the sign on I2.

To find Rth start by suppressing all the voltage sources, then look for opportunities to combine serial and parallel combinations of resistors. I.e., reduce the circuit. Don't be fooled by the geometrical layout of the drawing --- pay attention to the actual connections of the components by identifying the nodes.
 
  • #6
thanks for the help gneill. I think I have it figured out. Rth = 1.884k and Vth = 826mV.

I have another conceptual question, if you dont mind. Thevenin vs Norton - I am comfortable with thevenin so I normally find the Vth and Rth and use Vth/Rth = In to then find the norton equivalent. Is there an advantage to one model over the other? Should I bother learning how to find Norton equivalents first?
 
  • #7
gneill
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thanks for the help gneill. I think I have it figured out. Rth = 1.884k and Vth = 826mV.
Those values look alright.

I have another conceptual question, if you dont mind. Thevenin vs Norton - I am comfortable with thevenin so I normally find the Vth and Rth and use Vth/Rth = In to then find the norton equivalent. Is there an advantage to one model over the other? Should I bother learning how to find Norton equivalents first?

Both equivalents have their advantages in certain situations. If you're given a circuit that already has current sources, direct reduction to a Norton equivalent may provide a more straightforward approach. Or, perhaps a circuit is just simpler to analyze with mesh equations, where one of the mesh currents is identified with the short circuit current that you're looking for. Then you may only need to find that single mesh current rather than solve for them all (Cramer's Rule rules!).

You should be able to work problems either way if for no other reason than you might be required to demonstrate it on an exam :wink:
 

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