Thevenin's Theorem -- Using transformations to find the equivalent resistance

  • #1

Homework Statement:

Calculating the Ix current using the Thevenin

Relevant Equations:

Thevenin's Theorem
Hello Guys, I really need help, I am trying to simplify this circuit to calculate Rth and I got stuck.

Screenshot 2020-03-29 at 18.46.00.png


Screenshot 2020-03-29 at 19.06.06.png
 
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Answers and Replies

  • #2
cnh1995
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In your first sketch, what can you say about R5 and R7? Do you really need a delta-star transformation?:wink:
 
  • #3
In your first sketch, what can you say about R5 and R7? Do you really need a delta-star transformation?:wink:
I really don’t know what you mean. Could you please illustrate more?
 
  • #4
cnh1995
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Delta-star transformation is used when you cannot use simple series/parallel reduction. Is that the case with R5 and R7?
 
  • #5
cnh1995
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Delta-star transformation is used when you cannot use simple series/parallel reduction. Is that the case with R5 and R7?
Your method is also correct.
You (correctly) discarded R57 in the vertical leg of the star and you have the final reduced circuit.
Finally, Rth is the resistance seen by a voltage source connected between the Thevenin terminals.
What do you get when you do that in your final circuit?
 
  • #6
Delta-star transformation is used when you cannot use simple series/parallel reduction. Is that the case with R5 and R7?
So, you mean that R5 & R7 are in parallel?
 
  • #7
DaveE
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That's an odd way to draw the schematic, but I guess I don't know the context of the book/chapter.

Ideal current sources have infinite resistance. They will make whatever voltage is required to get the current indicated, so it doesn't matter to the rest of the circuit what value the series resistor(s) is. You could make R2 and R6 be 0 ohms (or any value, like ∞) and get the same effect on the rest of the circuit, 10A (10A no matter what is in that branch of the circuit).

Current meters will have zero resistance. They can't have a voltage drop or that would change the current in the circuit when it is inserted to make a measurement. If that is a current meter instead of a source, then you can replace it with a short (0 ohm resistor). It's the same as if they just told you the "current through R2 is 10A".

There is a similar issue with resistors in parallel with voltage sources. That doesn't apply to your problem, but I think it's easier to see intuitively. If you put a resistor across an ideal voltage source, you will make more current flow in the battery and the added resistor, but that has no effect on whatever else is connected to the source, that stuff will just see the applied voltage.
 
  • #8
cnh1995
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So, you mean that R5 & R7 are in parallel?
No, they are in series. This saves you the trouble of delta-star transformation. But your method also works.

If you are asked to find the equivalent resistance of a circuit between any two terminals, say x and y, what you are actually doing is find the resistance "seen" by a voltage source connected between x and y.
Connect a voltage source of emf (E) between x and y, measure the current (I) supplied by the voltage source, then E/I gives you the equivalent resistance between x and y.

What do you get when you do that in your final circuit schematic?
 
  • #9
guys, I came up with this answer I don't know if it's correct or not. Tell me your opinions!!
Screenshot 2020-04-01 at 05.00.13.png
 
  • #10
Joshy
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##R_{th}## looks right to me although you're really drawing a lot of attention to ## R_4 \parallel (R_5+R_7)##
 
  • #11
##R_{th}## looks right to me although you're really drawing a lot of attention to ## R_4 \parallel (R_5+R_7)##
but I'm stuck with Vth. Do you have any idea?
 
  • #12
Joshy
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I don't see any obvious shortcuts. I tried it with a mesh you have two loops and you already know the current of one of them. Get the voltages across each element and call that node between ##R_4## and ##E##... call it your reference (ground).

circuit.png
 
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