Find the resistance R in this network

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  • #1
Edy56
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Homework Statement:
I am supposed to find the equivalent to R2 and I got 2, but I am supposed to get 3
Relevant Equations:
None
IMG_20230209_020632.jpg
IMG_20230209_020635.jpg
 

Answers and Replies

  • #2
berkeman
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Welcome to PF. :smile:

It's pretty hard for me to read your uploaded pictures. Can you post a better picture of the original circuit, and type some of your work into the forum so it's readable? Thanks.

(you should consider learning to post math using LaTeX -- see the "LaTeX Guide" link below the Edit window)
 
  • #3
phinds
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It's pretty hard for me to read your uploaded pictures. Can you post a better picture of the original circuit, and type some of your work into the forum so it's readable? Thanks.
what he said (very small).jpg
 
  • #4
Edy56
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Welcome to PF. :smile:

It's pretty hard for me to read your uploaded pictures. Can you post a better picture of the original circuit, and type some of your work into the forum so it's readable? Thanks.

(you should consider learning to post math using LaTeX -- see the "LaTeX Guide" link below the Edit window)
the circuit:
ed.png

here is what I did:
(english is not my first language so sorry if i get any terms wrong)
I thought R3, R4 and R5 form a triangle so I turned it into a star for easier calculation and I did the same for R6 R7 and R1.
Then I thought R34 and R35 are parallel so I calculated that too and the same I did for R61 and R71.
Then I thought all of them were connected in order meaning I could add them all together.
That is:
R3435 +R45 + R67 + R6171.
I got the answer 2, I need to get 3.
Where did I go wrong?
 
  • #5
phinds
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How could we possibly know where you went wrong when you have not shown your work ???

SHOW EVERY STEP
 
  • #6
Edy56
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How could we possibly know where you went wrong when you have not shown your work ???

SHOW EVERY STEP
It's at the beginning of the thread
 
  • #7
phinds
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It's at the beginning of the thread
Yes, and it got this response:
It's pretty hard for me to read your uploaded pictures. Can you post a better picture of the original circuit, and type some of your work into the forum so it's readable? Thanks.
TYPE your work into the thread.
 
  • #8
Edy56
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the circuit:
View attachment 321978
here is what I did:
(english is not my first language so sorry if i get any terms wrong)
I thought R3, R4 and R5 form a triangle so I turned it into a star for easier calculation and I did the same for R6 R7 and R1.
Then I thought R34 and R35 are parallel so I calculated that too and the same I did for R61 and R71.
Then I thought all of them were connected in order meaning I could add them all together.
That is:
R3435 +R45 + R67 + R6171.
I got the answer 2, I need to get 3.
Where did I go wrong?
That's what I did here?
 
  • #9
phinds
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That's what I did here?
No, you listed the steps you did not show any of the calculations.

THIS:
"I thought R3, R4 and R5 form a triangle so I turned it into a star for easier calculation and I did the same for R6 R7 and R1."

is a set of steps. It is not calculations.
 
  • #10
Edy56
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No, you listed the steps you did not show any of the calculations.

THIS:
"I thought R3, R4 and R5 form a triangle so I turned it into a star for easier calculation and I did the same for R6 R7 and R1."

is a set of steps. It is not calculations.
I am not sure as to how write that. I can draw it, but I don't know how to write those steps.

If you mean calculating new Rs it's just basic multiplication and I have in the picture.
 
  • #11
phinds
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I'm sure you could figure out how to type numbers and equations, but I can see that it would be a lot of work. Perhaps someone else here will be willing to look at your hand-written work and follow it to see if they can find your error.
 
  • #12
Edy56
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I'm sure you could figure out how to type numbers and equations, but I can see that it would be a lot of work. Perhaps someone else here will be willing to look at your hand-written work and follow it to see if they can find your error.
Okay
 
  • #13
Baluncore
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I cannot read the component values.
R_Values.png
 
  • #15
Baluncore
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You do not need the Y to Δ transform.
Current source Ig is infinite resistance, so remove it from the circuit.
Voltage sources, E1 and E2 are zero resistance, so short them.
R1 = R5 = R6 = 2k;
R3 = R4 = 1k;
R7 = 4k.
Series; +
Parallel; //
R16 = R1 + R6 =
R34 = R3 + R4 =
R167 = R16 // R7 =
R345 = R34 // R5 =
R2 = R167 + R345 = 3k.
 
  • #16
Edy56
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You do not need the Y to Δ transform.
Current source Ig is infinite resistance, so remove it from the circuit.
Voltage sources, E1 and E2 are zero resistance, so short them.
R1 = R5 = R6 = 2k;
R3 = R4 = 1k;
R7 = 4k.
Series; +
Parallel; //
R16 = R1 + R6 =
R34 = R3 + R4 =
R167 = R16 // R7 =
R345 = R34 // R5 =
R2 = R167 + R345 = 3k.
I am not sure I understand...
The way I do it, I imagine that power is flowing from a to b and so when it's going from a the power splits in two because of R4 and R5 and thus making them parallel. Why isn't it like that?
 
  • #17
Baluncore
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Why isn't it like that?
Yes, you can do it in the longer way using the inverse Y to Δ transform;
https://en.wikipedia.org/wiki/Y-Δ_transform

Your mistake is that you have the current source, Ig = 0 ohms, as a short circuit, which is wrong. A current source has infinite resistance, which is an open circuit.
You do not need the Y to Δ transform if you make the current source = ∞ ohms.
 

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