- #1
Edy56
- 38
- 5
- 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
Find the resistance R in this network
- Engineering
- Thread starter Edy56
- Start date
-
- Tags
- Network Resistance
In summary, the conversation is about a circuit problem and the person asking for help has not shown their work or calculations. They have described their steps, but have not provided any numbers or equations. Another user has suggested removing some components and provided the values for the resistors. They also have simplified the circuit and calculated the final equivalent resistance to be 3k. However, the original user has not responded with their calculations or whether this solution worked for them.
- #2
berkeman
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Welcome to PF. 
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)
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
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berkeman said:
- #4
Edy56
- 38
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the circuit:berkeman said:Welcome to PF.
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)
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
- 19,013
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How could we possibly know where you went wrong when you have not shown your work ???
SHOW EVERY STEP
SHOW EVERY STEP
- #6
Edy56
- 38
- 5
It's at the beginning of the threadphinds said:
- #7
- 19,013
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Yes, and it got this response:Edy56 said:
TYPE your work into the thread.berkeman said:
- #8
Edy56
- 38
- 5
That's what I did here?Edy56 said:
- #9
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- 14,580
No, you listed the steps you did not show any of the calculations.Edy56 said:
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
- 38
- 5
I am not sure as to how write that. I can draw it, but I don't know how to write those steps.phinds said:
If you mean calculating new Rs it's just basic multiplication and I have in the picture.
- #11
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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
- 38
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Okayphinds said:
- #13
Baluncore
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I cannot read the component values.
- #14
Edy56
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Baluncore said:
- #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.
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
- 38
- 5
I am not sure I understand...Baluncore said:
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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Yes, you can do it in the longer way using the inverse Y to Δ transform;Edy56 said:
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.
FAQ: Find the resistance R in this network
1. How do I find the equivalent resistance in a series circuit?
In a series circuit, the equivalent resistance (R_eq) is the sum of all individual resistances. You can find it using the formula: R_eq = R1 + R2 + R3 + ... + Rn.
2. How do I find the equivalent resistance in a parallel circuit?
In a parallel circuit, the reciprocal of the equivalent resistance (1/R_eq) is the sum of the reciprocals of each individual resistance. You can find it using the formula: 1/R_eq = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn. Then, take the reciprocal of the result to get R_eq.
3. What is the difference between series and parallel circuits when finding resistance?
In series circuits, resistances add up directly, making the total resistance higher. In parallel circuits, the total resistance is lower than the smallest individual resistance because the current has multiple paths to flow through.
4. How do I combine series and parallel resistances in a complex network?
To find the equivalent resistance in a complex network with both series and parallel components, break down the circuit into simpler parts. First, calculate the equivalent resistance of series and parallel sections separately, then combine them step by step until you reach the overall equivalent resistance.
5. Can I use Ohm's Law to find the resistance in a network?
Yes, Ohm's Law (V = IR) can be useful. If you know the total voltage (V) across the network and the total current (I) flowing through it, you can find the total resistance (R) using the formula: R = V/I. This gives you the equivalent resistance of the entire network.