- #1

Edy56

- 8

- 0

- 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

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- Engineering
- Thread starter Edy56
- Start date

- #1

Edy56

- 8

- 0

- 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

- #2

berkeman

Mentor

- 64,441

- 15,791

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

- 18,206

- 11,219

- #4

Edy56

- 8

- 0

the circuit:

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

- 18,206

- 11,219

SHOW EVERY STEP

- #6

Edy56

- 8

- 0

It's at the beginning of the thread

SHOW EVERY STEP

- #7

- 18,206

- 11,219

Yes, and it got this response:It's at the beginning of the thread

TYPE your work into the thread.

- #8

Edy56

- 8

- 0

That's what I did here?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?

- #9

- 18,206

- 11,219

No, you listed the steps you did not show any of the calculations.That's what I did here?

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

- 8

- 0

I am not sure as to how write that. I can draw it, but I don't know how to write those steps.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.

If you mean calculating new Rs it's just basic multiplication and I have in the picture.

- #11

- 18,206

- 11,219

- #12

Edy56

- 8

- 0

Okay

- #13

Baluncore

Science Advisor

- 12,339

- 6,408

I cannot read the component values.

- #14

Edy56

- 8

- 0

I cannot read the component values.

View attachment 321988

- #15

Baluncore

Science Advisor

- 12,339

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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

- 8

- 0

I am not sure I understand...

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.

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

Science Advisor

- 12,339

- 6,408

Yes, you can do it in the longer way using the inverse Y to Δ transform;Why isn't it like that?

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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