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What is the equivalent resistance of this network?

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Yes, Between which point - all depend on that question.
 
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I know that it's delta ot wye. I just need to check my answer. Just tell me what you get.

Many were asking between what. I need the resistance looking into R4.
 

iVenky said:
I know that it's delta ot wye. I just need to check my answer. Just tell me what you get.

Many were asking between what. I need the resistance looking into R4.
Click to expand...
its 0.6 , seen by R4.
 

I will be clear this time. I need to replace all the resistances by a single resistance.
 

What is the answer? Is it 487 Ohms(+/- 2%)
 
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jeet_asic said:
its 0.6 , seen by R4.
Click to expand...

diwasverma said:
eq. resistance acroos R4 = .6 K
Click to expand...

Two voices against one... it seems you win... But...
Being lazy, I just drew the network on LTspice and I supplied R4 by a current source of 1 A. I got on R4 500V which means that the current source sees 500R.

So now, the contest is between you two and LTspice :twisted:
 
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SPICE is a woman: she's always right ;-) .
No current through R2 .
 
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    KerimF

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I agree with KerimF

R_NetW_01.jpg
 
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What is the answer? Is it 487 Ohms(+/- 2%)
Click to expand...
Reminds me to a joke about an engineer, a physicist and a mathematician solving the problem "How much is two times two?"

The engineer doesn't think much and says: "It's about four".

The physicist starts to calculate, then after a minute: "It's 3.9998 +/- 0.0015".

The mathematician thinks for a long time, than finally states: "The solution exists". And after another long while: "and it's finite".
 
FvM said:
Reminds me to a joke about an engineer, a physicist and a mathematician solving the problem "How much is two times two?"

The engineer doesn't think much and says: "It's about four".

The physicist starts to calculate, then after a minute: "It's 3.9998 +/- 0.0015".

The mathematician thinks for a long time, than finally states: "The solution exists". And after another long while: "and it's finite".
Click to expand...


Ha ha that's refreshingly hilarious.

---------- Post added at 01:36 ---------- Previous post was at 01:36 ----------

I am getting 2/9 (ie) 0.22222.
 
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What would be the resistance if R3 and R6 were 2k instead?
 

I put 10V across R4.

My simulator converges on 20 mA current flow.

Therefore the equivalent resistance is 500 ohms.

I can confirm no current flows through R2. Both its ends are at 5V.
 

I am getting 2/9 (ie) 0.22222.
Click to expand...
The fault would be more instructive, if you tell, how you calculated the result.

The most simple method is utilizing the symmetry of the circuit, first mentioned by erikl and shown verbosely by kak111. It allows to solve the problem, by simple mental arithmetic, the best engineer's method in my opinion

Without this symmetry, standard network calculation methods have to be used. For a stepwise solution, wye-delta conversion is useful, as said.

Solving a system of linear equations is your last ressort anyway.
 
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I agree with kak111 and with its method for rearranging the network. The solution should br 500 ohms.
 

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