Modification of Wheatstone Bridge

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The discussion centers on calculating the total resistance of a modified Wheatstone Bridge when R2 is replaced with an ideal wire. Participants agree that R1 and R3 will be in parallel, forming R6, which is then in series with R4 to create R7. R7 is ultimately in parallel with R5. A key point of contention arises regarding the impact of replacing both R1 and R2 with ideal wires, leading to the conclusion that R3 becomes irrelevant, leaving R4 and R5 in parallel. The conversation concludes with appreciation for the clarification provided.
songoku
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Homework Statement
This is not homework, the question just crossed my mind when I was reviewing about Wheatstone bridge. Please see the picture below
Relevant Equations
Series and Parallel circuit
1622590517073.png


I want to ask what the total resistance of the circuit will be if R2 is removed and being replaced with ideal wire.

I think R1 and R3 will be parallel → let this be R6

R6 will be series with R4 → let this be R7

Then R7 will be parallel to R5

Is that correct? Thanks
 
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I also think you are correct.
 
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If both R1 and R2 are replaced by ideal wire, would I need to use Delta - star transformation?

Something like this:
1622621474247.png


where:
$$R_A=\frac{R_3 \times R_4}{R_3+R_4+R_5}$$
$$R_B=\frac{R_3 \times R_5}{R_3+R_4+R_5}$$
$$R_C=\frac{R_4 \times R_5}{R_3+R_4+R_5}$$

After that, ##R_A## will be parallel to ##R_B## and the result will be series with ##R_C##

Is that correct? Thanks
 
No I don't think that's correct , if ##R_1## and ##R_2## are replaced by ideal wires then we shall have ##V_D=V_A=V_B## which means that effectively ##R_3## will also be out of the game. So you ll just have ##R_4## and ##R_5## in parallel. That's what I think.
 
Delta2 said:
No I don't think that's correct , if ##R_1## and ##R_2## are replaced by ideal wires then we shall have ##V_D=V_A=V_B## which means that effectively ##R_3## will also be out of the game. So you ll just have ##R_4## and ##R_5## in parallel. That's what I think.
Ah I see

Thank you very much for the help and explanation Charles Link and Delta2
 
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Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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