Modification of Wheatstone Bridge

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SUMMARY

The discussion centers on the modification of a Wheatstone Bridge circuit when resistors R1 and R2 are replaced with ideal wires. Participants confirm that R1 and R3 will be in parallel, forming R6, which is then in series with R4 to create R7. R7 is subsequently in parallel with R5. A key point raised is the necessity of using Delta-Star transformation, but it is clarified that if R1 and R2 are ideal wires, R3 becomes irrelevant, leaving only R4 and R5 in parallel. This conclusion is supported by the voltage equality at points A, B, and D.

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  • Understanding of Wheatstone Bridge circuits
  • Knowledge of series and parallel resistor combinations
  • Familiarity with Delta-Star transformation in circuit analysis
  • Basic principles of electrical resistance
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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 believe you have it correct. :)
 
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.
 
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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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