Engineering Circuit Analysis: Problem 4 - Obtain Current Through Galvanometer

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The discussion revolves around calculating the current through a Galvanometer in a Wheatstone bridge circuit, with a focus on a resistance of 14Ω. The user applied a delta-Y transformation and calculated an equivalent resistance of 32.4675Ω, leading to a current of 0.4927 A. However, another participant questions the validity of this approach, suggesting that the transformation may eliminate the resistor through which the current is being measured. This indicates a potential misunderstanding of the circuit analysis method used. The accuracy of the current calculation remains in doubt due to the concerns raised about the transformation applied.
cooper607
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Homework Statement



problem---4
(5 marks) Obtain the current through the Galvanometer G, having a resistance of 14Ω, in
the Wheatstone bridge shown in Fig-4

Homework Equations



i know the delta-Y transformation and applied it, i found out the R (eq) to be 32.4675ohm

The Attempt at a Solution


I=V/R= 0.4927 A
someone please check the answer of problem no 4 and let me know if its right or wrong
regards
 

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cooper607 said:

Homework Statement



problem---4
(5 marks) Obtain the current through the Galvanometer G, having a resistance of 14Ω, in
the Wheatstone bridge shown in Fig-4

Homework Equations



i know the delta-Y transformation and applied it, i found out the R (eq) to be 32.4675ohm



The Attempt at a Solution


I=V/R= 0.4927 A
someone please check the answer of problem no 4 and let me know if its right or wrong
regards

Doesn't look right to me.

If you apply a Δ-Y transformation, won't you transform-away the very resistor you wish to find the current through?
 

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