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Image credit: Byjus

This is the typical experimental setup to determine the unknown resistance of a given wire. As we move the sliding contact of the rheostat, we get different values of voltage and current. a plot of V vs i gives a straight line, and the slope of this line is the wire's resistance. Here, why doesn't the resistance of the rheostat impact the experiment?

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    $\begingroup$ What is K in your diagram? $\endgroup$ Feb 27 at 15:55
  • $\begingroup$ I actually took K to represent a set of contacts that can be used to complete the circuit, however, that is a good question. $\endgroup$ Feb 27 at 16:01
  • $\begingroup$ @AlbertusMagnus Going by the diagram only, it's an open circuit $\endgroup$ Feb 27 at 16:10
  • $\begingroup$ K is the key in the circuit diagram $\endgroup$
    – Navya
    Feb 29 at 15:19

1 Answer 1

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a plot of V vs i gives a straight line, and the slope of this line is the wire's resistance. Here, why doesn't the resistance of the rheostat impact the experiment?

I think you have this wrong. The slope equal to $V/I$ is the resistance of the resistor between Voltage meter terminals.

$V/I$ is not the resistance of the mystery wire.

If we measure the voltage $V_x$ across a resistor $R_x$ which has a current $I_x$ passing through it, then the ratio $V_x/I_x$ is equal to the resistance of $R_x$ (depicted as R in your diagram) and not some other resistor in the circuit.

The total resistance of the circuit is $R_{total} = (R + R_{wire} + f \ R_{rheostat})$ where f is the fraction of the rheostat that is 'live'.

The resistance of the mystery wire is $$R_{wire} = R_{total} - f \ R_{rheostat} - V/I$$ which is more complicated to work out and not independent of the rheostat resistance and the position of the rheostat slider f.

If $R_{wire} = V/I$ and $R_{test} = V/I$ then it is only by chance and not normally the case.

It seems sensible that if you want to measure the resistance of a particular wire, to connect the terminals of the Volt meter to either end of the test wire. You can then ignore the resistance of the rheostat and the connecting wires of the test circuit.

This analysis ignores the small current that flows through the Amp meter shunt resistor which is designed to be negligible and the internal resistance of the battery. The rheostat is just there to limit the total current through the circuit to reasonable values.

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    $\begingroup$ @KDP Totally in agreement with you, you must know the value of the rheostat if you want the resistance of the wire. $\endgroup$ Feb 28 at 16:46
  • $\begingroup$ @StevanV.Saban Nonsense. You should retract your post. $\endgroup$ Feb 28 at 16:51
  • $\begingroup$ I suppose the diagram was not clear... the wire of unknown resistance is connected parallel to the voltmeter ( R is the wire) $\endgroup$
    – Navya
    Feb 29 at 16:10
  • $\begingroup$ So you agree my answer is the correct answer and deserves a tick? :-) $\endgroup$
    – KDP
    Feb 29 at 23:17

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