0
$\begingroup$

I was going through my physics laboratory manual. In the Ohm's Law Experiment, the book states a few precautions without any reasoning.

1.The wire whose resistance has to be determined should ideally be made up of alloys such as Manganin and not a metal. 2. A low resistance rheostat should be used.

Any reason why this is so?

Also the book says that the resistance wire must be wound on its self before it is wound on bobbin or reel to avoid induction effect.I did not understand this.

Thank You.

PS:I am connecting my sample resistance wire(with its length constant) and taking different values of V and I by altering the rheostat resistance.

Improve this question
$\endgroup$
2
  • $\begingroup$ Please describe the experiment. It sounds like you are connecting a rheostat in series with a voltage source, connecting your sample across that, and measuring the current through the sample as a function of the rheostat resistance. Are you using different lengths of sample? Are you making a measurement very quickly? It would appear so as the induction effect will die away with time. $\endgroup$
    – Ross Millikan
    Commented Feb 7, 2016 at 15:37
  • $\begingroup$ resistance of manganin and constantan do not vary much with temperature. Also for higher resistance. $\endgroup$
    – Anubhav Goel
    Commented Feb 7, 2016 at 15:49

2 Answers 2

Reset to default
1
$\begingroup$

The reason to use the alloy is because it has a much higher resistivity than copper and so the alloy wire will have a higher resistance which with standard laboratory apparatus can be measured more accurately. I also seem to remember that the temperature of coefficient is lower for some of these alloys ie for a given increase in temperature the resistance of the alloy will change less.

The rheostat having a low resistance possibly means that it is used as a potential divider. If that is the case then you would have more control (use a greater range of the slider) on setting the voltage.

The only reason to worry about self inductance is if you are using alternating voltages and currents.

Improve this answer
$\endgroup$
2
  • $\begingroup$ but the rheostat wire is made up of constantan which has a pretty high resistivity.So how can we possibly have a low resistance rheostat by using such a wire? $\endgroup$
    – Karan Singh
    Commented Feb 7, 2016 at 15:58
  • $\begingroup$ If you look at the windings you would probably find that they are of fairly large diameter for a low resistance rheostat. $\endgroup$
    – Farcher
    Commented Feb 7, 2016 at 16:00
0
$\begingroup$

The choice of a non-metallic sample is to get the resistance high enough to measure. The resistance of the rheostat, the voltage supply, and the hookup wires can make the measurement inaccurate if you do not account for them. If they are much smaller than the resistance of the sample they will not matter much. You also will not try to draw too much current for your voltage supply to source.

If you make a coil of wire, it will have a certain amount of inductance. This will resist changes in the current by making a back emf of $L\frac {di}{dt}$. If you allow the transient to die away before you measure the current this will not matter. If you measure quickly after establishing the circuit, you want to keep $L$ small to minimize the effect.

Improve this answer
$\endgroup$
3
  • $\begingroup$ so does wounding the resistance wire on itself help me to keep L small?How? $\endgroup$
    – Karan Singh
    Commented Feb 7, 2016 at 15:54
  • $\begingroup$ Depending on how you wind it, you can cancel out the magnetic field it makes, reducing $L$. You did not provide any details of the winding approach, so it is hard to tell. $\endgroup$
    – Ross Millikan
    Commented Feb 7, 2016 at 16:00
  • 1
    $\begingroup$ The standard way is to fold the wire in half like a squashed U. Then the currents in adjacent bits of the wire flow in opposite directions and so the magnetic field due to wires cancel out. $\endgroup$
    – Farcher
    Commented Feb 7, 2016 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.