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Question about voltage, current and resistance

Mrs.Z Avatar
6y, 5m agoPosted 6 years, 5 months ago
Hi all, quick question.

We did an experiment measuring voltage and current (and using these results to calculate resistance), increasing the voltage for each result but using the same resistor.

In my results the resistance increased very slightly with each increase of voltage, is this what 'should' have happened? Or should it have remained constant?

I don't know if the slight increase is down to the calculations and the fact that the voltmeter and ammeter could only read to 2 decimal places.

Thanks x
Mrs.Z Avatar
6y, 5m agoPosted 6 years, 5 months ago
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Comments/page:
#1
You might find that as the voltage increases and the resistor remains the 'same', then the current will increase. This may have led to an increase in temperature of the resistor which in turn can lead to an increased resistance.

Alternatively, depending on the magnitude of your results, you may be correct that the tolerances of your components and measuring instruments become more significant as you increase the voltage.

Was that nerdy enough!?
#2
resistance increases with voltage increase but decreases in proportion to current increase.

R=pla
#3
V=IR

pd = current x resistance

therefore YES it is expected.
However this is only true provided current remains constant or changes at a certain proportion
#4
Thanks everyone, this is my results to show what i mean:

Voltage (V) Current (A) Resistance (Ω)
0.00 * 0.00 * 0.00
2.02 * 0.07 * 28.86
4.05 * 0.15 * 27
6.03 * 0.22 * 27.41
7.98 * 0.29 * 27.52
10.04 * 0.36 * 27.89

Then I have to explain these results with reference to Ohms Law

If the resistance remains constant then the current will increase along with the voltage, but can I say that the resistance is remaining constant or could it be increasing slightly?
The reason I think it may be due to the limitations of my equipment is due to the 2nd resistance being the highest
#5
Shouldn't you measure resistance with no power flowing? ie the multimeter puts its own power in, any other power will disrupt that?
#6
Benjimoron
Shouldn't you measure resistance with no power flowing? ie the multimeter puts its own power in, any other power will disrupt that?


I dunno! I just did as my 'not quite with it' maths teacher told me :w00t:
#7
Mrs.Z
I dunno! I just did as my 'not quite with it' maths teacher told me :w00t:


I just re-read above, my comment is pretty irrellevant. Temperature is the likely answer I would've thought.
#8
Thanks everyone :)
#9
If you take the second set of measurements If the current measurement is slightly out this has a big effect on the calculated resistance,
For example adding or subtracting 0.01 amp would give:
2.02 / 0.06 = 33.67
2.02 / 0.08 = 25.25

That's quite a big change in the calculated resistance for such a small change in current.

Increasing the voltage and current will improve the accuracy of the result. By the time you get to the last set of measurements adding or subtracting 0.01 amp has less effect:
10.04 / 0.35 = 28.69
10.04 / 0.37 = 27.14

You can also argue that the voltage measurement may by out by 0.01 volt. So the worst case figures are:
10.05 / 0.35 = 28.71
10.03 / 0.37 = 27.11

Without better measurements all you can say is that the resistance is somewhere between 27.11 and 28.71 ohms.

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