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Simple effective resistance of a circuit at low voltages

  1. Apr 30, 2017 #1
    I passed small voltages (0-5V) through a known resistance, measured the current each time, and got different "effective resistances" for each case. However, when graphed, the effective resistances appear to approach the known resistance of the circuit as a limit at higher voltages.

    What is the reason for this? Nobody I ask can give me a satisfactory response. I've been told resistance never changes .... is it due to the inner workings of the ammeter I'm measuring it with?

    Thanks!
     
  2. jcsd
  3. Apr 30, 2017 #2
    It's not entirely true that resistance can't vary. Resistance of a wire depends on the temperature of the wire as well as it's temperature coefficient of resistance (which indeed depends on the material of the wire).
    Rt=R0(1 + αt)

    where Rt is the resistance at t°C, R0 is the resistance at 0°C,
    t is the temperature of the wire when the current of known potential difference is applied
    α is the temperature coefficient of resistance depending on the material of the wire.

    So, as the potential difference varies, the temperature of the wire changes slightly and deviates slightly from the experimental value
     
  4. Apr 30, 2017 #3

    jim hardy

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    You've asked us to evaluate data that you haven't shown..

    "I've got a secret" is a game, not a request for help.
     
  5. Apr 30, 2017 #4
    Sorry didn't mean to be secretive, lol. Here's a pic of the data and the graph I constructed from it (I just uploaded it here as a jog)
     

    Attached Files:

  6. Apr 30, 2017 #5
    Hmm, so the resistance depends linearly on temperature...but this doesn't match my results, I'm still confused
     
  7. Apr 30, 2017 #6
    Wait a second I might have stumbled on part of my answer. P=I2R, so if temperature itself is dependent on the current (as I think it should be, since the temperature increase in a resistor arises from the resistance to current flow), then this means that the temperature is proportional to power, but itself approaches an upper temperature limit. But this would only work for a nonlinear resistor, so I'm still in the dark lol. Thoughts?
     
    Last edited: Apr 30, 2017
  8. Apr 30, 2017 #7

    sophiecentaur

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    Where are you measuring the voltage across the resistor? PD should always be measured independently across the terminals and 'inside' the connections to the power supply. You have to eliminate contact resistance and other such nasties.
    What sort of temperatures are you experiencing? A tungsten filament can increase in resistance by a factor a ten, when it is at full power. What metal is the wire made of?
    In fact, it would improve your chances of a good answer if you gave a full description of your experiment (circuit diagram and even a photo of the layout). You can't assume we know exactly what you have ben doing and the details can be very important.
     
  9. Apr 30, 2017 #8
    The circuit was built on this breadboard by Digilent Inc (http://store.digilentinc.com/electr...-one-usb-oscilloscope-multimeter-workstation/). It was simply a few resistors in series, the voltage was measured from one resistor leg to the other, across the whole resistor chain. Since I was only working on 0-5V with only 0.02-0.18 mA current, the temperature rise must be so tiny its insignificant. I know the explanation must be something fundamental that I'm just missing.

    Again, I'll post the pic of my results, which has the resistance graph on it.
     

    Attached Files:

  10. Apr 30, 2017 #9
    Here's a plot of your data with a polynomial degree 1 fit:

    p1_fit.png

    Doesn't look that bad. Fit gives an effective resistance of 30.85 kohm.

    I'd take a guess that your measurements are a bit inaccurate down at those microamp levels. Have a look at the datasheet for your equipment for its accuracy at whatever range setting you're using.

    Alternatively, lower the values of your resistors.
     
  11. Apr 30, 2017 #10
    Yes, that is the plot I got as well ... and I do understand that we get resistance from the slope. I guess what I'm failing to grasp is why the Input Resistance varies, while the real resistance remains constant. My professor could not really explain it to me, he studied mechanical, not electrical lol. Quite simply, here is my confusion. Take the first data point, 0.1 volts, 0.02 milliamps. Then according to V=IR, the resistance should be 5000 ohms. But obviously the resistance was 31 kohms, so ... what is the concept I'm failing to comprehend?
     
  12. Apr 30, 2017 #11
    The equipment you're using to measure this isn't perfect. Typically, the lower the value of something you're measuring, the more inaccurate your equipment gets. That's only 20 microamps!

    Try redoing your experiment with lower values of resistors. Aim for an effective resistance of a couple hundred ohm or something like it. Make sure you don't go too low and fry your resistors, although this is a rite of passage that you'll inevitably go through at some point.
     
  13. Apr 30, 2017 #12
    Okay thanks, I will retry the experiment with lower resistors and higher current. The thing is though, I think what I'm describing is an attribute of Linear Resistors. Ohms law is just a description of linear resistors, right? V=IR, this can be looked at in the basic linear form y=mx+b, where b=0 and m=R. Then y=Rx. We both graphed voltage on the x-axis, and current on the y-axis, so the equation of our line is actually x = y*(1/R), or I=V*(1/R). So no matter what value resistor I measure, I should always get the linear VI relationship, right? Because that's why we call them linear resistors.

    Is it a property of linear resistors to have a very small resistance at low voltages, but at high enough voltages, they approach their rated resistance value? That's the question that's bugging me. I've used the fitted equation that we both got from the data to extrapolate the current out to about 100 volts, and I've attached the resulting graph. As you can see, resistance starts at zero, but quickly climbs to the rated value. Thoughts on this?
     

    Attached Files:

  14. Apr 30, 2017 #13
    No, there are significant nonlinear effects that depend on temperature and (high) voltage, but none you should be worried about here. You're probably just seeing the effects of measuring a very low level of current with a piece of equipment that's not designed for measuring that range accurately. If you had such a piece of equipment, I'd wager you'd get the results you initially expected.

    There doesn't actually seem to be a datasheet with metering accuracies listed available for the kit you're using, which is a bit disappointing. I guess they expect you to go through the schematics and components yourself, or maybe I'm just not looking in the right place.

    Edit: I see there's a voltmeter function in the kit. How are you measuring current?
     
  15. Apr 30, 2017 #14
    With a fluke (on ammeter setting) in series with the resistors. I'll build a circuit with about 500mA of current, that should be more than enough to minimize any errors that the instrument gives. Will keep you posted.
     
  16. Apr 30, 2017 #15
    That's a lot of current. Make sure you calculate the power you'll be dissipating in each resistor before you turn anything on (they have a maximum power rating).

    What Fluke model would that be and what range are you using it on when measuring the 20 microamps?
     
  17. Apr 30, 2017 #16
    Yeah good point, I'll recalculate it for close to the max power that the resistors can handle. Don't want to fry anything. I'll have to get back to you on the Fluke model, its a good one though, part of the campus equipment.
     
  18. May 1, 2017 #17

    sophiecentaur

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    You should be aware that the breadboard system was never intended to be a 'measurement system'. It's a way of testing circuit designs in which the range of values of circuit components are assumed to swamp any of the incidental (parasitic) components. You can't tell what sort of contacts that your components / wires are making. (We still need to see what your actual connections and layout consisted of.) Half of your measuring system is hidden underneath the plastic cover. This is the sort of experiment that would probably best be carried out with wires and good-ol' crocodile clips. I have had classes of A level students, 'verifying' that Ohm's Law applies to out-of-the-drawer resistors (using that crude equipment). They were getting 'good' straight line graphs over a range of 10:1 supply volts. So it's quite possible to do a lot better than your results suggest. Doing it with visible connections would give you a chance to verity that there are no unexpected voltage drops anywhere, due to bad contacts. Fault finding can be a very satisfying exercise and can really help with gaining a deeper level of understanding.
     
  19. May 1, 2017 #18

    rbelli1

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    Were you using the mA or uA setting on the meter?

    This Fluke 87 meter http://support.fluke.com/find-sales/Download/Asset/2161164_6116_ENG_B_W.PDF which is fairly common as far as flukes go can't measure many of the values you are trying to measure on the mA setting. The first one is for example will have an accuracy of +-0.2% +-0.04mA. This is more than double the value you measured. The last digit has nearly no meaning except for comparative measurements.

    BoB
     
  20. May 1, 2017 #19
    The fluke has an automatic current range selector, there is no way to select milliamps or microamps... This was a class experiment that was checked by staff. I don't know the exact model of the fluke, but I will let you know as soon as I have access to it again.
     
  21. May 1, 2017 #20

    jim hardy

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    That's a pet peeve of mine. Beginners should start out with simple analog instruments.

    This will measure 20 microamps with ease. With no ambiguity as to what range it's on.

    TFlipflop2.jpg

    old jim
     
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