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Homework Help: Time Constant ≠ Time constant from experiment

  1. Dec 9, 2015 #1
    1. The problem statement, all variables and given/known data
    In a lab experiment, we wanted to compare the time constant in a simple RC circuit by comparing the value of Resistance * Capacitance that we measure directly and the value of RC that we get from the procedure. The procedure entailed using a wave generator and an oscilloscope to graph the charging and discharging patterns of the capacitor. We did four trials using different capacitors and resistances, but I will only use one as an example to get the idea of what we did across.
    So in the we had a single capacitor (with capacitance C = 25.2 nF) and two resistors in series (which we calculate to be R = 245.5 kΩ) that was connected in a loop with the wave generator terminals. The wave generator was set to produce square waves and cause the capacitor to charge and discharge constantly. The oscilloscope terminals were attached to the circuit on each side of the capacitor and after some fine adjusts on the range of voltage and time, a nice charging pattern was produced. In this case, the max voltage reached is 10 volts.
    After filming the charging pattern, we plotted some easy to determine points for (t, V) to replicate the curve in Excel. We then graphed the smooth curve that is generated from the theory that the voltage across the capacitor is V = Vo(1 - e^(-t/RC)), where we take Vo to be 10 volts and R and C to be the values that we measured before the experiment began. We were told by our instructor that Vo would be the max values for voltage that your data is heading to, and since a good portion of our graph our data values are nearly at 10V, we take Vo to be 10V.
    Our objective is to compare the time constant that our experiment data is indicating to the theoretical time constant we get from measure R and C directly. The resistance R and capacitance C was measured using an multimeter, where R was measured on the ohmmeter setting and the capacitance was measured by first discharging it and then inserting it into slits in the multimeter (for this specific use) and using the capacitance setting. To determine the time constant indicated by our data, we then changed the capacitance C to change the time constant until the curve mapped to the data points.

    2. Relevant equations
    Voltage over a capacitor is: V = Vo(1 - e^(-t/RC))
    Equivalent resistance for series resistors: Req = R1 + R2
    Time constant (tau): tau = R*C

    3. The attempt at a solution
    The theoretical time constant I calculate is R*C = 245.5 kΩ * 25.2 nF = 6.19 ms. However, according to the theoretical curve the time constant must be close to 4.89 ms. To figure out what might be causing this discrepancy I have looked at if any possible error involved in measuring R or C could make the graph be off so much. To get the theoretical curve to map to the data points I either have to decrease C by 5.3 nF or decrease the resistance R by 52 kΩ. Both of these seem seem like impossible errors especially considering that the multimeter should have been able to read accurately 245.x kΩ and 25.x nF, so I figure that there has to be some other factor that is causing the discrepancy, but I cannot figure out what that factor might be.
    I have been reading that measuring capacitance is not the easiest thing to do, but our meter seemed to read it just fine. However, maybe this is not a very accurate method? Changing Vo does not make the graph any better either and I doubt the resistances being off by 50kΩ. For the three other trials we did (another single capacitor, two capacitors in series, and two capacitors in parallel), we have the same issue. In all four, the theoretical curve is always lower than the curve indicated by the data points, which I assume is the effect of some issue that happened in all four trials. Any help is greatly appreciated.
  2. jcsd
  3. Dec 9, 2015 #2

    Simon Bridge

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    ... you mean the TC by T=RC disagrees with the TC obtained from the curve fitted to the data ...

    Your first thought is that the experiment does not support the relationship T=RC ... but you have to start looking at your assumptions.
    You have checked the assumption that the meter you have accurately reads R and C values ... you can test the meter against known values or check the user-manual that comes with the device to see how good it is expected to be for the kind of value measured. Generally, you can trust the manufacturer to get it right though.

    If you believe your data plots too - then it indicates there is another source of resistance and capacitance besides the physical components contributing to the time constant.
  4. Dec 9, 2015 #3
    I did check the data points (t, V) that we estimated from the curve (we took a slow motion of the pattern) by changing the value for V by an amount that would seem to be the maximum deviation we are off by. It allowed me to adjust the curve somewhat and it did help (I was able to decrease the amount of change that we were applying to C), but it did not solve the issue at hand. As for the known components in the circuit that might contribute to the resistance or capacitance sources, we had to use two separate sections of copper wiring to connect the circuit involving the wave generator, resistors, and capacitors. According to the theoretical curve, if there is a component causing resistance, it needs to decrease the value for R that we use in the time constant RC component by 52 kΩ. If the resistance is in series, the resistance would add, but that is not what we need. For each trial, we need the time constant to be smaller. I wouldn't think that the resistors and copper wire would be acting in parallel, but maybe this happens somehow?
    Besides the wiring, there was the wave generator. Could the wave generator be making the resistance in the circuit less or inadvertently decreasing the capacitance of the capacitor? Something odd I noticed is that for one of the single capacitors (the one I have been talking about this entire time) and for the parallel capacitor trial, if I pretend there was another capacitor in series with capacitance C = 800 nF, the theoretical matches the plotted points, albeit at higher values of C it took larger and larger values to see any noticeable change in the graph. However, for the other single capacitor or series trial it does not (the series trial indicates that the third capacitance would be 6 nF and for the single capacitor it would only take 50 nF). So unless the wave generator is significantly altering this "ghost" capacitor by this much, could it be doing something else? Are the capacitors somehow changing their capacitor when placed in the circuit?
  5. Dec 9, 2015 #4

    Simon Bridge

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    The data value is the value that the theory needs to explain.
    T=RC is giving a high value then the effect of unaccounted for resistances and capacitances in the circuit must be to reduce the effective value.

    The wave generator can be modelled as an AC power source in series with a small resistance ... you'll observe that the peak voltage output depends on the size of the load. Also look up "stray capacitance".
  6. Dec 9, 2015 #5
    So it is possible that resistors are acting like capacitors as well? I haven't been able to find a range for what type of capacitance resistors of 98 kΩ would carry, but would it be safe to assume that whatever capacitance they contribute that it would be in series with the equivalent capacitance of the resistors that we have in the circuit (considering a single capacitor, two capacitors in series, and two in parallel?
    Something to note is that when I consider the ghost capacitances to be in parallel (thus I am simply adding any stray capacitance) to the equivalent resistance, the capacitance value that make the two single capacitor trials good is around -6 nF, the series capacitor trial requires that I add 13 nF, and the parallel capacitor trial requires that the I must add in -9 nF. Now, -6 nF and -9 nF are comparable, but +13 nF does not seem so much and I wouldn't think that the resistors would act as they were in parallel with the capacitors. Also, if the wave generator has a small resistance that acts in series with the other resistors in the circuit, that would increase my time constant I think.
  7. Dec 10, 2015 #6


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    The voltmeter (osscilloscope) and generator can contribute to the capacitance and resistance in the circuit - as you suggest.
    Maybe you can assume a constant value for these in all three sets of data.
    Also maybe you get the specs on these?
    Be sure to use data during the time that the generator voltage is zero (check that there is no DC offset).
    You did adjust the oscilloscope beam to the zero voltage line?
  8. Dec 10, 2015 #7
    Which part of the charging/discharging pattern are you talking about, the beginning of the discharge pattern? What do you mean by DC offset and how do I check it through my data?

    The instructor did not mention any zero voltage line so we only focused on adjusting the height and length of the pattern so that we could have a whole number for the max voltage and get enough data points from the initial half of the charging process. The oscilloscope screen did have a coordinate plane setup with grid lines and I can tell you that we had the zero voltage line set 3 grids below the x-axis, where 1 vertical grid equates to 2 volts. The instructor did not seem to care if they were below the axis, only that you had a good shape.

    In my previous post I list the results of trying to find a constant value in each. It would help if I knew for sure what type of relationship the wave generator has to the capacitors, would it be considered in series or parallel? I attach picture below.

    Okay, so I had a thought about the issue in finding a constant capacitance that is contributed in some way. In each trial, we had a different Vo such that we are using Vo = 12V, 10V, 8V, and 10V for the single, single, series, and parallel capacitors. Now, if the wave generator and capacitors are in series, the charge Q over each each capacitor (the generator and then whatever config of C we are using) is the same, but the voltage over the generator does not equal to the voltage over the capacitor because there is the resistor in the circuit, right? Thus, since C = Q/V, the capacitance wouldn't be the same either when we change the voltage max of each trial, since that implies that the wave generator also has an increased voltage. Is this reasonable and can someone walk me through how we might account for this when we try to reason why the ghost capacitance changes?

    Also, a big note on this lab is that the instructor gave us the purpose, "To investigate the behavior of capacitors in a DC circuit." I guess the instructor does not believe there should be any sort AC power supply present?

  9. Dec 11, 2015 #8


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    It does not matter the cpacitor discharges as soon as the supply voltages drops to zero.
    The square wave generator might have a setting where a DC offset is added to the signal, but I would think that your instructor would have mentioned it , otherwise he is just plain mean (you also might have missed his mentioning it). This offset can lift the square wave up or pull it down by a constant voltage, that is like adding aconstant positive or negative voltage to the whole of the square wave signal.

    Ok that takes care of that possibility of messing things up then. As long as you were aware where the zero voltage line is located on the screen. This might also have made you aware if an DC offset was added to the signal - that is you would have noticed it, the signal not coming back to the zero voltage level on the oscilloscope display.

    No idea on that one, just guessing - maybe an additional resistor and capacitor in series to the RC circuit. As I say the generator specs might clear that up. I was thinking maybe you could allow for such in the maths or analysis of the data.

    You are dealing with DC since the current flows in only one direction:smile:
    It might be a good idea to make V/Vo (dimensionless!) your dependent variable.
    Last edited: Dec 11, 2015
  10. Dec 11, 2015 #9
    Oh, he did state that we had to adjust the wave generator and oscilloscope so that the pattern would not be moving up and down, I guess he just didn't say what caused it.

    Okay, interestingly enough all of our data agrees with the theoretical curves if we simply subtract 6 nF from the equivalent capacitance's in the time constant. This in turn lowers the time constant and raises the theoretical curve up to the plotted data. Out of all the combinations of things I have tried on changing the time constant, this seems to not be such a coincidence to me that - 6 nF makes all trials work well. I am inclined to think that whatever is causing the discrepancy, it causes that the experimental capacitance be 6 nF lower for each configuration of capacitors, no matter whether we first use the parallel series formula to find Ceq from C1 and C2 or the series formula to find Ceq.
    Noting that we are using subtraction for the capacitance, this would mean that the ghost capacitance would be in parallel with the configuration of capacitors. Looking at our circuit, the oscilloscope actually looks to be in parallel with the capacitor. I was also reading that the inputs of the oscilloscope cause there to something called input capacitance, which interacts in parallel with Ceq. At first I thought input capacitance explained it because all we needed to do was subtract nF from each Ceq, but then I realized that if there is more capacitance in the experimental circuit, we should be adding 6 nF to the theoretical curve.
    Nonetheless, it makes me think that it has to be the oscilloscope that is doing something to the capacitance in the circuit such that it acts in parallel with Ceq and decreases it. As far as I know, only geometry affects the capacitance of capacitors and I can't think of why some capacitor in the circuit would actually decrease the capacitance of the other capacitors. I am hoping that someone knows what could cause this, as it seems to me that this is the cause of the discrepancy because -6 nF made all trials look good even though we switched oscilloscopes for two of the trials and it would make sense that some component (capacitor, inductor, comparator . . . I have no clue now after I have read so much online) would causes there to be - 6 nF.
  11. Dec 12, 2015 #10


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    Aaah, yes the equation you mention is for charging up, so its the graph on the left, the hump.

    As far as I know a capacitance is added by the oscilloscope to eliminate DC voltages. Usually via a selection switch on the oscilloscope marked AC/DC. That being said I seem to remember the probe itself adds an capacitance? As for the adjustment "so that the signal or graph does not go up or down" I think that might have been the trigger adjustment of the oscilloscope.
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