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enter image description here

In the shown circuit, I was measuring how the voltage on capacitor $V_C$ changes with time. The theory says that $V_C(t) = V_T(1- e^{-t/R_1C})$, where $V_T$ is voltage on battery. If I plot my experimantally obtained measurments on graph $U_C(t)$, I get the following diagram: enter image description here We can see that the voltage on capacitor over time gets to $V_{C_{max}} = 3.1 V$. And the curve looks theoretically corrent - no problems yet. But our goal is to get the time of relaxation $T = RC$. We can obtain it on two different ways:

(1) We can obtain it graphically: $\log(1- \frac{V_C(t)}{V_T}) = (-\frac{1}{T}) t$

(2) Or we can obtain it theoretically - calculation from formula: $T = R_1C$

From (2) we get $T = R_1C = 2.7 MΩ \times 1000pF = 0.0027s$, But from graph (1) ... enter image description here

We have to remove the minus form slope and multiply it with 1000, beacuse it is in milisecond. It follows that $1/T = 1129 \frac{1}{s}$ and $T = 0.000886 s$.

Our two values for relaxation time do not match, so i figured that $R_1$ is quite big compared to volmeters resistance $R_2$ which we have to use in our further calculations. Our new circuit is... enter image description here

From formula $V_{C_{max}} = V_T \frac{R_1}{R_1 +R_2}$ we can calculate the resistance of volmeter $R_2$ if we substitute $V_T = 12V, V_{C_{max}} = 3.1 V$ and $R_1 = 2.7 MΩ$ we can calculate $R_2 = 7.75 MΩ$.

So if we put this in formula for the new circuit Charging capacitor, with series and parallel resisters found on this link, we can calculate a relaxation time of this new circuit as a $T = \frac{R_1R_2C}{R_1+R_2}$ and we get $T = 2.00 s$ WHICH DOES NOT MATCH WITH THE EXPERIMENTALY OBTAINED VALUE.

My questin: Why dos not tehroy match with the experimental value? Is my theory wrong? If it is, why? Is it possible that I just have the wrong value for the capacitance of my Capacitor?

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  • $\begingroup$ In the line: "From (2) we get T=R1C=2.7MΩ×1000pF=2.7s ..." I think that is a mistake in units, Mega-Ohm = 10^6 Ohm and pF = 10^{-12} F, so your theory gives a time T=0.0027 s, however is still far from the experimental value. Could you please add your data to your question to do some tests? $\endgroup$ – user239504 May 21 at 18:32
  • $\begingroup$ The voltage at the meter should be proportional to the resistance of the meter. $\endgroup$ – R.W. Bird May 21 at 18:41
  • $\begingroup$ @R.W.Bird what do you mean? $\endgroup$ – Edward Henry Brenner May 21 at 18:46
  • $\begingroup$ Am I missing something here? Theoretically after a long time the current falls to zero and the voltage across the capacitor equals the battery emf. Your graph shows it leveling of way less than that. $\endgroup$ – Bob D May 21 at 19:15
  • $\begingroup$ After the capacitor is charged 3.1 = 12 [${R_2}/(R_ 1+ R_2)$] $\endgroup$ – R.W. Bird May 21 at 19:46
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There are several errors in your analysis:

  1. The equivalent circuit of a voltmeter is not the one you represented: a real voltmeter should be represented as an ideal voltmeter, with infinite input resistance, in parallel with the voltmeter's input resistance ($R_2$ in your case), not in series.

  2. The correct value of $V_{C_\mathrm{max}}$ is $$V_{C_\mathrm{max}} = V_T \frac{R_2}{R_1 +R_2}$$ from which $R_2\approx 940\,\mathrm{k}\Omega$ and $T\approx 0.697\,\mathrm{ms}$. If you then take into account the tolerances of the capacitor and of the resistor, and the measurement uncertainty, you will find that this value is compatible with your measurement result.

  3. Finally, please do not call $T$ relaxation time: the more standard name for that parameter is time constant.

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  • $\begingroup$ 940 kOhms is a suspiciously crappy impedance for a multimeter (which I assume the OP has used instead of some weird and heavy instrument from the 1960's). Secondly, there is nothing particularly wrong with calling the time constant "relaxation time". See en.wikipedia.org/wiki/Relaxation_(physics) $\endgroup$ – oliver May 21 at 19:01
  • $\begingroup$ @oliver 1 MΩ is the typical input impedance of an oscilloscope and of many data acquisition boards, so 940 kΩ doesn't come as a surprise. Note that if for "weird and heavy instrument from the 1960's" you mean an electromechanical instrument, it wouldn't have enough speed to respond to such a transitory response. $\endgroup$ – Massimo Ortolano May 21 at 19:06
  • $\begingroup$ Well... that's true :-) I missed that the time axis is milliseconds, so an oscilloscope seems reasonable. $\endgroup$ – oliver May 21 at 19:08

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