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If there is a power supply charging a capacitor (e.g. 4 μF) through a resistor (3.2 MΩ), the time constant can be calculated with the capacitance and resistance.

If I, however, measured the voltage across the capacitor, then that voltmeter in parallel adds a loading effect. How significant is this to the time constant assuming the internal resistance has the same order of magnitude as the resistor?

enter image description here

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  • \$\begingroup\$ An ideal voltmeter does not "load" the circuit. If it is not ideal, then the equivalent model should be looked at (it depends on the voltmeter of course). \$\endgroup\$
    – Eugene Sh.
    Feb 8, 2022 at 16:39
  • \$\begingroup\$ A very warm welcome to the site. Please note that it's not a free design house, homework-answering service or online personal tutorial. People will help you take the next step if your question shows you've already done as much as you possibly could - which yours doesn't, I'm afraid. Please edit your question and greatly improve it. Show all your own work and own findings so far in considerable detail with any schematic. The schematic tool here is easy to use. The better the quality of your question, the better the quality of the answers it will attract. Again, welcome. \$\endgroup\$
    – TonyM
    Feb 8, 2022 at 16:40
  • \$\begingroup\$ This does read like a homework question with no attempt made... \$\endgroup\$
    – TonyM
    Feb 8, 2022 at 16:40

2 Answers 2

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If the voltmeter's internal resistance is comparable to the 3.2 MOhm, then the loading effect is important and worth considering. We can form a Thévenin equivalent for the circuit attached to the capacitor (i.e. the voltage source, its series impedance, and the loading impedance of the voltmeter):

schematic

simulate this circuit – Schematic created using CircuitLab

where Vth and Rth are a consequence of both resistances. We can show that:

$$R_{th} = \left(\frac{1}{3.2\,[\text{M}\Omega]} + \frac{1}{R_{\text{voltmeter}}}\right)^{-1} $$

and

$$V_{th} = \frac{R_{\text{voltmeter}}}{3.2\,[\text{M}\Omega] + R_{\text{voltmeter}}}$$

You might recognize the second equation as being a simple voltage divider. However, it's worth noting that the capacitor charges to a lower voltage, but its time constant is shorter since it sees a smaller effective impedance across its terminals.

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  • \$\begingroup\$ I am not sure this is the right way to make the equivalent. In the original circuit with non-ideal voltmeter there will always be current from the source, even with the capacitor fully charged. In your circuit there won't be any current in the steady state. Maybe I missing something though, as it looks correct otherwise... \$\endgroup\$
    – Eugene Sh.
    Feb 8, 2022 at 16:44
  • \$\begingroup\$ @EugeneSh. The current from the source flows to the voltmeter only. If we form a Thévenin equivalent of the voltage source, resistor, and voltmeter (i.e. everything but the capacitor), then that current is internal to the network being transformed, so we don't expect it to necessarily exist after the transform is complete. \$\endgroup\$
    – nanofarad
    Feb 8, 2022 at 16:51
  • \$\begingroup\$ UPD: I guess it is because the equivalent as it is "seen" by the capacitor, and it does not care about the current in the "original" circuit as long as there is no current through itself. \$\endgroup\$
    – Eugene Sh.
    Feb 8, 2022 at 16:51
  • \$\begingroup\$ Right, just figured it out. I should brush off the rust from this stuff.. \$\endgroup\$
    – Eugene Sh.
    Feb 8, 2022 at 16:52
  • \$\begingroup\$ I edited the sentence that sets up the transform in order to clarify what's in the linear network being collapsed to Vth+Rth \$\endgroup\$
    – nanofarad
    Feb 8, 2022 at 16:53
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If you assume that the multimeter is a cheap entry level model, and it has same 3.2 Mohm load impedance as the source impedance, then the multimeter will eventually discharge the capacitor to equilibrium and it will read exactly 5V which half of the source voltage.

If the capacitor is assumed to be charged to 10V before connecting the multimeter, then the RC time constant of the new circuit is 1.6 Mohm * 4 uF is 6.4 seconds. It takes approximately 5 time constants, or 32 seconds, to discharge the capacitor to within 1% of the final result, and of course longer to discharge it to within 0.1% of the final result.

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