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enter image description here

The circuit above was given and I am supposed to measure the voltage across the last capacitor. Without C1 initially, knowing that this is a capacitive divider, I am expecting 1/10 of 5V, i.e. 0.5V to be measured.

To my surprise, the measured value is 0.0521V.

The digital multimeter used is a Fluke 179 and I suspect that the discrepancy might have something to do with it but I was not able to find any reasonable explanation.

Can anyone explain what happened here?

Measuring the voltage across the first capacitor gives 450.2 mV, which is somewhat close to 0.5V but I couldn't explain why there's a 0.05V discrepancy.

Also interesting is that I observed the voltage across the 1M resistor is 4.58V, which was not expected. Are there any explanations for the two phenomena here?

Edit

Say a capacitor C1 is added across the circuit, will it help to solve the issue of the measured capacitor discharge to the DMM and allow me to measure 0.5V?

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  • \$\begingroup\$ "Say a capacitor C1 is added across the circuit[...]" No. The new C1 capacitor would just quickly charge to 5V and remain at that voltage while the source is connected. \$\endgroup\$
    – devnull
    Commented May 19, 2021 at 17:11
  • \$\begingroup\$ @vangelo so there is no way/changes I can make on the circuit to make me measure 0.5V across the capacitor? \$\endgroup\$
    – cZe99
    Commented May 19, 2021 at 17:15
  • \$\begingroup\$ It depends on what is "changeable" in the circuit. Different voltage source? Other capacitors? Using an opamp with FET inputs to buffer the multimeter? A suggestion would be to open another question (you may reference this one for context) stating what you want to achieve (e.g. proving that the DC voltage is split among the capacitors) and others may offer quite different approaches. \$\endgroup\$
    – devnull
    Commented May 19, 2021 at 20:33
  • \$\begingroup\$ Check the meter, it adds a 1nf plus the leads etc. I did not find its current loading (resistance) which will also change your readings. Just saying this sounds like a trick question. \$\endgroup\$
    – Gil
    Commented May 20, 2021 at 3:26

2 Answers 2

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Assuming:

  • identical capacitor values
  • all capacitor initially discharged
  • leakage resistance much higher than 1 M \$\Omega\$

yes, after the transient (connection of the 5V DC source) and without the multimeter, the voltage should reach the expected value.

But, as soon as you connect the multimeter, it discharges that lower capacitor (5 times RC = 50 ms, assuming 10 M\$\Omega\$ input resistance).

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  • \$\begingroup\$ Thank you for the answer. Will it be possible to "recharge" the capacitor by connecting a capacitor parallel to the 10 capacitors? So that I will measure 0.5V for the last capacitor. Or such discharge is inevitable? And could you give a possible explanation for the measured voltage drop across the resistor as well? \$\endgroup\$
    – cZe99
    Commented May 19, 2021 at 16:41
  • \$\begingroup\$ I'm not sure I fully understood this question. If the following is wrong interpretation on my part, please edit your question to add a diagram with your new suggestion: if you connect a resistor across a capacitor, in the absence of DC current from other parts of the circuit, that specific capacitor will discharge across this resistor (the other 9 capacitors were not involved in this discharge process). \$\endgroup\$
    – devnull
    Commented May 19, 2021 at 16:52
  • \$\begingroup\$ I have edited my question. \$\endgroup\$
    – cZe99
    Commented May 19, 2021 at 17:03
  • \$\begingroup\$ "And could you give a possible explanation for the measured voltage drop across the resistor as well?" You could repeat the experiment (with initially discharged capacitors and without measuring their voltage after applying the voltage) and different resistor values for and estimation on the leakage current. \$\endgroup\$
    – devnull
    Commented May 19, 2021 at 17:09
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Caps transformers or voltage dividers are insulators with high leakage resistance, often higher than a DMM.

So this must be done with AC.

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