# Physically measuring the voltage of a capacitor at a series capacitor circuit

#### Doros

Joined Dec 17, 2013
137
Dear all hello,

Is it possible to measure the voltage across a capacitor physically on a circuit, probably using an electrometer or a very high impedance voltmeter?
For example having the circuit attached is a way to measure the voltage across the capacitor C4? It should read 4 volts.

Thank you very much for your support

#### Dodgydave

Joined Jun 22, 2012
9,855
The throry says yes, but on practice the capacitors will have internal resistance and leakage, and so may give a different voltage.

#### Doros

Joined Dec 17, 2013
137
The throry says yes, but on practice the capacitors will have internal resistance and leakage, and so may give a different voltage.
Thank you Dodgydave for the response.

What you mean is that if we wanted physically to measure the voltage across C4, instead of 4V that we should measure, we would see another value, i.e 3.34V due to the reasons you mentioned above?

thanks again

#### Jony130

Joined Feb 17, 2009
5,211
All voltmeter will have some internal resistance, typical 10MΩ but the cheap one will have 1MΩ. So the capacitor will finally be discharged.
So to be able to measure a "steady" voltage across the capacitor in your circuit you need to replace DC voltage source with a AC voltage source.

#### Doros

Joined Dec 17, 2013
137
Thanks Jony130, I could do that, and make sense

So even with an electrometer, with very high imedance, I would see the measurement result slowly (depending on the impedance) but sreadily diminishing

#### Jony130

Joined Feb 17, 2009
5,211
Time constant is the key here T = R*C and what "slowly" means to you.

#### Doros

Joined Dec 17, 2013
137
I understand and thanks’ Jony130. "Slowly" is a matter of speaking. Depending on the parameters you mention above you see the discharging or charging of the cap at a quicker or slower rate

#### MrAl

Joined Jun 17, 2014
7,888
Hello,

With all capacitors exactly equal we get this:
Vc=e^(-(2*t)/(3*C1*R3))/3

and that is a decreasing exponential. It will discharge to within 1 percent of original value in the time 5*C1*R3.
R3 is the parallel resistance.
With all caps 5uf and the parallel resistance (meter) of 10megohms it should take about 250 seconds to discharge to within 1 percent.
The initial discharge rate will be about 6.6mv per second, so if you do the measurement fast it might work, depending on the leakage of the caps which will also upset the balance.

But because of the leakage, the voltage may stabilize too. For example with that one cap leakage low and the others around 10 megohms (equivalent resistance) the voltage would stay at 1/3 of Vcc.

To do this right, you would connect three meters, one across each capacitor, all the same. The capacitors must also be of the same exact value, and the meters all exactly the same, so there will be some difference in real life.

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#### Doros

Joined Dec 17, 2013
137
Thank you all,

I will do it on the workbench and share my results

#### ErnieM

Joined Apr 24, 2011
8,082
Why is this measurement importaint to you? Is this a learning exercise or is there a Practably use for this?

#### Doros

Joined Dec 17, 2013
137
Actually I am experimenting with discharging capacitors (plates one) through ionization, and I needed to be somehow certain of the charges kept on them. To know them since capacitance is known I need to be certain about the voltage of them

#### ErnieM

Joined Apr 24, 2011
8,082
If you are investigating the discharge due to ionization why are you leaving them connected to a charging source? The caps will be recharged as they are discharged.

Once you open the voltage source the need for three caps in series comes into question. Why series? They all see different conditions due to tolerance in the other caps they are in series with. Why not three (or any number) caps in parallel so each sees the same conditions as the others.

Then one may be exposed as the other are left unexposed.

#### DickCappels

Joined Aug 21, 2008
6,796
If your voltages are high enough and stable with respect to groung you can probably measure the voltages on the plates with respect to ground directly without discharging the capacitor by using an electric field mill.

http://a-tech.net/ElectricFieldMill/

#### Doros

Joined Dec 17, 2013
137
Thank you ErnieM with your valuable input.

Actually I should put the actual set-up, where I need a voltage drop in order to measure lower voltages. The set up includes high voltage for charging and then lower voltages in order to measure them.
I would like to see calculations in paper be valid in real conditions.
I will come back when I will develop further the set up.

#### Doros

Joined Dec 17, 2013
137
That's what I am doing now DickCappels. I would like to replace it with contact measurement through an electrometer

#### Doros

Joined Dec 17, 2013
137
For example, what is the voltage with respect to ground at point A? (schematic attached)

#### Jony130

Joined Feb 17, 2009
5,211
The equation that will describe the voltage at point A with respect to ground (voltage across the capacitor) looks something like this:

$$V(t) = Vo\left ( 1-e^{\frac{-t}{RC}} \right )$$

$$Vo = 20000V$$
And
$$R*C = 10M\Omega* 1\mu F = 10s$$

#### DickCappels

Joined Aug 21, 2008
6,796
If you wait long enough the voltage from point A to ground will approach 20KV. Depends on how long you want to wait.

By the way, since you are working at 20 kv you might be able to use electrostatic voltmeters which would not load down the capacitor except for some small unavoidable leakage.