Does this even make sense? Measuring voltage of a capacitor

[flyingpig]

Homework Statement

This is a lab problem and basically I am supposed to explore the properties of a capacitor.

Charge capacitor C₁ up to some voltage V1 (less than 5V) and disconnect the capacitor from the supply. Measure the voltage with the bench voltmeter and record.

C₁ = 3300uF

V₁ = 4.50V

Supply Voltage = 4.50V

Here is the question that confused me

Is the voltage reading steady? Why might you expect this voltage to vary?

My DC supply 4.50V was the same as I measured it. But I forgot what I was measuring. Does it make sense if I clamp my capacitor with the voltmeter and measure the voltage of a capacitor? I am thinking it doesn't because I have a capacitance, not voltage.

If so what was I measuring? I don't think it would make sense to measure the DC Supply with a voltmeter...

Why would the reading not be steady?

 

[SammyS]

There is voltage across the capacitor, so YES it makes sense to measure the voltage of (across) the capacitor. A voltmeter does not measure capacitance.

 

[berkeman]

Homework Statement

This is a lab problem and basically I am supposed to explore the properties of a capacitor.



Here is the question that confused me

Is the voltage reading steady? Why might you expect this voltage to vary?

My DC supply 4.50V was the same as I measured it. But I forgot what I was measuring. Does it make sense if I clamp my capacitor with the voltmeter and measure the voltage of a capacitor? I am thinking it doesn't because I have a capacitance, not voltage.

If so what was I measuring? I don't think it would make sense to measure the DC Supply with a voltmeter...

Why would the reading not be steady?

What is the fundamental difference between an "ideal" and a "real" voltmeter?

Pull up the datasheet for your lab DVM. Which specification do you think might apply to this part of the lab? And based on that spec, how long do you think you should watch the clamped DVM before you see a voltage variation?

 

[flyingpig]

An ideal voltmeter is one that does not have resistance. A real one has. Oh dear...that means my data is wrong

 

[berkeman]

An ideal voltmeter is one that does not have resistance. A real one has. Oh dear...that means my data is wrong

What does your data look like?

 

[flyingpig]

Supply Voltage = 4.50V

V₁ = 4.50V

 

[SammyS]

What happened when/if you left your voltmeter connected to the capacitor for an extended period of time? (without the supply)

 

[flyingpig]

It dropped really quickly! I remember now! We were supposed to record that instantaneous voltage! So the data is right...?

 

[SammyS]

It dropped really quickly! I remember now! We were supposed to record that instantaneous voltage! So the data is right...?

No way I can say for sure! LOL!

However, the behavior you describe, is what I would expect.

 

[flyingpig]

I am just writing out the procedure. lol. Hopefully I got it right...

 

[flyingpig]

I have a followup question for myself. Okay because the real voltmeter dropped really quickly, I dn't understand.

Does that mean current stored in the capacitor flow flew the voltmeter? I thought the voltage would be less, but not dropped to 0.

 

[Dick]

No offense, but were you really supposed to connect a capacitor to a voltmeter with no theory background on what you might expect or what the meter was actually measuring? If so, being baffled is understandable.

 

[flyingpig]

No offense, but were you really supposed to connect a capacitor to a voltmeter with no theory background on what you might expect or what the meter was actually measuring? If so, being baffled is understandable.

Yes and no...

All of our labs are a few topics ahead of what we are learning in class. So the first labs was about connecting wires in circuits while we were introduced to charges and electric field.

Yeah it's messed up, but the Physics Lab Manager has made it so marking is lenient (can't say the same for my chemistry lab manager...)

 

[Dick]

Yes and no...

All of our labs are a few topics ahead of what we are learning in class. So the first labs was about connecting wires in circuits while we were introduced to charges and electric field.

Yeah it's messed up, but the Physics Lab Manager has made it so marking is lenient (can't say the same for my chemistry lab manager...)

Ok, but no idea what a capacitor does when it's connected to a voltage source?

 

[flyingpig]

It charges up the capacitor.

 

[Dick]

There is a potential difference across it.

True. But that has no information content beyond the initial question. So you had no idea what to expect? Ok, then just report your observations. That should work. What was the initial voltage level you found after you removed the power supply? Roughly how long did it take to fall to half the initial value? I think that's the best they can expect.

 

[Dick]

It charges up the capacitor.

If you don't know anything about what to expect then just follow my previous advice. Report what you observed. With hopefully a number or two in it. If you do then tell me why what you observed makes sense and why you might expect it.

 

[flyingpig]

True. But that has no information content beyond the initial question. So you had no idea what to expect? Ok, then just report your observations. That should work. What was the initial voltage level you found after you removed the power supply? Roughly how long did it take to fall to half the initial value? I think that's the best they can expect.

it was right at 4.50V, the voltage that I charged, but it dropped almost instantaneously. But they are asking why might I expect it to be not steady? I don't think saying the internal resistance of the voltmeter is an answer. Resistance doesn't make the voltage to 0, it just reduces it right?

 

[Dick]

it was right at 4.50V, the voltage that I charged, but it dropped almost instantaneously. But they are asking why might I expect it to be not steady? I don't think saying the internal resistance of the voltmeter is an answer. Resistance doesn't make the voltage to 0, it just reduces it right?

If the voltmeter didn't allow charge to flow through it (if its resistance were infinity) the capacitor would remain charged forever and the voltage would always be 4.50V. But the voltmeter is a real world instrument. It discharged the capacitor. How's that?

 

[flyingpig]

But I thought you could only discharge a capacitor by joining the leads, does the voltmeter suck out all the voltage from the capacitor? When we are hooking up the voltmeter with the capacitor, what is exactly measuring the capacitor's voltage?

 

[Dick]

But I thought you could only discharge a capacitor by joining the leads, does the voltmeter suck out all the voltage from the capacitor? When we are hooking up the voltmeter with the capacitor, what is exactly measuring the capacitor's voltage?

The voltmeter is measuring the voltage. But every instrument affects the thing it's measuring. It detects the voltage by sucking some (hopefully only a little bit) of current out of the thing it's measuring. That's why Berkeman suggested checking the specs on the measuring device. You are wrong to think connecting a voltmeter to the capacitor isn't to some degree shorting it out.

 

[flyingpig]

Does that mean say when an ammeter measures current, it sucks out current too? But how does a voltmeter measure voltage by sucking out current? Does sucking out voltage make sense?

 

[SammyS]

The analogy would be closer to: An ammeter drops a little voltage across it, thus reducing the current which would pass through the circuit, if the ammeter were not there.

An ideal voltmeter would have infinite resistance.

An ideal ammeter would have zero resistance.

 

[flyingpig]

The confusion for me is that does that mean the current from the capacitor the voltmeter suck out has some kind of a special message that tells the voltmeter "hey I was charged with x Volts"

In other words, does current have some other magic properties?

 

[ileacus]

An ideal voltmeter "sucks out" zero current. Its infinite resistance would prevent any current from flowing through it, and it would measure an exact voltage difference (potential) between its two leads. Because a real world voltmeter cannot have infinite resistance, some of the current "leaks" across it, and the measured voltage will drop without some other current source (e.g., a battery) attached to the circuit.

So, if you have charged a capacitor, then disconnected it from the current source, and then measure its voltage with a real voltmeter, the capacitor will "slowly" lose some of it charge - the capacitor in effect becomes the only current source in the circuit. Were the voltmeter ideal, it would prevent any current flow, and simply measure the potential difference between the two leads of the capacitor.

Hope that helps.
[edit]
Perhaps an analogy will help. Consider measuring the gravitational potential energy of a rock on the edge of a cliff of a certain height (analogue of electric potential energy, or voltage) . The "ideal" potential energy measuring device just measures the height of the cliff, without moving the rock. A "real world" device would allow the rock to drop ever so slightly to make its measurement, and if it were kept there, the rock would continue to fall until its potential energy were 0, i.e., it hit the ground. Obviously, a real world device for measuring gravitational potential energy doesn't have the same limitations as a voltmeter - a measuring stick tells you exactly the height of the rock without moving it. In the world of electrics, the device has to touch the rock, providing a ramp for it to roll down. The current that "rolls down" through the voltmeter isn't special any more than the motion of the rock in the hypothetical gravitational potential energy measuring device. In fact, it's unwanted and ultimately results in all of the potential energy being expended - as kinetic energy (down the ramp) for the rock and as current (through the voltmeter) for the capacitor.

 

[flyingpig]

But if a voltmeter has infinite resistance, then how can it even read in the first place? No current can go through, no "message" is being send from the capacitor to the voltmeter.

 

[berkeman]

But if a voltmeter has infinite resistance, then how can it even read in the first place? No current can go through, no "message" is being send from the capacitor to the voltmeter.

You are basically correct. For a real voltmeter to operate, you at least have to charge up the gate of the input MOSFET stage. That takes a small current to charge it up, assuming the gate capacitance and interconnect parasitic capacitance is in the low pF range.

Just think of the "ideal" voltmeter having a super-tiny input MOSFET stage, so the input current to charge up that capacitance is negligible, and the MOSFET gate leakage current is also negligible...

 

[flyingpig]

No our voltmeter is like a "clock", it isn't digital. You have three inputs. It kinda looks like this

 

[berkeman]

Yeah, that's an analog voltmeter, and it probably has a pretty low input resistance (1MegOhm maybe? or lower?). No wonder the capacitor discharged relatively quickly.

 

[flyingpig]

But does that voltmeter have this thing called "MOSFET"? It doesn't feel natural to include the word "MOSFET" in my answer.