Why Did My Capacitance Measurement Exceed Expectations?

In summary, the conversation was about a lab on charges, capacitors, and voltages. The experiment used a 2200uF capacitor and a 10kΩ resistor connected in series with a DC supply. The computer collected data on the voltage across the resistor and the voltage across the capacitor, using Ohm's Law and integration to find the charge on the capacitor. The resulting graph was Charge vs Voltage, with a slope representing the capacitance of the capacitor. The question was raised about why the determined capacitance was slightly higher than the specified value, leading to a discussion about tolerances and potential sources of error in the experiment. The conversation ended with a final question about whether a lower tolerance level would have improved the data.
  • #1
flyingpig
2,579
1

Homework Statement




Basically we did a lab on charges, capacitors and voltages. We used a 2200uF capacitor and we hooked it up with a 10kΩ Resistor connected in series with a DC Supply. The DC Supply was turned down to 0V and all the other senors were "zeroed" out.

Simultaneously, when we clicked on our computer data collecting device we turned the DC Supply on to raise the voltage to 1.0V steadily. The voltage had to keep going up steadily so we could get a straight line, y = mx + b, where m is the slope of the graph. The graph was Charge vs Voltage.

At the end, we did a line that best fits curve and our slope was 0.002262, which translates to 2262uF

Now my question, how come I got something big...?
 
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  • #2


What components was the computer actually monitoring?
 
  • #3


The voltage and resistance. The "current" data was calculated using it. Then it was graphed against capacitance.
 
  • #4


Was the voltage that of the power supply, or was it the voltage across the capacitor?

You said the graph was Charge vs. Voltage, which makes sense. current vs capacitance makes no sense, since the capacitance is a constant.
 
  • #5


I think it is voltage across the capactor. Basically we had to hold the voltage "constant" by steadily increasing it (by changing the power supply's voltage), if we lose the rhythm we basically would screw up our own graph (we want a voltage vs time graph to be linear)
 
  • #6


No I meant Charge vs Potential of the capacitor
 
  • #7


Sammy stay with me...my TA has given up on me
 
  • #8


In order to explain experimental results, it's important to understand what's being measured and also have a good idea about how those quantities are measured.

So, before trying to answer your basic question, I wanted to get some sense of what all was involved.

Having worked with these systems which use a computer to monitor and sometimes control an experiment, I now think I have a good sense of what you were monitoring.
The computer was monitoring the voltage across the resistor and also was monitoring the voltage across the capacitor. Knowing the resistance of the resistor, the voltage across the resistor could be converted to current via Ohm's Law. Then the current was integrated by the computer to find the charge on the capacitor - no doubt, assuming the capacitor started out in an uncharged state. The computer could then produce a Charge vs Voltage graph for the capacitor. This graph should be linear, the slope being the capacitance of the capacitor.

Why was this set up so the power supply voltage was linear in time? So that current current would flow at a reasonable rate throughout the data collection time - not be too large at the beginning - not tapering off too much over time.​

Now for your basic question, which I think is: Why was the capacitance that was determined by the experiment, 2262μF, larger than the actual capacitance, 2200μF ?
Components like resistors and capacitors are manufactured to a given tolerance. Your capacitance measurement is a little more than 3% higher than the specified capacitance. Not bad! Moreover, unless you measured the resistance of the resistor, that's probably also in error. If its resistance was higher than the 10kΩ assumed in the computer program, the current and thus the charge would have been less than that computed, so that the capacitance might actually be less than 2262μF, perhaps even less than 2200μF.​
 
  • #9


I am a freshman, I am not sure if the word "tolerance" of a capacitor exists in our course...although I can understand what that means.
 
  • #10


flyingpig said:
Sammy stay with me...my TA has given up on me

LOL !

I came close a couple of times, but your persistence is amazing, (probably also annoying to your TA).
 
  • #11


SammyS said:
LOL !

I came close a couple of times, but your persistence is amazing, (probably also annoying to your TA).

=( Don't give up on me...
 
  • #12


Oh darn it...why do people hate me?
 
  • #13


Wait how does the "tolerance" relate to the 62μF?
 
  • #14


It's over for me. Thanks for helping me anyways (and always!) Sammy, I don't know what I'll do without this forum
 
  • #15


flyingpig said:
It's over for me. Thanks for helping me anyways (and always!) Sammy, I don't know what I'll do without this forum

What do you mean.

I'm not giving up on you.

You do seem jump to conclusions.
 
  • #16


Sammy, I've learned more about a capacitors in the last two days than ever lol
 
  • #17


flyingpig said:
Wait how does the "tolerance" relate to the 62μF?

Have you learned to read the code on resistors?

The 4th stripe: gold or silver or none mean ±5% or ±10% or ±20% of the specified value of resistance.
 
  • #18


Yeah I have! But i just haven't even thought about that. Wait does that mean resistance = tolerance?
 
  • #19


Although we only did it once...
 
  • #20


You tested a 2200uF capacitor and found it to be 2262uF?

Get the manufacturer's name and part number off the side of the capacitor. Enter it into a search engine and pull up the data sheet for it. Find the specification called "tolerance" expressed as a +/- percentage. Most capacitors are either 5, 10 or 20 percent.

It would seem your capacitor is ok.

Edit: oops. I made a stale post. Tge tolerance issue is addressed.
 
  • #21


It is a "nichicon" 2200uF

Unfortunatly the results thta came up are beyond me...
 
  • #23


Can I just ask one last question? If the tolerance level were lower, would that have make my data better?
 
  • #24


flyingpig said:
Can I just ask one last question? If the tolerance level were lower, would that have make my data better?
Sorry about the delay in answering.

The only way I can think of to determine whether your data were (The word "data" is actually plural.) good or not --- is to measure the capacitance by some other means. You also would need to measure the resistance of the resistor. I'm pretty sure the computer program assumes the resistance was 1 kΩ. I suspect that if the resistance were 1% lower than specified, then the capacitance you determined would be 1% higher than the true capacitance, etc.

When I checked out your link to the capacitor's manufacturer, the tolerance was ±20%.

Your result showed the capacitance to be about 3% higher than the specified value.

All that is beside the point.

If the capacitor AND resistor each had a tolerance of ±1%, then finding the Capacitance to be 3% high MIGHT mean your data were not too good.
 

FAQ: Why Did My Capacitance Measurement Exceed Expectations?

How is it possible to get more capacitance without violating the First Law of Thermodynamics?

In order to understand how this is possible, it is important to first understand what capacitance is. Capacitance is a measure of an object's ability to store an electric charge. It depends on the physical characteristics of the object, such as its size, shape, and material. Therefore, by changing these physical characteristics, it is possible to increase the capacitance without violating the First Law of Thermodynamics, which states that energy cannot be created or destroyed.

Can capacitance be increased without adding any external energy?

Yes, capacitance can be increased without adding any external energy. This is because capacitance is dependent on the physical characteristics of the object, not on the amount of energy it contains. By changing the physical characteristics, such as increasing the surface area or changing the distance between the plates, the capacitance can be increased without adding any external energy.

How does the addition of dielectric material increase capacitance?

Dielectric materials are insulators that are placed between the plates of a capacitor. When a dielectric material is added, it increases the capacitance by reducing the electric field between the plates. This allows for more charge to be stored on the plates without increasing the voltage, resulting in an increase in capacitance. However, this process does not violate the First Law of Thermodynamics, as no energy is being created or destroyed.

Is it possible to decrease capacitance without violating the First Law of Thermodynamics?

Yes, it is possible to decrease capacitance without violating the First Law of Thermodynamics. This can be achieved by changing the physical characteristics of the object, such as decreasing the surface area or increasing the distance between the plates. This will result in a decrease in capacitance without violating any laws of thermodynamics.

Can the First Law of Thermodynamics be violated by manipulating capacitance?

No, the First Law of Thermodynamics cannot be violated by manipulating capacitance. As mentioned earlier, capacitance is dependent on the physical characteristics of the object and not on the amount of energy it contains. Therefore, by changing the physical characteristics, it is possible to manipulate capacitance without violating any laws of thermodynamics.

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