LAB - capacitor discharge and time constant

In summary, the student's experiment did not work correctly and they are not sure why. The student is trying to find a solution to the problem.
  • #1
Trista
33
0
OK, here is my frustration: I go to my lab, exchange equipment several times because some equipment is bad or wrong, do an experiment to show the time constant. First we charge the capacitor. Trial and error to find a good time increment. THEN, its time to discharge. AFTER 10 minutes of the capacitor not discharging (or shall we say, after 10 minutes it only discharged 1/2 volt, then 20 minutes another 1/2 volt gone...) we ask the teacher what could be wrong. She gives us a lecture on how we should have done the calculations and pre guessed what should be happening. Well, its not happening even with experimentor bias... which wouldn't have made a bit of difference. :grumpy:

Can ANYONE give me a clue as to why the capacitor didn't discharge correctly? It charges at about 63% and SHOULD discharge at 37%, but ours would have taken days, I'm afraid. She didn't have an answer, except to sit there for the rest of the day to see how long it takes. :cry:
 
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  • #2
You've told us virtually nothing at all about your experimental setup, so there's really nothing at all that we can say. It does sound, however, like you have no load resistor for your capacitor to discharge through. Perhaps the "discharge" is really just happening through the resistance of the voltmeter, which is in the ballpark of a few megohms.

- Warren
 
  • #3
Hopefully this schematic I have attached will help you see the circuit we made. Our capacitor was attached to a 'board' and the whole circuit was ok'd by our instructor. The R (of course ) stands for resistor.

The first C1 = 1000 microF
The first R1 = 10 kOhms

The 2nd C2 = 2200 microF
The 2nd R2 = 4.7 kOhms

I hope the schematic is attached...? I tried to attach it.
 

Attachments

  • lab_capacitor prob.doc
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  • #4
With a 1000 uF cap and a 10 kOhm resistor, the voltage on the cap should decrease to about 37% of its initial value in about 60 seconds.

I suspect that you may have some bad connections somewhere in your discharge path, and the cap is only discharging through the voltmeter. Did you solder the connections yourself? Are you using a proto-board? Have you made sure the proto-board's connections are good, and that you've assembled it correctly?

Just because your teacher "ok'd" the circuit doesn't mean you made no mistakes. You just made a mistake she didn't catch.

- Warren
 
  • #5
well said, Warren. The connections were alligator clips and this board that you shove the wires in. Very, well, rinki dink, or something. All of our labs have been poorly done like this. One group out of the 4 was able to get the experiment to work correctly. I intend to use their data (the group that got it to work) along side ours, since ours was not successful.

I do appreciate your feedback. I think there was something wrong with the board because we had some problems on the parallel circuit lab the week before. The biggest problem is that the teacher just told us to find the stuff without any primer on how to use it. She wants us to 'discover' how to use this stuff, as if it isn't frustrating enough just to learn it.

again, I thank you for your feedback. You've given me something to think about for my conclusion.
 
  • #6
The boards you're using (which you "shove wires in") are normally called proto-boards. Even a properly functioning proto-board can be hard to understand. Typically, all of the holes in a vertical line are connected together, while holes next to each other horizontally are not connected together. There are often horizontal buses which run along the top and bottom of the proto-board, too.

It's possible that you just wired something up incorrectly.

Note that if the voltmeter had an input resistance of 10 megohm, which is not unreasonable, the discharge circuit would have a time constant of more than 17 hours.

- Warren
 

What is a capacitor discharge?

In simple terms, a capacitor discharge is the release of electrical energy stored in a capacitor. This can happen quickly in a single burst or gradually over time.

What is a time constant in relation to capacitor discharge?

The time constant is a measure of how quickly a capacitor discharges. It is determined by the resistance and capacitance of the circuit and can be calculated using the formula RC, where R is the resistance and C is the capacitance.

How does a capacitor discharge affect the voltage in a circuit?

As a capacitor discharges, the voltage in the circuit decreases. This is because the capacitor is releasing its stored energy, which results in a decrease in voltage over time.

What is the purpose of a capacitor discharge in a circuit?

A capacitor discharge can be used to control the timing of a circuit or to provide a sudden burst of energy. It can also be used to smooth out fluctuations in voltage and prevent damage to sensitive electronic components.

What are some common applications of capacitor discharge?

Capacitor discharge is commonly used in flash photography, to power the ignition systems in cars, and in electronic circuits for timing and energy storage purposes. It is also commonly used in medical devices, such as defibrillators and pacemakers.

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