E&M lab(Finding the capacitance in an RC circuit from the time constant)

In summary, the conversation is about a student seeking help with a lab experiment involving an RC circuit and a capacitor. They explain the setup of the lab and their data collection methods. They also mention the use of an equation to model the potential difference in the capacitor and the linearization process used to find the time constant. The student shares their data and notes that the values they obtained are not matching with the theoretical value. They also mention potential sources of error and ask for advice. The conversation ends with a suggestion to measure for leakage and the possibility that the capacitor may be non-ideal.
  • #1
Raziel2701
128
0

Homework Statement


sigh* Well, I have more than a week to do this lab, and usually I don't resort for help here, but on this lab there are so many things I am unsure of.

So here's the setup of the lab:

I have a one Farad capacitor, a DC power supply, a multimeter, a bunch of banana-cables as well as a decade box and a multimeter. I connect them to make an RC circuit with the resistance set at 75ohms in the decade box and the multimeter set to measure the voltage across the capacitor. I charge the capacitor up to 5V, and then disconnect the capacitor from the power supply in order to discharge the capacitor.

Every ten seconds, I measure the voltage recorded on the multimeter. I run the experiment two more times so I have three data sets to analyze. By graphing this data and doing some fitting that I'll explain later, I am to find the time constant and the capacitance and see if it matches the theoretical value.


Homework Equations



[tex]V(t)=V_{0}e^{\frac{-t}{RC}}[/tex]

This is the given equation that models the potential difference in the capacitor as it discharges. Notice that RC in the exponent is the time constant, and since I know that the circuit has a resistance of 75 ohms and the capacitor is one Farad, the time constant, as obtained from the experiment, should be somewhere around 75.

The linearization used to supposedly find the time constant was to plot the natural log of the recorded voltages against time. I found this by simply trying to solve for RC from the previous equation and then noticed that the slope of the line would correspond to the inverse of RC. Thus the slope of the line should correspond to [tex]\frac{1}{RC}[/tex].



The Attempt at a Solution



I've played around with my data, I figured that when t=RC, the voltage in the capacitor must have decreased to 37% of its initial value. My three data sets kind of follow this trend. The data sets present a percentage of 33, 34 and 39.8 percent. I can share this data sets if requested. I can't gather any useful information from this fact, but at least it was refreshing to see something come out as theorized.

I've plotted V versus time and it is an exponentially decaying function. I'm using Graphical Analysis to plot my data, and I've been using the exponential fit to see how well it looks. It looks great. And the constant that Graphical Analysis gives me. Hold on, here is what graphical analysis does with my data under an exponential fit. It tries to make it fit this equation:

[tex]A^{-Cx} +B[/tex]

Where A, C and B are constants and x is my dependent variable. The software gives me a value to all of the constants and I've also used this "C" constant to be equal to the inverse of RC.

I don't know if this last one is true though. I've calculated the time constant through both the linearization of ln(V) vs time, and from the constant given through plotting software. Both answers are off from the mark, but especially so when using the linearization process.

When using this method I obtain values of 219, 119 and 128. When using the constant calculated by the software I get values of 23, 38 and 36. Much closer to 75 but still rather off.

I've yet to reproduce the experiment again. I will this Friday. In the meantime it's penance time for me since I need to come up with a better intuition for this experiment. The instrumentation is accurate and working properly as far as I'm concerned. Errors in measurement seems unlikely since my three data sets have consistent, reproducible values. The capacitor could be leaking charge, but would that account for such obscene discrepancies between the expected and actual results?

The connection to make the RC circuit has been verified by the professor, so I know there's no fault there.

Am I to receive no advice since it looks like I simply need to redo the experiment?

Please excuse any faults in my formatting. I just joined and I have no experience posting in here.

I appreciate any comments.
 
Physics news on Phys.org
  • #2
What is the constant for A that the exponential fit gives you? If it isn't e you have to adjust the time constant. [itex] a^x = e ^{x ln a} [/itex]. Of course B should be 0. If it isn't, something is wrong.

Since you did get a voltage of about 1/e after about 1 time constant, and 5V after 0 time constants (I hope) the slopes of voltages against time can't be that far off. They did look like straight lines?

to find out leakage, you can wait a few minutes and measure the voltage again

Those large capacitors can be quite non-ideal, with capacitance more than 10% off and
dependent on temperature, voltage. I wouldn't expect errors this large.
 
  • #3


Hello,

It sounds like you have put a lot of effort into this lab and I commend you for that. It is always frustrating when results do not match with expectations, but that is part of the scientific process.

Firstly, it is important to make sure that all of your equipment is functioning properly. As you mentioned, the capacitor could be leaking charge which could affect your results. It may be worth checking this before redoing the experiment.

Secondly, it is important to consider sources of error in your experiment. Are there any factors that could be affecting your results? For example, is there any external electromagnetic interference that could be affecting the circuit? Are there any temperature changes in the room that could be affecting the capacitor? Considering these factors may help explain any discrepancies in your results.

Additionally, it may be helpful to gather more data points and possibly run the experiment more times to get a more accurate average. This could also help identify any outliers in your data.

Lastly, it may be worth discussing your results with your professor or lab partners. They may have some insights or suggestions that could help you better understand the experiment and your results.

I hope this helps and good luck with your experiment!
 

FAQ: E&M lab(Finding the capacitance in an RC circuit from the time constant)

What is an RC circuit?

An RC circuit is a type of electrical circuit that contains both a resistor (R) and a capacitor (C). It is commonly used in electronic devices to control the flow of current and create a time delay.

What is the time constant in an RC circuit?

The time constant in an RC circuit is a measure of how quickly the capacitor charges or discharges. It is calculated by multiplying the resistance (R) and capacitance (C) values in the circuit.

How do you find the capacitance in an RC circuit from the time constant?

To find the capacitance in an RC circuit from the time constant, you can use the formula C = T/R, where C is the capacitance, T is the time constant, and R is the resistance. This formula can be rearranged to solve for C.

What does capacitance represent in an RC circuit?

Capacitance represents the ability of a capacitor to store electrical charge. In an RC circuit, the higher the capacitance, the longer it takes for the capacitor to charge or discharge.

How can you measure the capacitance in an RC circuit?

There are several ways to measure capacitance in an RC circuit, including using a multimeter or an oscilloscope. These devices can display the voltage across the capacitor and the time it takes for the capacitor to charge or discharge, allowing you to calculate the capacitance using the formula mentioned in question 3.

Similar threads

Replies
4
Views
741
Replies
2
Views
2K
Replies
4
Views
1K
Replies
1
Views
1K
Back
Top