# Physics Lab - Electrical Circuits with Resistors and Capacitors

• mintsnapple
In summary: Notice that as C decreases, the time constant also decreases. The same goes for R. So, you can conclude that tau is inversely proportional to both C and R.
mintsnapple

## Homework Statement

Hello, I am having trouble with a physics lab report I'm working on. I'm going to try and be as clear as possible. Basically, I'm using different values of R and C in determining tau, the time scale in a circuit's behavior. I used a circuit with a resistor and capacitor and hooked it to an oscilloscope, and used the OpenChoice Desktop program to get 2500 data points in the capacitor voltage curve. My question is...I'm not really sure how to start plotting I guess. I have constant values of R and varying values of C and vice versa. How do I best start plotting when I have 2500 data points for each? And should I use a ln(V) vs T plot to find out what tau is? I'm sorry if this seemed vague, I can try and clarify more, but I really need help if possible! Thank you!

## The Attempt at a Solution

You should start off by plotting voltage against time. Which voltage? All 3! This is how the waveshape will occur in real life, so you need to get very familiar with it and its finer details so that you can henceforth picture it in your mind without needing to see it.

When that is all done, why not plot log(V) as you suggest!

Maybe also make another plot, you decide what results are worth presenting/summarizing in graphical form.

2,500 data points is a little over the top.

1 person
NascentOxygen said:
You should start off by plotting voltage against time. Which voltage? All 3! This is how the waveshape will occur in real life, so you need to get very familiar with it and its finer details so that you can henceforth picture it in your mind without needing to see it.

When that is all done, why not plot log(V) as you suggest!

Maybe also make another plot, you decide what results are worth presenting/summarizing in graphical form.

2,500 data points is a little over the top.
Thanks for the response. However I still don't quite understand...
Let's say I have datapoints for 11k ohms/.21 microfarads and another set for 16k ohms/.21 microfarads. What would be the best way to plot this?

mintsnapple said:
Thanks for the response. However I still don't quite understand...
Let's say I have datapoints for 11k ohms/.21 microfarads and another set for 16k ohms/.21 microfarads. What would be the best way to plot this?
Plot voltages vs. time as the capacitor charges up.

Have you thought about the useful measurements that you hope to be able to make on the plots?

1 person

Plotting ln v versus time is one of the graphs you can construct. Do this and see what the slope is.

But if you just plot voltage versus time, the slope of the steepest tangent to that exponential curve should also give time constant, tau.

1 person
NascentOxygen said:
Plotting ln v versus time is one of the graphs you can construct. Do this and see what the slope is.

But if you just plot voltage versus time, the slope of the steepest tangent to that exponential curve should also give time constant, tau.

I have lots of negative voltage points...Wouldn't plotting ln v be less effective then?

After plotting all these different graphs with different R and C values, how could I then make a conclusion on how tau is affected?

mintsnapple said:
I have lots of negative voltage points...Wouldn't plotting ln v be less effective then?
If all your points are of negative voltage, change their sign before taking the log. If only some are negative, then on the voltage axis plot ln(v - vi) where vi is the initial voltage. None of this will change the slope of the curve, and it's from the slope that you can read off ##\Large\tau##.

After plotting all these different graphs with different R and C values, how could I then make a conclusion on how tau is affected?
You observe how the graphs change when C (and/or R) changes.

Last edited:

## 1. What is Ohm's law and how does it apply to electrical circuits?

Ohm's law states that the current flowing through a conductor is directly proportional to the voltage applied, and inversely proportional to the resistance of the conductor. In other words, as the voltage increases, the current increases, and as the resistance increases, the current decreases. This law applies to all types of electrical circuits, including circuits with resistors and capacitors.

## 2. How do resistors and capacitors affect the flow of electricity in a circuit?

Resistors limit the flow of current in a circuit by converting electrical energy into heat. The resistance of a resistor is measured in ohms and can be increased or decreased by changing the material or size of the resistor. Capacitors, on the other hand, store electrical energy and release it when needed. They can affect the flow of current in a circuit by storing or releasing energy, depending on the voltage and capacitance of the capacitor.

## 3. What is the difference between series and parallel circuits?

In a series circuit, all components are connected in a single path, meaning the current flowing through one component must also flow through all the others. The total resistance in a series circuit is the sum of all individual resistances. In a parallel circuit, components are connected in multiple paths, meaning the current can split and flow through different branches. The total resistance in a parallel circuit is less than the individual resistances, making the overall current higher.

## 4. How do I calculate the total resistance in a series or parallel circuit?

In a series circuit, the total resistance is simply the sum of all individual resistances (R1 + R2 + R3...). In a parallel circuit, the total resistance is calculated using the formula 1/Rt = 1/R1 + 1/R2 + 1/R3..., where Rt is the total resistance and R1, R2, R3... are the individual resistances. This can then be inverted to find the total resistance (Rt = 1/ (1/R1 + 1/R2 + 1/R3...)).

## 5. How do I calculate the total capacitance in a series or parallel circuit?

In a series circuit, the total capacitance is calculated using the formula Ct = C1 + C2 + C3..., where Ct is the total capacitance and C1, C2, C3... are the individual capacitances. In a parallel circuit, the total capacitance is simply the sum of all individual capacitances (Ct = C1 + C2 + C3...).

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