The author asked me to calculate XC and fill in the table based upon some voltage readings.
I cannot figure out how to do the calculations and and I cant find any explanation of the formula.
Here is the link to the project
https://electrojets.blogspot.com/p/objective-objective-of-this-project-is_10.html
Here is the table:


and the schematic.


Secondarily, I thought that an AC signal would pass right by a capacitor, untouched.The test signal that I am using is an audio tone so I guess that it is not "pure" AC since it never goes negative.

 

Your circuit could be treated as a voltage divider where the role of the variable resistance is taken by the capacitor.
At low frequencies, the capacitor has a high reactance Xc so, almost all input voltage is "drop" across Xc.
But when input frequency increases the reactance Xc of a capacitor decreases and thus causes the output voltage to increase.
At frequency at which reactance of a capacitor (Xc) is equal to the value of resistor the output voltage is equal Vout=0.707*Vin
And this happens at frequencies equals

The voltage ratio Vout/Vin of a voltage divider that contains two resistors doesn't depend on frequency because the resistance of resistors does not change with frequency. But is this case the divider's voltage ratio change with frequency.
Because R is unchanged but capacitor reactance Xc is changing with frequency Vou/Vin = R/(Xc + R)

As for Xc use this equation


Xc = 1/(2Π*F*C) or C = 1/(2Π*F*Xc)

 

Here is some reading that should help you:

https://www.allaboutcircuits.com/te...rrent/chpt-4/ac-resistor-circuits-capacitive/

https://en.wikipedia.org/wiki/Electrical_reactance

This is more advanced:

https://en.wikipedia.org/wiki/Electrical_impedance#Deriving_the_device-specific_impedances

 

As I was reading more I came across articles on capacitive reactance and it appears that that is what I have. And
Xc is the symbol of capacitive reactance in a high pass filter. Maybe the article author wanted me to figure it out.

Thank you for adding to my reading list, no snarikiness intended.

 

Thanks folks and I am drawing a complete blank in how to breadboard that circuit. I keep getting Vout and the ground symbol confused. Could one of you kind gentle persons, please show me how to wire this. I pick up a wire then put it back. This is too easy to be so confusing.

 

Is this enough?


As a side note, the AC signal does not have + or minus sign.

 

Is this enough?
View attachment 149415

As a side note, the AC signal does not have + or minus sign.

Thank you Jony but I have some questions.
Are the V-IN, +, and the V-OUT, +, the signal and ground wires from my audio signal generator cable? Or are they the same wire?
I am not used to seeing a schematic without a battery somewhere.
What I am using is a head phone cable, cut, and stripped exposing the shielding and the left and Right channel's.
I use the shielding and either the right or left channel.
When I hook it up to a scope it looks OK.

 

Vin - is a signal from a generator. And Vout - is a place where you connect your AC voltmeter or oscilloscope.
And the ground must be connected together
https://learn.sparkfun.com/tutorials/how-to-use-a-breadboard

 

Vin - is a signal from a generator. And Vout - is a place where you connect your AC voltmeter or oscilloscope.
And the ground must be connected together
https://learn.sparkfun.com/tutorials/how-to-use-a-breadboard

Thanks again Jony, I am going to master this, I swear it. I have put together far tougher ones, so this should not be so difficult.
The part that confuses me is because the other circuits that I did had a V in, (the battery + terminal), V out, (the battery - terminal, and Gnd, (the battery - terminal).
I think that I need to study some more.

 

Thanks again Jony, I am going to master this, I swear it. I have put together far tougher ones, so this should not be so difficult.
The part that confuses me is because the other circuits that I did had a V in, (the battery + terminal), V out, (the battery - terminal, and Gnd, (the battery - terminal).
I think that I need to study some more.

This is a passive filter, so there is no separate power. The power in the output signal comes from the input signal.

 

So, do I leave the ground wire from the signal generator ( the coax shielding ) just hanging ?
I am thinking yes. When I leave it unconnected I get a voltage reading, whereas if I plug it into GND, the voltage reading drops to zero.
I will never understand this stuff. It hertz !
Sorry for the over used joke.