# AC - frequency affect on RC voltage

1. Apr 25, 2005

### StonieJ

We observed that whenever the frequency of an AC circuit was increased, the voltage across an RC component changed shape. Specifically, 1) the trace lost its exponential growth and decay and instead had linear growth and decay, and 2) the voltage amplitude was much smaller.

My guess as to the reason this occurs is because an increase in the frequency means that the voltage, and therefore current, in the circuit changes directions more rapidly. Therefore, the current does not travel long enough in a certain direction to allow the capacitor to build up charge. This would explain the smaller amplitude part. I'm also guessing this is the reason it appears linear. Since it is not given enough time to build, you are only seeing the very beginning segment of an exponential growth and decay trace, which appears linear when this time interval is small enough. That is, even a sine wave would look linear if you took a small enough segment.

I don't have the background to answer this question with concrete math formulas (well, I might, but I can't recall them), so I hope this more verbose answer still works. Thanks.

2. Apr 25, 2005

### learningphysics

Given a sinusoidal source... there will be a transient response that quickly goes to zero. And a steady - state response which should be a sine wave.
I'm assuming that you're looking at the steady-state response. Also, I'm assuming the R and C are in parallel. What are the other parts of the circuit?

The impedance of a capacitor(when dealing with the steady-state sinusoidal response) is $$\frac{1}{jwC}$$, so as frequency gets higher (w gets higher)... the capacitor's impedance drops and it gets more and more like a short circuit. So the voltage across the capacitor drops towards zero as frequency gets higher... so that would explain the amplitude drop.

3. Apr 25, 2005

### StonieJ

Thanks for the feedback.

Yes, we are only looking at the steady-state response. Also, I believe the RC is in series. If you're really curious, here's a graphic:

https://webspace.utexas.edu/youngba2/www/AC circuit.jpg

I'm assuming your answer is still correct, but let me know if it being in series changes anything.

4. Apr 26, 2005

### learningphysics

Yes... looking at the diagram, it looks to me like you're measuring the voltage across the capacitor (ch2)... and as frequency increases the impedance of the capacitor definitely drops. So what I said above is still correct.

Assuming the source is of the form Vcos(wt+theta)...you can calculate the amplitude (if you wish) by getting the magnitude of:
$$V\times\frac{\frac{1}{jwC}}{\frac{1}{jwC}+R}$$

I just used voltage divider to get the above.

You can see from the above that the expression will drop towards 0 as w increases.

Last edited: Apr 26, 2005