# AC - frequency affect on RC voltage

• StonieJ
In summary, when the frequency of an AC circuit is increased, the voltage across an RC component changes shape. This is because the voltage and current in the circuit change directions more rapidly, not allowing enough time for the capacitor to build up charge. This results in a smaller amplitude and a linear growth and decay trace instead of an exponential one. Furthermore, the impedance of the capacitor drops as frequency increases, causing the voltage across it to drop towards zero.
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.

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.

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%20circuit.jpg

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

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StonieJ said:
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%20circuit.jpg

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

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 by a moderator:

## 1. How does the frequency of an AC current affect the voltage in an RC circuit?

The frequency of an AC current refers to how many cycles or repetitions the current completes in a given period of time. In an RC circuit, the voltage across the resistor and capacitor is affected by the frequency of the AC current. As the frequency increases, the voltage across the capacitor decreases and the voltage across the resistor increases. This is due to the capacitor's ability to store and release charge at different frequencies, resulting in a phase shift between the voltage across the resistor and capacitor.

## 2. What is the relationship between frequency and voltage in an RC circuit?

The relationship between frequency and voltage in an RC circuit can be described by the impedance equation: Z = √(R^2 + (1/ωC)^2), where Z is the impedance (or total resistance), R is the resistance, ω is the angular frequency (2πf), and C is the capacitance. As the frequency increases, the impedance decreases, resulting in a decrease in voltage across the capacitor and an increase in voltage across the resistor.

## 3. How does the value of the resistor and capacitor affect the voltage in an RC circuit at different frequencies?

The value of the resistor and capacitor in an RC circuit affects the voltage at different frequencies through the impedance equation mentioned above. A larger resistance or capacitance will result in a larger impedance, therefore decreasing the voltage across the capacitor and increasing the voltage across the resistor. The opposite is true for smaller resistance or capacitance values.

## 4. Can the frequency of an AC current affect the overall voltage output in an RC circuit?

Yes, the frequency of an AC current can affect the overall voltage output in an RC circuit. This is because the voltage across the resistor and capacitor are in phase at lower frequencies, but as the frequency increases, a phase shift occurs causing the overall voltage output to decrease.

## 5. How does the frequency response of an RC circuit differ from a DC circuit?

A DC circuit has a constant voltage output, regardless of the frequency of the current. However, in an RC circuit, the voltage output varies with the frequency of the AC current. At lower frequencies, the voltage output is higher due to the capacitive reactance being low. As the frequency increases, the voltage output decreases due to the increase in capacitive reactance. This results in a frequency-dependent voltage output in an RC circuit, unlike a DC circuit.

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