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I am graphing a 1 uF capacitor's frequency response in a circuit like the one below with a 100k ohm resistor.

I am using a 10V (peak to peak) sine wave as the input.

I've calculated the -3 db down point of this circuit to be at about 1.59 Hz, and noticed that this is also when the reactance of the capacitor is 100k ohms as well.

Given that this is similar to a voltage divider circuit I would expect the waveform from Vout (yellow) to be half of the original waveform (pink), however I found that not to be the case.

I'm not sure what is going on here as neither the mMax or the RMS voltages are equal to half of their originals (it doesn't visually look like half either.)

LTspice gave similar results, if not even farther away from halfway due to it assuming ideal components.

What is my issue here?

enter image description here enter image description here

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    \$\begingroup\$ Do the voltage divider math with complex numbers where \$X_C=-j100k\$. \$\endgroup\$ Commented Jun 21, 2021 at 1:37
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    \$\begingroup\$ -3dB in power is how much in voltage? \$\endgroup\$ Commented Jun 21, 2021 at 2:51
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    \$\begingroup\$ Can you simultaneously display and compare the voltage waveform across C1 with the voltage waveform across R1. \$\endgroup\$ Commented Jun 21, 2021 at 3:35
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    \$\begingroup\$ Good question. Just because the magnitudes of their impedances are equal does not mean that Vout = Vin/2. It means |Vcap| = |Vres| where |Vcap| is the magnitude of the capacitor voltage and similarly for the resistor. I think that is why James suggests that you compare voltage across C1 with voltage across R1. You can't do KVL on AC unless you use phasor arithmetic and account for relative phase. \$\endgroup\$ Commented Jun 21, 2021 at 6:03
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    \$\begingroup\$ FWIW, it's not just "similar to a voltage divider circuit", it is a voltage divider circuit. But AC voltages are somewhat different from DC voltages, so the intuition you've developed in dealing with DC circuits will need to be adjusted. \$\endgroup\$ Commented Jun 21, 2021 at 12:49

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The current through a capacitor leads the voltage across it by 90 degrees. The current through and the voltage across a resistor are in phase. Because the components are in series it means that the current through them must be identical. This all means that the voltage across the resistor is forced to lead the voltage across the capacitor by 90 degrees.

So the voltages across the two components peak at different times to each other in each cycle.

The peak of the output voltage is 0.707 times the peak of the input voltage at the frequency where R=Xc. This is called the cut-off frequency where the output voltage is 3dB down on its low frequency value. This is also the half power frequency where the power is also down 3dB.

The instantaneous voltages of the two components, when added, must equal the instantaneous value of the input voltage.

To calculate the total impedance draw a right angled impedance triangle. The reactance and resistance can then be added vectoraly by using pythagoras as the total impedance is represented by the hypotenuse. A similar triangle can be drawn using the triangle's sides to represent the voltages across the components. The length of the hypotenuse then represents the input voltage.

So, for instance, if the input voltage is 1V and the voltages across the two components are each 0.707V then sqrt(0.707^2 + 0.707^2) = sqrt(0.5 + 0.5) = sqrt(1) = 1.

As an aside, if you are wondering why at the cut-off frequency, both the power and voltage are down 3dB when one is at half and the other is at 0.707 it's because the voltage is calculated using 20log(0.707/1) dB and the power is calculated using 10log(0.5/1) dB.

Incidently, if you were to create a high pass filter by swapping the positions of the two components with each other, the voltages and phases of the three waveforms (input, across R and across C) are identical to the low pass circuit it's just that for a high pass filter the output is taken from across the resistor instead of across the capacitor.

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