Trouble with RC Circuit

schaefera

Hello, I'm trying to analyze the RC circuit... specifically a low-pass filter, so the set up is voltage source to resistor to capacitor, and the voltage source provides an oscillating voltage Vin, whereas I measure Vout across the capacitor.

I think I'm getting confused between two different possible analyses: first, comparing Vin and Vout and second, comparing the current and the voltage in the circuit. Please tell me where I'm going wrong.

If I consider the ratio Vout/Vin, I have the complex number: 1/(1+iwRC) which has the phase phi1= -arctan(wRC). This would imply that at very high w, the output voltage is 90 degrees behind the input voltage. But since at very high w the capacitor drops very little voltage, this circuit is essentially just a resistor dropping all the voltage. So why wouldn't we expect the output and input voltages to be IN phase in this case? Similarly, at very low w the capacitor drops most of the voltage and despite the fact that the phase angle approaches 0 in this limit. Since a purely capacitive circuit has a 90 degree phase shift, shouldn't we expect this shift to be present in that case?

I think it would be easier if I could draw a phasor diagram, but I can only draw a phasor diagram to compare Vout (which would be the sum of capacitive voltage and resistive voltage, the two being at right angles to each other) to the current (which is parallel to the resistive voltage)... I wouldn't know where to place Vin on such a diagram would I?

Again, I think I'm getting confused between looking at Vout/Vin and Vout/Total current. Please tell me where my errors are! Thank you!

cepheid

The current is (almost) in phase with V_in in this limit (ϕ_I = arctan(1/ωRC)). However, the voltage on the capacitor depends on how much charge it has. q = CV therefore dq/dt = i = Cdv/dt, where v is the capacitor voltage here. So what's happening is that, when V_in, and therefore the current, are at a maximum, the rate at which charge is accumulating on the plates is a maximum. So the capacitor voltage is only just ramping up to its maximum value. Hence the 90 degree phase lag. Once the capacitor voltage reaches a maximum value, dv/dt = 0, and hence i = 0, which is consistent with what is happening with the voltage source, which has already decreased in voltage down to 0, and is now going negative. Once it goes negative, the capacitor voltage is now greater than the source voltage, and the current reverses direction. The current i becomes large and negative, which means that the capacitor is discharging and its voltage is ramping down to its minimum (trailing the current in phase by 90 deg). Repeat this cycle over and over again.

No, because here you have i = dv/dt = 0, and therefore V_out = V_in.

I see no reason why you can't draw all three phasors on the same diagram.

schaefera

Let's say I draw Vr, the voltage across the resistor, and some angle up from the horizontal axis. Vc, the capacitor's voltage, is 90 degrees behind that. Their sum, which depends on relative lengths, falls somewhere between those two and that would be Vout correct? Where do I draw Vin?

sophiecentaur

Is your problem with where you are putting your voltmeter so that you are apparently getting a phase shift in one case and not in the other?
Relate everything to Earth, to make it easier. If you draw the circuit with the C, connected between Earth and the Resistor and you then connect the source PD to the resistor. The voltage you measure (to Earth) across the capacitor will always be in quadrature with the volts of the supply - just getting less and less as the frequency increases. The phasor diagram can be drawn with Vr along the x axis, Vc along the y axis and Vin will be the diagonal joining those two vectors. Note, I have not used "Vout" because that would either be Vc or Vr, depending on your choice of which one is the 'output'

schaefera

Aha! So let me see if I got this right: what I've been calling Vout is really Vc, and Vin is the total of all voltages, what I've called Vtot. So, that's why I've been having the issue with phase shifts-- if phi is the angle by which current and voltage are out of phase, that's the same as the angle between Vin and Vr (because Vr is in phase with current). But the difference between Vin and Vc (or Vout, as I've used before) is actually 90 degrees MINUS phi, since we're measuring from Vc, and not from Vr, to Vtot?