Phase Shift due to Capacitor

Consider the following circuit, Running an AC simulation, Now consider this, Running an AC simulation, How come with the RC circuit, we have a constant -90 phase shift. I don't understand how the "voltage lags current by 90 degrees" comes into play here. Surely, we're just plotting the transfer function right (i.e Vout/Vin) so who cares about current? And even if a capacitor provides a constant -90 degree shift, why doesn't it happen with just the capacitor circuit?

• You are dealing with a theoretical voltage source in (1) with Rs=0 which cannot exist. – Tony Stewart Sunnyskyguy EE75 Jan 28 at 21:45
• @SunnyskyguyEE75 Ooo. very good point. Didn't think of that. – AlfroJang80 Jan 28 at 21:49

"Voltage lags current by 90°" is correct — but you never actually measure the current through the capacitor. Try adding that measurement to both of your experiments, and you find that this is always true about the voltage and current associated with the capacitor.

In your second circuit, you've essentially turned the voltage source into a current source (by making the resistor so much larger than the capacitor's impedance), so now you can see the phase lag.

The complex transfer function of the RC network is (voltage devider, frequency domain):

$$\frac{V_{out}}{V_{in}} = \frac{\frac{1}{sC_1}}{\frac{1}{sC_1} + R_1} = \frac{1}{sC_1R_1 + 1}$$

s is the complex frequency, s = jω , j is the complex unit, ω = 2×π×frequency, no switching on/off is considered here.

The transfer function is of type

$$\frac{1}{sK + 1}$$

which is clearly a complex function delivering different phases for different frequencies. The current does play a role if the resistor R1 <> 0.

If R1 is 0 , the transfer function yields 1/1 = 1, which is a real constant number, with constant phase zero, in other words the phase does not depend on the frequency in that case. Only in this case (R1 = 0) the current through C1 does not play any role since C1 is directly parallel to the voltage source.