How to Calculate Phase Angle Between Current and Voltage in Series RC Circuit?

Hello. Doing some homework. Want to understand, how to calculate angle between current and voltage. Have got tested R (active resistance) and C (capacitor) circuit in series. I know voltage across both of them, I know amperage and I can calculate Xc (resistance on capacitor) and I need that angle..AC connected. If you need schematics, I can draw Thank You!

Paul...OK... some basic definitions of terms, you are not always using the right ones:In a capacitor, current leads the voltage, in an inductor current lags the voltage, by 90 degrees in a perfect component.XC - the "resistance" of a capacitor, is called capacitive REACTANCE, same with inductors (XL).The combination of reactance (from a capacitor or inductor) and resistance (from a resistor) is called IMPEDANCE (Z).You can work out problems like this by drawing a diagram representing reactance and resistance at right angles to each other.Let's say we have a 1 uF capacitor and a 10,000 ohm resistor in seriesThe reactance of our capacitor will be 1 / 2.pi.f.CSo at 50Hz (sorry, I'm British   the reactance of our capacitor will be1/ 2 . 3.14 . 50 . 1 x 10-6 = 3183 ohmsSo we can draw these values as below:
The impedance is the result of adding the resistance and the reactance, but as they are at right angles we cannot just add them.  There are two ways of doing this to get the resulting impedance:1. Use trigonometry: Tan Theta = 3183/10000 = 0.3183, so theta = 17 degreesCos Theta =  resistance / impedance  so Impedance =  resistance / cos theta Z = 10,000 / 0.952 = 10494 ohms.Now you have the impedance, you can calculate the current if you have the voltage.For example if the voltage across the circuit is 250V AC (again, I'm British! :-)then I = V/Z = 250 / 10494 = 0.0238A (23.8 mA).Has this made it any easier?It helps if you have in mind your trigonometric functions:                

Sorry Paul your original question was: "Want to understand, how to calculate angle between current and voltage." In the above calculations the calculated angle, in this case 17 degrees, is the angle you want.  If you put a voltage across the impedance (you regard that at the reference or 0 degrees) the current will lead the voltage by 17 degrees.Sorry, I had not specifically addressed this point so thought I would make it clear.