Exercise:
Calculate the impedance and the phase shift for an ac circuit with a resistor resistance RO and a capacitor capacitive reactance XCO in series.

Solution:
Since the resistor and the capacitor are connected in series the same current I flows through both elements; we use it as the reference phasor. The voltage across the resistor V_R RI is in phase with the current while the voltage across the capacitor V_C X_C I lags the current by degree. The total voltage V is the phasor of V_R and V_C. center tikzpicturestealth scale. draw- - -- noderight mathrmRe; draw- - -- nodeabove mathrmIm; draw- blue thick -- nodemidway above I; draw- red thick -- nodemidway below V_R; draw- red thick -- - nodemidway right V_C; draw- red ultra thick -- - nodemidway below left V; draw arc :-.:; node at - varphi; tikzpicture center Dividing every phasor in the diagram by the real in-phase current I turns the voltage triangle o the impedance triangle since ZV/I RV_R/I and X_CV_C/I. From this right triangle we read off the magnitude of the total impedance and the phase angle varphi between the total voltage V and the current I: Z fracVI sqrtR^+X_C^ sqrtR^+XC^ Z approx resultZP varphi -arctanleftfracV_CV_Rright -arctanleftfracX_CRright -arctanleftfracXCRright ph approx resultphP medskip Alternatively we can assign a complex quantity to each element. The resistor's contribution is purely real R while the capacitor's complex reactance is tilde X_C -iX_C since the current through a capacitor leads the voltage across it by degree. As the elements are in series the complex impedances add: tilde Z R + tilde X_C R - iX_C The magnitude and the argument of this complex number reproduce exactly the impedance and phase shift found above: Z |tilde Z| sqrtR^+X_C^ Z approx resultZP varphi argtilde Z -arctanleftfracX_CRright ph approx resultphP confirming the result obtained with the phasor diagram.