Exercise:
Show that the impedance and phase shift of an ac circuit with a resistor resistance R a capacitor capacitance C and a coil inductance L in series are given by * Z sqrtR^+leftfracomega C-omega Lright^ and * tanDeltaphi fracomega L-dfracomega CR H: Since the current is the same through all three elements in the series circuit you can start with the corresponding phasor in a phasor diagram and add the voltage phasors with their respective phaseshift.

Solution:
In the phasor diagram the voltage phasor for the resistor is in phase with the current while the voltage phasors for the capacitor and the coil have a phase shift of pm fracpi respectively see figure. center includegraphicswidthmm#image_path:phasor-diagram-series-circuit# center The phasor for the total voltage V can be found by adding the phasors for the partial voltages V_R V_C and V_L as vectors. The amplitude of the total voltage is V sqrtV_R^+leftV_C-V_Lright^ Using the relations V_R R I V_C X_C I V_L X_L I it follows for the impedance Z fracVI fracsqrtleftR Iright^+leftX_C I-X_L Iright^I sqrtR^+leftX_C-X_Lright^ With the expressions for the reactances X_C and X_L we find Zomega sqrtR^+leftfracomega C-omega Lright^ The phase shift is given by see figure tanDeltaphi fracV_L-V_CV_R fracX_L I-X_C IR I fracX_L-X_CR fracomega L-fracomega CR Alternatively the expressions can be derived using the complex reactances tildeX_L and tildeX_C. For a series circuit the total complex impedance corresponds to the of the partial values: tildeZ R+tildeX_L+tildeX_C R+jomega L+fracjomega CR+jomega L-fracjomega C R+jleftomega L-fracomega Cright The real impedance is given by the modulus of the complex impedance: Z sqrtRetildeZ^+ImtildeZ^sqrtR^+leftomega L-fracomega Cright^ The phase shift corresponds to the argument: tanDeltaphi fracImtildeZRetildeZ fracomega L-fracomega CR