Step 1: Set up the circuit:
The source is \( V = V_m \sin \omega t \). In a series circuit the current i is common, so we draw all the voltage phasors relative to the current phasor.
Write \( X_L = \omega L \) and \( X_C = 1/\omega C \) for the two reactances.
Step 2: Add the phasors to get the impedance (I):
Along the current: \( V_R = i_m R \). At 90 degrees ahead: \( V_L = i_m X_L \). At 90 degrees behind: \( V_C = i_m X_C \).
The L and C phasors are opposite, so they combine to \( i_m (X_L - X_C) \) at right angles to \( V_R \).
Pythagoras gives \( V_m = i_m \sqrt{R^2 + (X_L - X_C)^2} \).
The impedance is the ratio \( Z = V_m / i_m = \sqrt{R^2 + (X_L - X_C)^2} \).
Step 3: Current expression and phase (II and III):
In the triangle, the voltage phasor makes an angle with the current phasor, where \( \tan \phi = \dfrac{X_C - X_L}{R} \).
So \( i = \dfrac{V_m}{Z} \sin(\omega t + \phi) \).
The current leads V when \( X_C > X_L \) (capacitive circuit), lags V when \( X_L > X_C \) (inductive circuit), and is in phase when \( X_L = X_C \).
Step 4: Power factor (ii):
Average power is \( P = V_{rms} I_{rms} \cos \phi \). The factor \( \cos \phi = R/Z \) is called the power factor.
Its largest value is 1. It occurs for a circuit with only R, or at resonance \( X_L = X_C \), where \( Z = R \).
Its smallest value is 0. It occurs when the circuit has only L or only C, so no power is used up and the current is wattless.
Final Answer:
Impedance \( Z = \sqrt{R^2 + (X_L - X_C)^2} \); the power factor is maximum (1) at \( \phi = 0 \) and minimum (0) at \( \phi = 90^{\circ} \).
\[ \boxed{Z = \sqrt{R^2 + (X_L - X_C)^2}} \]