Q.Derive an expression for the impedance of an LCR circuit connected to an AC power supply.

[!TLDR] Adding the perpendicular phasors $e_R=i_0R$ (in phase) and $(e_L-e_C)=i_0(X_L-X_C)$ (net reactive) by Pythagoras gives $e_0=i_0\sqrt{R^2+(X_L-X_C)^2}$, so $ …

Consider a resistor R, inductor L and capacitor C connected in series, carrying a common current $i=i_0\sin\omega t$ (Fig. 13.12). Each element's own voltage drop has a fixed phase relationship to this common current: the resistor's voltage $e_R=i_0R$ is exactly IN PHASE with i; the inductor's voltage $e_L=i_0X_L$ LEADS i by $\pi/2$; and the capacitor's voltage $e_C=i_0X_C$ LAGS i by $\pi/2$.\n\nRepresenting each as a phasor (Fig. 13.13): draw $e_R$ along the same direction as the current phasor $i_0$ (along OA on the reference axis); draw $e_L$ perpendicular to $i_0$, rotated $90^\circ$ anticlockwise; and draw $e_C$ perpendicular to $i_0$ but rotated $90^\circ$ CLOCKWISE, i.e. exactly opposite to $e_L$. Since $e_L$ and $e_C$ point in exactly opposite directions along the same (perpendicular) line, they combine by simple subtraction into a single net reactive phasor of magnitude $(e_L-e_C)$ (taking $e_L>e_C$, i.e. OB' in the figure) along that perpendicular direction.\n\nThe resultant applied voltage phasor $e_0$ is then the vector sum of the two remaining MUTUALLY PERPENDICULAR phasors, $e_R$ (along OA) and $(e_L-e_C)$ (along OB', perpendicular to O …

Add the resistor's in-phase voltage phasor and the net (L minus C) perpendicular reactive voltage phasor using the Pythagorean t …

Adding $e_R$, $e_L$ and $e_C$ as plain (scalar) magnitudes instead of accounting for their different phases -- the correct combination requires first subtracting $e_C$ from $e_L$ (since they are exactly opposite in phase) an …