Derive an expression for the impedance of a series LCR, circuit, when an AC voltage is applied to it.

Explanation

Let an inductor of inductance L, capacitor of capacitance C, resistor of resistance R are connected in sieres with an AC source which supplies an AC voltage `V = V_(0) sin omega t`. Let I be the current in the circuit. The voltages across the three elements are shown in the phasor.

`V_(R) = IR, V_(R)` is in phase with I, `V_(L) = I X_(L) V_(L)` leads I by `90^(@)` & `Y_(C) = I X_(C), V_(X)` lags behind I by `90^(@)`.
If `V_(C) gt V_(L)` then `(V_(C) - V_(L))` is the resultant of `V_(L)` and `V_(C)` and is in the direction of `V_(C)`. Let OA and OB represents the resultant `(V_(C) - V_(L))` and `V_(R)` respectively. The rectangle OACB is completed. The diagnal .OC. represents the resultant voltage V. Let `phi` be the phase angle between V and I.
From Triangle OCB, `OC^(2) = OB^(2) + CB^(2)`.
`V^(2) = V_(R)^(2) + (V_(C) - V_(L))^(2)`
`V^(2) = (IR)^(2) = [I X_(C) - I X_(L)]^(2)`
`V^(2) = Isqrt(R^(2)+(X_(C) - X_(L))^(2))`
`V = I sqrt(R^(2) + (X_(C) - X_(L))^(2))`
`I = (V)/(sqrt(R^(2) + (X_(C) - X_(L))^(2)))`
`I = (V)/(Z)`
where `Z = sqrt(R^(2) + (X_(C) - X_(L))^(2))` is called impedance