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

Explanation

Consider a resistance R, an inductor of self inductance L and a capacitor of capacitance C connected in series across an AC source.The applied voltage is given by, `v=v_0 sin omegat` …(1)

where, `v_0` is the instantaneous value, v. is the peak value and `omega = 2pif`, f being the frequency of AC.
If i be the instantaneous current at time t, the instantaneous voltages across R, L and Care respectively iR, `iX_L` and `iX_C`. The vector sum of the voltage amplitudes across R, L, C equals the amplitude `v_0` of the voltage applied.
Let `v_R,v_L` and `v_C` be the voltage amplitudes across R, L and C respectively and `I_0` the current amplitude.
Then `v_R=i_0R` is in phase with `i_0`.
`v_L=i_0 X_L=i_0 (omegaL)` leads `i_0` by `90^@`
`v_c=i_0X_C=i_0 (1/(omegaC))` lags behind `i_0` by `90^@` .

The current in a pure resistor is phase with the voltage, whereas the current in a pure inductor lags the voltage by `pi/2` rad. The current in a pure capacitor leads the voltage by `pi/2` rad. For `V_L > V_C`, phase angle `phi` between the voltage and the current is positive.
From the right angled triangle OAP,
`OP^2 = OA^2 +AP^2 =OA^2 + OB^2 (because AP=OB)`
`=v_R^2 +(v_L-v_C)^2`
`=(iR)^2 + (iX_L - iX_C)^2`
`=i^2 (R^2 + (X_L -X_C)^2)`
`therefore i=v/sqrt(R^2 + (X_L -X_C)^2)=v/Z`
and `Z=sqrt(R^2 + (X_L -X_C)^2)`
Where Z is the impedance of the circuit
Phase angle between v & i.
`tan phi =(v_L-v_C)/v_R =(X_L-X_C)/R_L`
`phi=tan^(-1) ((X_L-X_C)/R)`