Obtain an expression for the impedance of a series LCR circuit. (using phasor diagram method).

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 is the instantaneous value, `v_0` is the peak value and `omega =2pif` , f being the frequency of AC.
If i be the instantaneous current at time t, the instantaneous votages across R,L and C are 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_0X_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^2=V_R^2 +(V_L-V_C)^2 =(iR)^2 +(iX_L-iX_C)^2`
`=i^2(R^2 + (X_L-X_C)^2)`
`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)`