Derive an expression for impedance of an a.c. circuit with an induct L, capacitor C and a resistor R in series. What is condition of resonance?

Explanation

Consider that a source of an alternating emfis connected to a series combination of an inductor of inductance L, a resistor of resistance R and a capacitor of capacitance Cas shown in Fig. (a).
Let E and I be the instantaneous values of emfand current in the LCR-circuit , and `V_L ,V_C and V_R` be the instantaneous values of the voltages across inductor L, capacitor and resistor R respectively. Then,

`V_L = 1X_L, V_C=I1 X_C and V_R=IR`
`X_L = omegaL and X_C =1/(omegaC)` are reactances due to inductor and capacitor respectively.
In order to find the relation between E, `V_L V_C and V_R`, let us draw the phasor diagram for current and voltages as explained below :
In an a.c. circuit containing R only, `V_R` and I are in phase with each other. Therefore, both the phasors VR and I are represented along OX [Fig. (b)].
In an a.c. circuit containing L only, `V_L` leads by phase angle `pi/2` and likewise the phasor `V_L` is represented along OY.
Finally, in an a.c. circuit containing C only, `V_C` lags behind I by phase angle. Therefore, `V_C` is represented along OY.:
Suppose that OA, OB and OC represent the magnitudes of the phasors `V_R, V_L and V_C` respectively. In case `V_L gt V_C` (as shown in Fig.), then OD represents `V_L-V_C` the resultant of `V_L and V_C` Finally, OE, the resultant of OA and OD represents E. Thus, from right angled AOAE, we have
`OE=sqrt(OA^2+AE^2)=sqrt(OA^2+OD^2)`
or `E= sqrt(V_R^2+(V_L-V_C)^2)`
Substituting the values of `V_R, V_L and V_C` we have
`E=sqrt((IR)^2+(IX_L-IX_C)^2)=1sqrt((R^2+(X_L-X_C)^2)`
or `I=E/(sqrt(R^2+(X_L -X_C)^2))" "...(1)`
The effective opposition offered by L, C and R to a.c. is called impedance of LCR-circuit.. If Z is impedance of LCR circuit, then
`I=E/Z" "...(2)`
From equations (1) and (2), we have
`Z=sqrt(R^2+(X_L-X_C)^2)=sqrt(R^2+(omegaL-1/(omegaC))^2)`
The equation gives impedance of LCR-circuit.
Condition for resonance : In fact, when `omegaL=1/(omegaC)` the impedance of the circuit becomes. just equal to R(minimum). In other words, the LCR-series circuit will behave as a purely resistive circuit. Due to the minimum value of impedance (equal to R), the circuit in LCR-series circuit will be maximum and this condition is known as resonance.