A series L-C-R circuit is connected …

Suppose resistance R, inductance L and capacitance C are connected in series and an alternating source of voltage V =V0sinωt is applied across it. (fig. a) On account of being in series, the current (i ) flowing through all of them is the same.   Suppose the voltage across resistance R is VR, voltage across inductance L is VL and voltage across capacitance C is VC. The voltage VR and current i are in the same phase, the voltage VL will lead the current by angle 90° while the voltage VC will lag behind the current by angle 90° (fig. b). Clearly VC and VL are in opposite directions, therefore their resultant potential difference =VC -VL (if VC >VC ). Thus VR and (VC -VL ) are mutually perpendicular and the phase difference between them is 90°. As applied voltage across the circuit is V, the resultant of VR and (VC -VL ) will also be V. From fig.  The phase difference (ϕ) between current and voltage ϕ is given by tanϕ = (XC - XL)/R The graph of variation of peak current im with frequency is shown in fig.  

Expression for Impedance in LCR series circuit: Suppose resistance R, inductance L and capacitance C are connected in series and an alternating source of voltage V =V₀sinωt is applied across it. (fig. a) On account of being in series, the current (i ) flowing through all of them is the same.

Suppose the voltage across resistance R is V_R, voltage across inductance L is V_L and voltage across capacitance C is V_C. The voltage V_R and current i are in the same phase, the voltage V_L will lead the current by angle 90° while the voltage V_C will lag behind the current by angle 90° (fig. b). Clearly V_C and V_L are in opposite directions, therefore their resultant potential difference =V_C -V_L (if V_C >V_C ). Thus V_R and (V_C -V_L ) are mutually perpendicular and the phase difference between them is 90°. As applied voltage across the circuit is V, the resultant of V_R and (V_C -V_L ) will also be V. From fig.

The phase difference (ϕ) between current and voltage ϕ is given by tanϕ = (X_C - X_L)/R The graph of variation of peak current im with frequency is shown in fig. 

With increase in frequency, current first increases and then decreases. At resonant frequency, the current amplitude is maximum.