Impedance of LCR circuit
Let an inductance L, capacitance C and resistance R be connected in series to an alternating source of e.m.f. E as shown in Fig. (a).
The alternating current at all points in a series circuit has the same amplitude and phase, but the voltage across each element will be of different amplitude and phase.
(i) If VR is the voltage across R, then
VR = IR
The voltage across the resistor in the phase with current.
(ii) If VL is the voltage across L, then
VL = I XL
The voltage across the inductor leads the current by a phase π/2.
(iii) If VC is the voltage across C, then
VC = IXC
The voltage across the inductor lags behind the current by a phase π/2.
Let us construct the phasor diagram for this circuit. A single phasor I is used to represent the current in each element and VR, VL and VC represent voltage phasors.
If VR is represented along OX by OL (current along OX), VL and VC will be perpendicular to x-axis. Let VL acts along Y-axis and represented by OM and VC acts along Y'-axis and represented by OP. So VL and VC all acting along the same line but in opposite directions.If VL > VC, then (VL - VC) is along VL represented by OQ and perpendicular to phasor VR as shown in Fig. (b). The sum of VR, VL and VC equal to E and makes an angle Φ with the current I
From the phasor diagram, we find that
where Z = \(\sqrt{R^2+(\omega L- \frac{1}{\omega C})^2}\) is the effective opposition offered by L, C and R and is called the impedance of LCR-circuit.
Since Φ is the angle made by the resultant of (VR - VC) and R with x-axis, so from phasor diagram, we have:
Now three cases are of interest:
(i) If ωL > 1/ωC i.e. the circuit is inductive dominant circuit and tan Φ is +ve, so the current will lag behind e.m.f. by a phase Φ.
(ii) If ωL < 1/ωC i.e. the circuit is capacitive dominant circuit and tan θ is -ve, so the current will lead the e.m.f. by a phase Φ.
(iii) If ωL = 1/ωC then tan Φ = 0 or Φ = 0 i.e. the current and e.m.f. will be in phase and the circuit is said to be in resonance.
