What is RLC Series Circuit? Circuit Diagram, Phasor Diagram, Derivation & Formula

Figure (1): RLC Series Circuit.

An RLC series circuit consisting of a resistor ‘R’ Ohm (Ω) in series with an inductance of ‘L’ Henrys and capacitance of ‘ C’ Farads connected to an A.C supply as shown in figure (1).

Applying Kirchhoff’s voltage law to the above circuit,

$$V={{V}_{R}}+{{V}_{L}}+{{V}_{C}}=IR+I{{X}_{L}}+I{{X}_{C}}$$

Where,

V = Source voltage

V_R = Voltage drop across resistor = I_R

V_L = Voltage drop across inductor = IX_L

V_C = Voltage drop across capacitor = IX_C

Figure (2): RLC Series Circuit Phasor Diagram when V_L > V_C.

The phasor diagram is drawn taking the current as the reference phasor as shown in figure (2).

From the phasors diagram

$$O{{E}^{2}}=O{{A}^{2}}+O{{D}^{2}}$$

$${{V}^{2}}=V_{R}^{2}+{{\left( {{V}_{L}}-{{V}_{C}} \right)}^{2}}$$

$$V=\sqrt{V_{R}^{2}+{{\left( {{V}_{L}}-{{V}_{C}} \right)}^{2}}}$$

$$V=\sqrt{{{\left( IR \right)}^{2}}+{{\left( I{{X}_{L}}-I{{X}_{C}} \right)}^{2}}}$$

$$V=I\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}$$

$$V=IZ$$

Where,

$$Z=\sqrt{{{R}^{2}}+{{X}^{2}}}\Omega$$

= Impedance of the circuit.

From the phasor diagram here, three cases arise.

1. X_L > X_C i.e., X_L – X_C is positive
2. X_C > X_L i.e., X_L – X_C is negative
3. X_L = X_C i.e., X_L – X_C is zero.

Case (i): X_L > X_C

From the phasor diagram shown in figure (2), the net reactance is inductive. Therefore, for the condition X_L > X_C the circuit behaves like an inductive circuit. The current lags behind the voltage by 90º.

Net reactance,

$$X={{X}_{L}}-{{X}_{C}}$$

Impedance,

$$Z=\sqrt{{{R}^{2}}+{{X}^{2}}}\Omega$$

The phase difference between voltage and current,

$$\phi ={{\tan }^{-1}}\left( \frac{X}{R} \right)={{\tan }^{-1}}\left( \frac{{{X}_{L}}-{{X}_{C}}}{R} \right)$$

The power taken by the circuit,

$$P=VI\cos \phi \text{    watts}$$

Since pure inductance and capacitance do not consume power, power factor of the given circuit is $\cos \phi$ lagging.

Case (ii): X_C > X_L

Figure (3): RLC Series Circuit Phasor Diagram when X_C > X_L.

From the phasor diagram shown in figure (3), the net reactance is capacitive. Therefore, for the condition X_C > X_L the circuit behaves like a capacitive circuit. The current leads voltage by 90º.

Net reactance,

$$X={{X}_{L}}-{{X}_{C}}$$

Impedance,

$$Z=\sqrt{{{R}^{2}}+{{X}^{2}}}$$

$$Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{C}}-{{X}_{L}} \right)}^{2}}}\Omega$$

The phase difference between voltage and current,

$$\phi ={{\tan }^{-1}}\left[ \frac{X}{R} \right]$$

$${{\tan }^{-1}}\left[ \frac{{{X}_{L}}-{{X}_{C}}}{R} \right]$$

The power taken by the circuit,

$$P=VI\cos \phi \text{    Watts}$$

Power factor of the given circuit is $\cos \phi$ leading.

Case (iii): X_L= X_C

Figure (4): RLC Series Circuit Phasor Diagram when X_L = X_C.

From the phasor diagram shown in figure (4), the net reactance of the circuit for the given condition X_L= X_C is zero. The circuit behaves like a resistive circuit.

Net reactance,

$$X={{X}_{L}}-{{X}_{C}}$$

$$X=0$$

Impedance,

$$Z=\sqrt{{{R}^{2}}+{{X}^{2}}}$$

$$Z=\sqrt{{{R}^{2}}+0}=R\Omega$$

Power taken by the circuit,

$$P=VI\cos \phi \text{    Watts}$$

The phase difference between voltage and current is zero. Thus,

$$\phi = 0$$

Since the current is in phase with voltage, the power factor $\cos \phi$ is unity.