Series RLC Circuit: Analysis and Example Problems

Consider the circuit consisting of R, L and C connected in series across a supply voltage of V (RMS) volts. The resulting current I (RMS) is flowing in the circuit. Since the R, L and C are connected in series, thus current is same through all the three elements. For the convenience of the analysis, the current can be taken as reference phasor. Therefore,





  • XL = jωL = Inductive Reactince,
  •  Xc = 1/jωC = Capacitive reactance.
  •  VR  is in phase with I.
  •  VL  is leading the current I by 90°.
  •  VC  is lagging the I by 90°.

The total voltage is the phasor sum of VR, VL and VC, i.e.,

$$\mathrm{\mathit{V} = \mathit{V}_{R}+\mathit{V}_{L}+\mathit{V}_{C}}$$






Where, $(\sqrt{(\mathit{R})^{2}+(\mathit{X}_{L}-\mathit{X}_{C})^{2}})$ is the opposition offered to the current flow and is known as Impedance of the circuit. It is denoted by Z, thus,

$$\mathrm{\mathit{Z}=\mathit{R}+\mathit{X}_{L}+\mathit{X}_{C}=\mathit{R}+\mathit{j}(\omega L-\frac{1}{\omega C})}$$



Circuit Power Factor

The power of an AC circuit is defined as the ration of active power to the total power. i.e.



Power Consumed

The power is consumed in the circuit only by the resistor, the inductor and capacitor does consume any power. Therefore,


Three cases of series RLC circuit

Case 1 – When XL > XC, i.e. (XL - XC) is positive, thus, the phase angle φ is positive, so the circuit behaves as an inductive circuit and has lagging power factor.

Case 2 – When XL < XC, i.e. (XL - XC) is negative, thus, the phase angle φ is negative, so the circuit behaves as an inductive circuit and has lagging power factor.

Case 3 – When XL = XC, i.e. (XL - XC) is zero, thus, the phase angle φ is zero, so the circuit acts as a purely resistive circuit and has unity power factor.

Now, if the applied voltage is given by,

u=\mathit{V}_{m}sin(\omega t)}$$

Then, the equation of the circuit current will be,

$$\mathrm{i=\mathit{I}_{m}sin(\omega t\:±\:\Phi)}$$

The value of φ will be positive or negative depending upon which reactance (XL or XC) predominates.

Series Resonance

The resonance occurs in a series RLC circuit, when the reactive component of the impedance becomes zero, i.e.


$$\mathrm{\Rightarrow(\omega L-\frac{1}{\omega \mathit{C}})=0}$$

$$\mathrm{\Rightarrow\:\omega L=\frac{1}{\omega \mathit{C}}}$$

Therefore, the resonant frequency is


Effects of series resonance

  • XL = XC,thus ω0 = $1/ \sqrt{LC}$
  • ZR = R = Minimum
  • Circuit current at resonance, Ir = V/R =Maximum.
  • Circuit power factor is unity. Hence, circuit is purely resistive.
  • The voltage across inductor and capacitor being equal, i.e. VL= VC.

Resonance Curve

The curve between current and frequency is known as resonance curve.




Therefore, the bandwidth of the circuit is


Q – Factor of Series Resonant Circuit

The Q-factor (Quality Factor) of the circuit is defined as the ratio of reactive power to the active power, i.e.

$$\mathrm{\mathit{Q}-factor=\frac{Reactive Power}{Active Power}}$$


$$\mathrm{\Rightarrow\:\mathit{Q}-factor=\frac{\omega L}{R}=\frac{1}{\omega CR}}$$

At resonance,



Numerical Example

A 240 V, 50 Hz AC supply is applied a coil of 0.08 H inductance and 4 Ω resistance connected in series with a capacitor of 8 μF. Calculate the following −

  • Impedance,
  • Circuit current,
  • Phase angle between voltage and current,
  • Power factor,
  • Power consumed,
  • Q-factor of the circuit at resonant frequency.



$$\mathrm{\mathit{X}_{L}=\omega L=2\pi \mathit{fL}=2\pi×50×0.08=25.12 Ω}$$

$$\mathrm{\mathit{X}_{C}=\frac{1}{\omega C}=\frac{1}{2\pi \mathit{fL}}=\frac{1}{2\pi×50×8×10^{−6}}=398.09\:Ω}$$


  • Impedance of the circuit


  • Circuit current


  • Phase angle between voltage and current


The negative sing of phase angle shows that current is leading the voltage.

  • Power Factor

$$\mathrm{cos\phi=\frac{R}{Z}=\frac{4}{372.99}=0.01072 (leading)}$$

  • Power consumed


  • Q-factor of circuit at series resonance