# RLC Series Circuit Configuration with example

### Phasor Notation

$$e = 70.7 \sin wt \Rightarrow E= 50V \angle 0^\circ$$ Refer to Fig. 1.
Fig. 1: RLC Series circuit Configuration.
Fig. 2: Applying phasor notation to the RLC Series circuit Configuration of Fig. 1.
where $$\begin{split} Z_T &=Z_1 + Z_2+ Z_3 \\ &=3 \angle 0^\circ + 7 \angle 90^\circ - 3 \angle -90^\circ\\ &=3 Ω + j7Ω - j3Ω \\ &=3 Ω + j4Ω \\ Z_T &= 5Ω \angle 53.13^\circ \end{split}$$ Impedance diagram: See Fig. 3.
Fig. 3: Impedance diagram for the RLC Series circuit Configuration of Fig. 1.
Current: $$I = {E \over Z_T} = {50A \angle 0^\circ \over 5Ω \angle 53.13^\circ} = 10A \angle -53.13^\circ$$ Ohms laws: $$\begin{split} V_R &= I Z_R = (10A \angle -53.13^\circ)(3Ω \angle 0^\circ)\\ &=(30V \angle -53.13^\circ)\\ V_L &= I Z_L = (10A \angle -53.13^\circ)(7Ω \angle 90^\circ)\\ &=(70V \angle 36.87^\circ)\\ V_C &= I Z_C = (10A \angle -53.13^\circ)(3Ω \angle -90^\circ)\\ &=(40V \angle -143.13^\circ)\\ \end{split}$$ Kirchhoff's voltage law: $$\begin{split} \sum V &= E - V_R -V_L-V_C = 0\\ E &= V_R +V_L+V_C\\ &= 30V \angle -53.13^\circ + 70V \angle 36.87^\circ + 30V \angle -143.13^\circ\\ &= (18-j24)+ (56 + j42)+ (-24 - j18)\\ &= 50V - j18 = 53V \angle -19.79^\circ \\ \end{split}$$ as applied.
Phasor diagram: Note that for the phasor diagram of Fig. 4, I is in phase with the voltage across the resistor and leads the voltage across the capacitor by $90^\circ$.
Fig. 4: Phasor diagram for the RLC Series circuit Configuration of Fig. 1.
Time domain: In the time domain, $$\begin{split} i &= \sqrt{2}(10) \sin(wt - 53.13^\circ) \\ &= 14.14 \sin(wt - 53.13^\circ)\\ v_R &= \sqrt{2}(30) \sin(wt - 53.13^\circ)\\ &= 42.42 \sin(wt - 53.13^\circ)\\ v_L &= \sqrt{2}(70) \sin(wt +36.87^\circ) \\ &= 98.98 \sin(wt + 36.87^\circ)\\ v_C &= \sqrt{2}(30) \sin(wt -143.13^\circ) \\ &= 42.42 \sin(wt -143.13^\circ)\\ \end{split}$$ A plot of all the voltages and the current of the circuit appears in Fig. 5.
Fig. 5: Waveforms for the RLC Series circuit Configuration of Fig. 1.
Power: The total power in watts delivered to the circuit is $$\begin{split} P_T &= EI \cos \theta_T\\ &= (50V)(10A) \cos 53.13^\circ\\ &= (500V) (0.6) = 300W\\ \end{split}$$ where E and I are effective values and $\theta_T$ is the phase angle between E and I, or $$\begin{split} P_T &= I^2 R \\ &= (10A)^2(3 Ω)=(100) (3) \\ &= 300W\\ \end{split}$$ where I is the effective value, or, finally $$\begin{split} P_T &= P_R + P_L + P_C\\ &= V_RI \cos \theta + V_LI cos \theta + V_CI cos \theta\\\\ &= (30)(10) \cos 0^\circ + (70)(10) \cos 90^\circ + (30)(10) \cos 90^\circ \\ &= 300 W +0+0 = 300W\\ \end{split}$$
Power factor: The power factor $F_p$ of the circuit is $\cos 53.13^\circ =0.6$ lagging, where $53.13^\circ$ is the phase angle between E and I.
If we write the basic power equation $P = EI \cos \theta$ as follows: $$\cos \theta = {P \over EI}$$ where E and I are the input quantities and P is the power delivered to the network, and then perform the following substitutions from the basic series ac circuit: $$\begin{split} \cos \theta &= {P \over EI} = {I^2 R \over EI} \\ &={I R \over E} = {R \over E/I} = {R \over Z_T}\\ \end{split}$$ we find $$\bbox[10px,border:1px solid grey]{F_p = \cos \theta = {R \over Z_T}}$$ Further supporting the fact that the impedance angle $\theta$ is also the phase angle between the input voltage and current for a series ac circuit. To determine the power factor, it is necessary only to form the ratio of the total resistance to the magnitude of the input impedance. For the case at hand, $$F_p = \cos \theta = {R \over Z_T} = {3Ω \over 5Ω} =0.6 \text{lagging}$$ as found above.