# RC Series Circuit Configuration with example   Whatsapp  ### Phasor Notation

$$i = 7.07 \sin (wt+53.13^\circ \Rightarrow I= 5V \angle 53.13^\circ$$ Refer to Fig. 1. Fig. 1: RC Series circuit Configuration. Fig. 2: Applying phasor notation to the RC Series circuit Configuration of Fig. 1.
where $$Z_T =Z_1 + Z_2 \\ =6 \angle 0^\circ + 8 \angle -90^\circ\\ =6Ω - j8Ω\\ Z_T = 10 \angle -53.13^\circ$$ Impedance diagram: See Fig. 3. Fig. 3: Impedance diagram for the RC Series circuit Configuration of Fig. 1.
Voltage: $$E = IZ_T = (5A \angle 53.13^\circ) (10Ω \angle -53.13^\circ) = 50A \angle 0^\circ$$ Ohms laws: $$\begin{split} V_R &= I Z_R = (5A \angle 53.13^\circ)(6Ω \angle 0^\circ)\\ &=(30V \angle 53.13^\circ)\\ V_C &= I Z_C = (5A \angle 53.13^\circ)(8Ω \angle -90^\circ)\\ &=(40V \angle -36.87^\circ)\\ \end{split}$$ Kirchhoff's voltage law: $$\begin{split} \sum V &= E - V_R -V_C = 0\\ E &= V_R +V_C\\ &= 30V \angle 53.13^\circ + 40V \angle -36.87^\circ\\ &= (18+j24)+ (32 - j24)\\ &= 50V + j0 = 50V \angle 0^\circ \\ \end{split}$$ 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 RC Series circuit Configuration of Fig. 1.
Time domain: In the time domain, $$e = \sqrt{2}(50) \sin wt = 70.70 \sin wt$$ $$v_R = \sqrt{2}(30) \sin(wt + 53.13^\circ)\\ = 42.42 \sin(wt + 53.13^\circ)$$ $$v_C = \sqrt{2}(40) \sin(wt -36.87^\circ) \\ = 56.56 \sin(wt -36.87^\circ)$$ A plot of all of the voltages and the current of the circuit appears in Fig. 5. Note again that $i$ and $v_R$ are in phase and that $v_C$ lags $i$ by $90^\circ$. Fig. 5: Waveforms for the RC Series circuit Configuration of Fig. 1.
Power: The total power in watts delivered to the circuit is $$\begin{split} P_T &= EI \cos \theta_T\\ &= (50V)(5V) \cos 53.13^\circ\\ &= (250V) (0.6) = 150W\\ \end{split}$$ where E and I are effective values and $\theta_T$ is the phase angle between E and I, or $$\begin{split} P_T &= I^2 R \\ &= (5A)^2(6 Ω)=(25) (6) \\ &= 150W\\ \end{split}$$ where I is the effective value, or, finally $$\begin{split} P_T &= P_R + P_C\\ &= V_RI \cos \theta + V_CI cos \theta\\ &= (30)(5) \cos 0^\circ + (40)(5) \cos 90^\circ \\ &= 150 W +0 = 150W\\ \end{split}$$
Power factor: The power factor $F_p$ of the circuit is $\cos 53.13^\circ =0.6$ leading, 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, $$F_p = \cos \theta = {R \over Z_T} = {6Ω \over 10Ω} =0.6 \text{leading}$$ as found above.

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