# RC Series Circuit

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In an RC series circuit, the resistor and the capacitor are connected in series configuration, which means that the current flowing through the circuit passes through both components one after the other. The resistor resists the current flow, while the capacitor stores energy in the form of an electric field.
Lets explore the RL series circuit in the following example.
##### Phasor Notation

Refer to [Fig. 1].
$$i = 7.07 \sin (wt+53.13^\circ \Rightarrow I= 5V \angle 53.13^\circ$$
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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