Regarding the RLC series circuit, this article will explain the information below.

- Power factor \({\cos}{\theta}\) of the RLC series Circuit
- Active power \(P\), Reactive power \(Q\), and Apparent power \(S\) of the RLC series circuit

## [RLC Circuit] Power factor & Active, Reactive, and Apparent power

Shown in the figure above is an RLC series circuit with resistor \(R\), inductor \(L\), and capacitor \(C\) connected in series.

As an example, the parameters of the RLC series circuit are as follows.

- Supply voltage: \({\dot{V}}=200{\;}{\mathrm{[V]}}\)
- Frequency of power supply voltage: \(f=60{\;}{\mathrm{[Hz]}}\)
- Resistance value of resistor: \(R=50\sqrt{3}{\;}{\mathrm{[{\Omega}]}}\)
- Inductance of inductor: \(L=265.4{\;}{\mathrm{[mH]}}\)
- Capacitance of capacitor: \(C=53{\;}{\mathrm{[μF]}}\)

The power factor \({\cos}{\theta}\), active power \(P\), reactive power \(Q\), and apparent power \(S\) of the RLC series circuit can be obtained by the following procedure (steps 1 to 4).

Procedure

- Calculate the magnitude \(Z\) of the impedance of the RLC series circuit
- Calculate the magnitude \(I\) of the current flowing in the RLC series circuit
- Calculate the power factor \({\cos}{\theta}\) of the RLC series circuit
- Calculate the active power \(P\), reactive power \(Q\), and apparent power \(S\) of the RLC series circuit

We will now describe each procedure in turn.

Supplement

There are three types of power in an AC circuit: active power \(P\), reactive power \(Q\), and apparent power \(S\).

- Active power \(P\)
- It is the power consumed by the resistor \(R\) and is also called power consumption. The unit is [W].

- Reactive power \(Q\)
- It is the power that is not consumed by the resistor \(R\). The power that an inductor or capacitor stores or releases is called reactive power. The unit is [var].

- Apparent power \(S\)
- The power is the sum of active power \(P\) and reactive power \(Q\). The unit is [VA].

### Calculate the magnitude \(Z\) of the impedance of the RLC series circuit

The impedance \({\dot{Z}}_R\) of the resistor \(R\), the impedance \({\dot{Z}}_L\) of the inductor \(L\), and the impedance \({\dot{Z}}_C\) of the capacitor \(C\) can be expressed by the following equations, respectively.

\begin{eqnarray}

{\dot{Z}_R}&=&R\tag{1}\\

\\

{\dot{Z}_L}&=&jX_L=j{\omega}L\tag{2}\\

\\

{\dot{Z}_C}&=&-jX_C=-j\frac{1}{{\omega}C}\tag{3}

\end{eqnarray}

, where \({\omega}\) is the angular frequency, which is equal to \(2{\pi}f\), and \(X_L\) is called inductive reactance, which is the resistive component of inductor \(L\) and \(X_C\) is called capacitive reactance, which is the resistive component of capacitor \(C\).

The inductive reactance \(X_L\) and capacitive reactance \(X_C\) can be obtained by the following equations.

\begin{eqnarray}

X_L&=&{\omega}L=2{\pi}fL=2{\pi}{\;}{\cdot}{\;}60{\;}{\cdot}{\;}265.4×10^{-3}{\;}{\approx}{\;}100{\;}{\mathrm{[{\Omega}]}}\tag{4}\\

\\

X_C&=&\frac{1}{{\omega}C}=\frac{1}{2{\pi}fC}=\frac{1}{2{\pi}{\;}{\cdot}{\;}60{\;}{\cdot}{\;}53×10^{-6}}{\;}{\approx}{\;}50{\;}{\mathrm{[{\Omega}]}}\tag{5}

\end{eqnarray}

The difference between inductive reactance \(X_L\) and capacitive reactance \(X_C\) is called the composite reactance \(X\). The composite reactance \(X\) can be obtained by the following equation

\begin{eqnarray}

X&=&|X_L-X_C|=|100-50|=50{\;}{\mathrm{[{\Omega}]}}\tag{6}

\end{eqnarray}

The impedance \({\dot{Z}}\) of the RLC series circuit is the sum of the respective impedance and is as follows.

\begin{eqnarray}

{\dot{Z}}&=&{\dot{Z}_R}+{\dot{Z}_L}+{\dot{Z}_C}\\

\\

&=&R+jX_L-jX_C\\

\\

&=&R+j(X_L-X_C)\\

\\

&=&50\sqrt{3}+j(100-50)\\

\\

&=&50\sqrt{3}+j50{\;}{\mathrm{[{\Omega}]}}\tag{7}

\end{eqnarray}

The magnitude \(Z\) of the impedance of the RLC series circuit is the absolute value of the impedance \({\dot{Z}}\) in equation (7).

\begin{eqnarray}

Z=|{\dot{Z}}|&=&\sqrt{R^2+(X_L-X_C)^2}\\

\\

&=&\sqrt{(50\sqrt{3})^2+(100-50)^2}\\

\\

&=&\sqrt{(50\sqrt{3})^2+50^2}\\

\\

&=&100{\;}{\mathrm{[{\Omega}]}}\tag{8}

\end{eqnarray}

Related article

### Calculate the magnitude \(I\) of the current flowing in the RLC series circuit

The magnitude \(V\) of the supply voltage is the following value.

\begin{eqnarray}

V=|{\dot{V}}|=|200|=200{\;}{\mathrm{[V]}}\tag{9}

\end{eqnarray}

From equations (8) and (9), the magnitude \(I\) of the current flowing in the RLC series circuit can be obtained by the following equation

\begin{eqnarray}

I=\frac{V}{Z}=\frac{200}{100}=2{\;}{\mathrm{[A]}}\tag{10}

\end{eqnarray}

Since it is a series circuit, "the magnitude \(I_R\) of the current through the resistor \(R\)", "the magnitude \(I_L\) of the current through the inductor \(L\)", and "the magnitude \(I_C\) of the current through the capacitor \(C\)" are equal to "the magnitude \(I\) of current through the RLC series circuit", and the following formula is valid.

\begin{eqnarray}

I=I_R=I_L=I_C=2{\;}{\mathrm{[A]}}\tag{11}

\end{eqnarray}

Therefore, "the magnitude \(V_R\) of the voltage across the resistor \(R\)", "the magnitude \(V_L\) of the voltage across the inductor \(L\)", and "the magnitude \(V_C\) of the voltage across the capacitor \(C\)" can be obtained by the following formula.

\begin{eqnarray}

V_R&=&I_RR=2{\;}{\cdot}{\;}50\sqrt{3}=100\sqrt{3}{\;}{\mathrm{[V]}}\tag{12}\\

\\

V_L&=&I_LX_L=2{\;}{\cdot}{\;}100=200{\;}{\mathrm{[V]}}\tag{13}\\

\\

V_C&=&I_CX_C=2{\;}{\cdot}{\;}50=100{\;}{\mathrm{[V]}}\tag{14}

\end{eqnarray}

The magnitude \(I_X\) of the voltage across the composite reactance \(X\) can be obtained by the following equation.

\begin{eqnarray}

V_X=IX=2{\;}{\cdot}{\;}50=100{\;}{\mathrm{[V]}}\tag{15}

\end{eqnarray}

As can be seen from equation (15), the magnitude \(V_X\) of the voltage across the composite reactance \(X\) is the difference (\(|V_L-V_C|\)) between "the magnitude \(V_L\) of the voltage across the inductor \(L\)" and "the magnitude \(V_C\) of the voltage across the capacitor \(C\)".

### Calculate the power factor \({\cos}{\theta}\) of the RLC series circuit

The power factor \({\cos}{\theta}\) of an RLC series circuit is the ratio of the impedance magnitude \(Z\) to the resistance \(R\) and can be obtained by the following equation

\begin{eqnarray}

{\cos}{\theta}=\frac{R}{Z}=\frac{50\sqrt{3}}{100}=\frac{\sqrt{3}}{2}\tag{16}

\end{eqnarray}

Supplement

The power factor \({\cos}{\theta}\) of the RLC series circuit can also be obtained by the ratio of "the magnitude \(V_R\) of the voltage across the resistor \(R\)" to "magnitude \(V\) of the supply voltage". The following equation can be calculated, which is equal to equation (16).

\begin{eqnarray}

{\cos}{\theta}=\frac{V_R}{V}=\frac{100\sqrt{3}}{200}=\frac{\sqrt{3}}{2}\tag{17}

\end{eqnarray}

### Calculate the active power \(P\), reactive power \(Q\), and apparent power \(S\) of the RLC series circuit

By finding "the magnitude \(V\) of the power supply voltage", "the magnitude \(I\) of the current flowing in the RLC series circuit", and "the power factor \({\cos}{\theta}\) of the RLC series circuit," the active power \(P\), reactive power \(Q\), and apparent power \(S\) can be calculated.

#### [RLC series circuit] Calculation of apparent power \(S\)

The apparent power \(S\) can be obtained by the following equation.

\begin{eqnarray}

S=VI=200{\;}{\cdot}{\;}2=400{\;}{\mathrm{[VA]}}\tag{18}

\end{eqnarray}

Another solution

The apparent power \(S\) can also be obtained by the following equation. The calculation results show that it is equal to equation (18).

\begin{eqnarray}

S&=&I^2Z=2^2{\;}{\cdot}{\;}100=400{\;}{\mathrm{[VA]}}\tag{19}\\

\\

S&=&\frac{V^2}{Z}=\frac{200^2}{100}=400{\;}{\mathrm{[VA]}}\tag{20}

\end{eqnarray}

#### [RLC series circuit] Calculation of active power \(P\)

The active power \(P\) can be obtained by the following equation

\begin{eqnarray}

P=VI{\cos}{\theta}=200{\;}{\cdot}{\;}2{\;}{\cdot}{\;}\frac{\sqrt{3}}{2}=200\sqrt{3}{\;}{\mathrm{[W]}}\tag{21}

\end{eqnarray}

Another solution

Since the effective power \(P\) is the power consumed by the resistor \(R\), it can also be obtained by the following equation. The calculation results show that it is equal to equation (21).

\begin{eqnarray}

P&=&{I_R}^2R=2^2{\;}{\cdot}{\;}50\sqrt{3}=200\sqrt{3}{\;}{\mathrm{[W]}}\tag{22}\\

\\

P&=&\frac{{V_R}^2}{R}=\frac{(100\sqrt{3})^2}{50\sqrt{3}}=200\sqrt{3}{\;}{\mathrm{[W]}}\tag{23}

\end{eqnarray}

#### [RLC series circuit] Calculation of reactive power \(Q\)

The reactive power \(Q\) can be obtained by the following equation

\begin{eqnarray}

Q=VI{\sin}{\theta}=VI\sqrt{1-{\cos}^2{\theta}}=200{\;}{\cdot}{\;}2{\;}{\cdot}{\;}\sqrt{1-\left(\frac{\sqrt{3}}{2}\right)^2}=200{\;}{\mathrm{[var]}}\tag{24}

\end{eqnarray}

Another solution

Reactive power \(Q\) can also be obtained by the following equation. The calculation results show that it is equal to equation (24).

\begin{eqnarray}

Q&=&{I}^2X=2^2{\;}{\cdot}{\;}50=200{\;}{\mathrm{[var]}}\tag{25}\\

\\

Q&=&\frac{{V_X}^2}{X}=\frac{100^2}{50}=200{\;}{\mathrm{[var]}}\tag{26}

\end{eqnarray}

The power factor \({\cos}{\theta}\) of the RLC series circuit can also be obtained by the ratio of "active power \(P\)" to "apparent power \(S\)". The calculation yields the following equation, which is equal to equations (16) and (17).

\begin{eqnarray}

{\cos}{\theta}=\frac{P}{S}=\frac{200\sqrt{3}}{400}=\frac{\sqrt{3}}{2}\tag{27}

\end{eqnarray}

#### Summary

This article described the following information about the "RLC series circuit".

- Power factor \({\cos}{\theta}\) of the RLC series Circuit
- Active power \(P\), Reactive power \(Q\), and Apparent power \(S\) of the RLC series circuit

Thank you for reading.

Related article

In AC circuits, articles related to power factor \({\cos}{\theta}\), active power \(P\), reactive power \(Q\), and apparent power \(S\) are listed below.

If you are interested, please check the link below.

- RL Series Circuit (Power Factor, Active and Reactive Power)
- RC Series Circuit (Power Factor, Active and Reactive Power)
- RL Parallel Circuit (Power Factor, Active and Reactive Power)
- RC Parallel Circuit (Power Factor, Active and Reactive Power)
- RLC Parallel Circuit (Power Factor, Active and Reactive Power)