# RL Series Circuit (Power Factor, Active and Reactive Power)

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

• Power factor $${\cos}{\theta}$$ of the RL series Circuit
• Active power $$P$$, Reactive power $$Q$$, and Apparent power $$S$$ of the RL series circuit

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

Shown in the figure above is an RL series circuit with resistor $$R$$, inductor $$L$$, and capacitor $$C$$ connected in series.

As an example, the parameters of the RL 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: $$L=132.7{\;}{\mathrm{[mH]}}$$

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

Procedure

• Calculate the magnitude $$Z$$ of the impedance of the RL series circuit
• Calculate the magnitude $$I$$ of the current flowing in the RL series circuit
• Calculate the power factor $${\cos}{\theta}$$ of the RL series circuit
• Calculate the active power $$P$$, reactive power $$Q$$, and apparent power $$S$$ of the RL 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 RL series circuit

The impedance $${\dot{Z}}_R$$ of the resistor $$R$$ and the impedance $${\dot{Z}}_L$$ of the inductor $$L$$ 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}
\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 $$L$$.

The inductive reactance $$X_L$$ can be obtained by the following equations.

\begin{eqnarray}
X_L={\omega}L=2{\pi}fL&=&2{\pi}{\;}{\cdot}{\;}60{\;}{\cdot}{\;}132.7×10^{-3}\\
\\
&{\approx}&50{\;}{\mathrm{[{\Omega}]}}\tag{3}
\end{eqnarray}

The impedance $${\dot{Z}}$$ of the RL series circuit is the sum of the respective impedance and is as follows.

\begin{eqnarray}
{\dot{Z}}&=&{\dot{Z}_R}+{\dot{Z}_L}\\
\\
&=&R+jX_L\\
\\
&=&50\sqrt{3}+j50{\;}{\mathrm{[{\Omega}]}}\tag{4}
\end{eqnarray}

The magnitude $$Z$$ of the impedance of the RL series circuit is the absolute value of the impedance $${\dot{Z}}$$ in equation (4).

\begin{eqnarray}
Z=|{\dot{Z}}|&=&\sqrt{R^2+{X_L}^2}\\
\\
&=&\sqrt{(50\sqrt{3})^2+50^2}\\
\\
&=&100{\;}{\mathrm{[{\Omega}]}}\tag{5}
\end{eqnarray}

### Calculate the magnitude $$I$$ of the current flowing in the RL series circuit

The magnitude $$V$$ of the supply voltage is the following value.

\begin{eqnarray}
V=|{\dot{V}}|=|200|=200{\;}{\mathrm{[V]}}\tag{6}
\end{eqnarray}

From equations (5) and (6), the magnitude $$I$$ of the current flowing in the RL series circuit can be obtained by the following equation

\begin{eqnarray}
I=\frac{V}{Z}=\frac{200}{100}=2{\;}{\mathrm{[A]}}\tag{7}
\end{eqnarray}

Since it is a series circuit, "the magnitude $$I_R$$ of the current through the resistor $$R$$" and "the magnitude $$I_L$$ of the current through the inductor $$L$$" are equal to "the magnitude $$I$$ of current through the RL series circuit", and the following formula is valid.

\begin{eqnarray}
I=I_R=I_L=2{\;}{\mathrm{[A]}}\tag{8}
\end{eqnarray}

Therefore, "the magnitude $$V_R$$ of the voltage across the resistor $$R$$" and "the magnitude $$V_L$$ of the voltage across the inductor $$L$$" can be obtained by the following formula.

\begin{eqnarray}
V_R=I_RR=2{\;}{\cdot}{\;}50\sqrt{3}=100\sqrt{3}{\;}{\mathrm{[V]}}\tag{9}\\
\\
V_L=I_LX_L=2{\;}{\cdot}{\;}50=100{\;}{\mathrm{[V]}}\tag{10}
\end{eqnarray}

### Calculate the power factor $${\cos}{\theta}$$ of the RL series circuit

The power factor $${\cos}{\theta}$$ of an RL 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{11}
\end{eqnarray}

Supplement

The power factor $${\cos}{\theta}$$ of the RL 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 (11).

\begin{eqnarray}
{\cos}{\theta}=\frac{V_R}{V}=\frac{100\sqrt{3}}{200}=\frac{\sqrt{3}}{2}\tag{12}
\end{eqnarray}

### Calculate the active power $$P$$, reactive power $$Q$$, and apparent power $$S$$ of the RL series circuit

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

#### [RL 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{13}
\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 (13).

\begin{eqnarray}
S&=&I^2Z=2^2{\;}{\cdot}{\;}100=400{\;}{\mathrm{[VA]}}\tag{14}\\
\\
S&=&\frac{V^2}{Z}=\frac{200^2}{100}=400{\;}{\mathrm{[VA]}}\tag{15}
\end{eqnarray}

#### [RL 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{16}
\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 (16).

\begin{eqnarray}
P&=&{I_R}^2R=2^2{\;}{\cdot}{\;}50\sqrt{3}=200\sqrt{3}{\;}{\mathrm{[W]}}\tag{17}\\
\\
P&=&\frac{{V_R}^2}{R}=\frac{(100\sqrt{3})^2}{50\sqrt{3}}=200\sqrt{3}{\;}{\mathrm{[W]}}\tag{18}
\end{eqnarray}

#### [RL 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{19}
\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 (19).

\begin{eqnarray}
Q&=&{I_L}^2X_L=2^2{\;}{\cdot}{\;}50=200{\;}{\mathrm{[var]}}\tag{20}\\
\\
Q&=&\frac{{V_L}^2}{X_L}=\frac{100^2}{50}=200{\;}{\mathrm{[var]}}\tag{21}
\end{eqnarray}

The power factor $${\cos}{\theta}$$ of the RL 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{22}
\end{eqnarray}

#### Summary

1. Power factor $${\cos}{\theta}$$ of the RL series Circuit
2. Active power $$P$$, Reactive power $$Q$$, and Apparent power $$S$$ of the RL series circuit
In AC circuits, articles related to power factor $${\cos}{\theta}$$, active power $$P$$, reactive power $$Q$$, and apparent power $$S$$ are listed below.