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).

We will now describe each procedure in turn.

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}

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}

[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}

[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}

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

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

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

Thank you for reading.