RL Parallel Circuit (Power Factor, Active and Reactive Power)

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

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

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

Shown in the figure above is an RL parallel circuit with resistor \(R\) and inductor \(L\) connected in parallel.

As an example, the parameters of the RL parallel circuit are as follows.

• Supply voltage: \({\dot{V}}=100{\;}{\mathrm{[V]}}\)
• Frequency of power supply voltage: \(f=60{\;}{\mathrm{[Hz]}}\)
• Resistance value of resistor: \(R=50{\;}{\mathrm{[{\Omega}]}}\)
• 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 parallel 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 parallel 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 sum of the reciprocals of each impedance is the reciprocal of the impedance \({\dot{Z}}\) of the RL parallel circuit. Therefore, it can be expressed by the following equation.

\begin{eqnarray}
\frac{1}{{\dot{Z}}}&=&\frac{1}{{\dot{Z}_R}}+\frac{1}{{\dot{Z}_L}}\\
\\
&=&\frac{1}{R}+\frac{1}{jX_L}\\
\\
&=&\frac{1}{R}-j\frac{1}{X_L}\tag{4}
\end{eqnarray}

From equation (4), by interchanging the denominator and numerator, the following equation is obtained:

\begin{eqnarray}
{\dot{Z}}&=&\frac{1}{\displaystyle\frac{1}{{\dot{Z}_R}}+\displaystyle\frac{1}{{\dot{Z}_L}}}\\
\\
&=&\frac{1}{\displaystyle\frac{1}{R}-j\displaystyle\frac{1}{X_L}}\tag{5}
\end{eqnarray}

The magnitude \(Z\) of the impedance of the RL parallel circuit is the absolute value of the impedance \({\dot{Z}}\) in equation (5).

\begin{eqnarray}
Z=|{\dot{Z}}|&=&\frac{1}{\sqrt{\left(\displaystyle\frac{1}{R}\right)^2+\left(\displaystyle\frac{1}{X_L}\right)^2}}\\
\\
&=&\frac{1}{\sqrt{\left(\displaystyle\frac{1}{50}\right)^2+\left(\displaystyle\frac{1}{50}\right)^2}}\\
\\
&=&25\sqrt{2}{\;}{\mathrm{[{\Omega}]}}\tag{6}
\end{eqnarray}

Calculate the magnitude \(I\) of the current flowing in the RL parallel circuit

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

\begin{eqnarray}
V=|{\dot{V}}|=|100|=100{\;}{\mathrm{[V]}}\tag{7}
\end{eqnarray}

From equations (6) and (7), the magnitude \(I\) of the current flowing in the RL parallel circuit can be obtained by the following equation

\begin{eqnarray}
I=\frac{V}{Z}=\frac{100}{25\sqrt{2}}=2\sqrt{2}{\;}{\mathrm{[A]}}\tag{8}
\end{eqnarray}

Since it is a parallel circuit, "the magnitude \(V_R\) of the voltage across the resistor \(R\)" and "the magnitude \(V_L\) of the voltage across the inductor \(L\)" are equal to "the magnitude \(V\) of the supply voltage", and the following formula is valid.

\begin{eqnarray}
V=V_R=V_L=100{\;}{\mathrm{[V]}}\tag{9}
\end{eqnarray}

Therefore, "the magnitude \(I_R\) of the current flowing through the resistor \(R\)" and "the magnitude \(I_L\) of the current flowing through the inductor \(L\)" can be obtained by the following formula.

\begin{eqnarray}
I_R&=&\frac{V_R}{R}=\frac{100}{50}=2{\;}{\mathrm{[A]}}\tag{10}\\
\\
I_L&=&\frac{V_L}{X_L}=\frac{100}{50}=2{\;}{\mathrm{[A]}}\tag{11}
\end{eqnarray}

Calculate the power factor \({\cos}{\theta}\) of the RL parallel circuit

The power factor \({\cos}{\theta}\) of an RL parallel 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{Z}{R}=\frac{25\sqrt{2}}{50}=\frac{1}{\sqrt{2}}\tag{12}
\end{eqnarray}

Calculate the active power \(P\), reactive power \(Q\), and apparent power \(S\) of the RL parallel circuit

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

[RL parallel circuit] Calculation of apparent power \(S\)

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

\begin{eqnarray}
S=VI=100{\;}{\cdot}{\;}2\sqrt{2}=200\sqrt{2}{\;}{\mathrm{[VA]}}\tag{14}
\end{eqnarray}

[RL parallel circuit] Calculation of active power \(P\)

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

\begin{eqnarray}
P=VI{\cos}{\theta}=100{\;}{\cdot}{\;}2\sqrt{2}{\;}{\cdot}{\;}\frac{1}{\sqrt{2}}=200{\;}{\mathrm{[W]}}\tag{17}
\end{eqnarray}

[RL parallel 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}}=100{\;}{\cdot}{\;}2\sqrt{2}{\;}{\cdot}{\;}\sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^2}=200{\;}{\mathrm{[var]}}\tag{20}
\end{eqnarray}

The power factor \({\cos}{\theta}\) of the RL parallel 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 (12) and (13).

\begin{eqnarray}
{\cos}{\theta}=\frac{P}{S}=\frac{200}{200\sqrt{2}}=\frac{1}{\sqrt{2}}\tag{23}
\end{eqnarray}

Summary

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

1. Power factor \({\cos}{\theta}\) of the RL parallel Circuit
2. Active power \(P\), Reactive power \(Q\), and Apparent power \(S\) of the RL parallel circuit

Thank you for reading.