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

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

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

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

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

As an example, the parameters of the RLC 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: \(L=66.4{\;}{\mathrm{[mH]}}\)
• 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 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 RLC parallel 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}{\;}66.4×10^{-3}{\;}{\approx}{\;}25{\;}{\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}

Here, the composite reactance \(X\) of the inductor \(L\) and capacitor \(C\) can be obtained by the following equation.

\begin{eqnarray}
X&=&\left|\frac{1}{\displaystyle\frac{1}{X_L}-\displaystyle\frac{1}{X_C}}\right|=\left|\frac{1}{\displaystyle\frac{1}{25}-\displaystyle\frac{1}{50}}\right|=50{\;}{\mathrm{[{\Omega}]}}\tag{6}
\end{eqnarray}

The sum of the reciprocals of each impedance is the reciprocal of the impedance \({\dot{Z}}\) of the RLC 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}{{\dot{Z}_C}}\\
\\
&=&\frac{1}{R}+\frac{1}{jX_L}+\frac{1}{-jX_C}\\
\\
&=&\frac{1}{R}-j\frac{1}{X_L}+j\frac{1}{X_C}\\
\\
&=&\frac{1}{R}+j\left(\frac{1}{X_C}-\frac{1}{X_L}\right)\tag{7}
\end{eqnarray}

From equation (7), 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}}+\displaystyle\frac{1}{{\dot{Z}_C}}}\\
\\
&=&\frac{1}{\displaystyle\frac{1}{R}+j\left(\frac{1}{X_C}-\frac{1}{X_L}\right)}\tag{8}
\end{eqnarray}

The magnitude \(Z\) of the impedance of the RLC parallel circuit is the absolute value of the impedance \({\dot{Z}}\) in equation (8).

\begin{eqnarray}
Z=|{\dot{Z}}|&=&\frac{1}{\sqrt{\left(\displaystyle\frac{1}{R}\right)^2+\left(\displaystyle\frac{1}{X_C}-\displaystyle\frac{1}{X_L}\right)^2}}\\
\\
&=&\frac{1}{\sqrt{\left(\displaystyle\frac{1}{50}\right)^2+\left(\displaystyle\frac{1}{50}-\displaystyle\frac{1}{25}\right)^2}}\\
\\
&=&25\sqrt{2}{\;}{\mathrm{[{\Omega}]}}\tag{9}
\end{eqnarray}

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

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

\begin{eqnarray}
V=|{\dot{V}}|=|100|=100{\;}{\mathrm{[V]}}\tag{10}
\end{eqnarray}

From equations (9) and (10), the magnitude \(I\) of the current flowing in the RLC 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{11}
\end{eqnarray}

Since it is a parallel circuit, "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 \(C\)" are equal to "the magnitude \(V\) of the supply voltage", and the following formula is valid.

\begin{eqnarray}
V=V_R=V_L=V_C=100{\;}{\mathrm{[V]}}\tag{12}
\end{eqnarray}

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

\begin{eqnarray}
I_R&=&\frac{V_R}{R}=\frac{100}{50}=2{\;}{\mathrm{[A]}}\tag{13}\\
\\
I_L&=&\frac{V_L}{X_L}=\frac{100}{25}=4{\;}{\mathrm{[A]}}\tag{14}\\
\\
I_C&=&\frac{V_C}{X_C}=\frac{100}{50}=2{\;}{\mathrm{[A]}}\tag{15}
\end{eqnarray}

The magnitude \(I_X\) of the current flowing in the composite reactance \(X\) can be obtained by the following equation

\begin{eqnarray}
I_X=\frac{V}{X}=\frac{100}{50}=2{\;}{\mathrm{[A]}}\tag{16}
\end{eqnarray}

As can be seen from equation (16), the magnitude \(I_X\) of the current flowing in the composite reactance \(X\) is the difference (\(|I_L-I_C|\)) between "the magnitude \(I_L\) of the current flowing in the inductor \(L\)" and "the magnitude \(I_C\) of the current flowing in the capacitor \(C\)".

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

The power factor \({\cos}{\theta}\) of an RLC 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{17}
\end{eqnarray}

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

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

[RLC 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{19}
\end{eqnarray}

[RLC 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{22}
\end{eqnarray}

[RLC 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{25}
\end{eqnarray}

The power factor \({\cos}{\theta}\) of the RLC 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 (17) and (18).

\begin{eqnarray}
{\cos}{\theta}=\frac{P}{S}=\frac{200}{200\sqrt{2}}=\frac{1}{\sqrt{2}}\tag{28}
\end{eqnarray}

Summary

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

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

Thank you for reading.