# 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).

Procedure

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

Related article

The following article explains "Impedance of RLC Parallel Circuits" in detail. If you are interested, please check the link below.

### 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}

Supplement

The power factor $${\cos}{\theta}$$ of the RLC parallel circuit can also be obtained by the ratio of "the magnitude $$I_R$$ of the current flowing through the resistor $$R$$" to "the magnitude $$I$$ of the current flowing through the RLC parallel circuit". The following equation can be calculated, which is equal to equation (17).

\begin{eqnarray}
{\cos}{\theta}=\frac{I_R}{I}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}\tag{18}
\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}

Another solution

The apparent power $$S$$ can also be obtained by the following equation. The calculation results show that it is equal to equation (19).

\begin{eqnarray}
S&=&I^2Z=(2\sqrt{2})^2{\;}{\cdot}{\;}25\sqrt{2}=200\sqrt{2}{\;}{\mathrm{[VA]}}\tag{20}\\
\\
S&=&\frac{V^2}{Z}=\frac{100^2}{25\sqrt{2}}=200\sqrt{2}{\;}{\mathrm{[VA]}}\tag{21}
\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}

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 (22).

\begin{eqnarray}
P&=&{I_R}^2R=2^2{\;}{\cdot}{\;}50=200{\;}{\mathrm{[W]}}\tag{23}\\
\\
P&=&\frac{{V_R}^2}{R}=\frac{100^2}{50}=200{\;}{\mathrm{[W]}}\tag{24}
\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}

Another solution

Reactive power $$Q$$ can also be obtained by the following equation. The calculation results show that it is equal to equation (25).

\begin{eqnarray}
Q&=&{I_X}^2X=2^2{\;}{\cdot}{\;}50=200{\;}{\mathrm{[var]}}\tag{26}\\
\\
Q&=&\frac{{V}^2}{X}=\frac{100^2}{50}=200{\;}{\mathrm{[var]}}\tag{27}
\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

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