# 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).

Procedure

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

Supplement

The power factor $${\cos}{\theta}$$ of the RL 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 RL parallel circuit". The following equation can be calculated, which is equal to equation (12).

\begin{eqnarray}
{\cos}{\theta}=\frac{I_R}{I}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}\tag{13}
\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}

Another solution

The apparent power $$S$$ can also be obtained by the following equation. The calculation results show that it is equal to equation (14).

\begin{eqnarray}
S&=&I^2Z=(2\sqrt{2})^2{\;}{\cdot}{\;}25\sqrt{2}=200\sqrt{2}{\;}{\mathrm{[VA]}}\tag{15}\\
\\
S&=&\frac{V^2}{Z}=\frac{100^2}{25\sqrt{2}}=200\sqrt{2}{\;}{\mathrm{[VA]}}\tag{16}
\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}

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 (17).

\begin{eqnarray}
P&=&{I_R}^2R=2^2{\;}{\cdot}{\;}50=200{\;}{\mathrm{[W]}}\tag{18}\\
\\
P&=&\frac{{V_R}^2}{R}=\frac{100^2}{50}=200{\;}{\mathrm{[W]}}\tag{19}
\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}

Another solution

Reactive power $$Q$$ can also be obtained by the following equation. The calculation results show that it is equal to equation (20).

\begin{eqnarray}
Q&=&{I_L}^2X_L=2^2{\;}{\cdot}{\;}50=200{\;}{\mathrm{[var]}}\tag{21}\\
\\
Q&=&\frac{{V_L}^2}{X_L}=\frac{100^2}{50}=200{\;}{\mathrm{[var]}}\tag{22}
\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

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