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}

Related article

### 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

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

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

Thank you for reading.

Related article

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

If you are interested, please check the link below.

- RL Series Circuit (Power Factor, Active and Reactive Power)
- RC Series Circuit (Power Factor, Active and Reactive Power)
- RLC Series Circuit (Power Factor, Active and Reactive Power)
- RC Parallel Circuit (Power Factor, Active and Reactive Power)
- RLC Parallel Circuit (Power Factor, Active and Reactive Power)