# RLC Parallel Circuit (Impedance, Phasor Diagram)

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

• Equation, magnitude, vector diagram, and impedance phase angle of RLC parallel circuit impedance

## Impedance of the RLC parallel circuit

An RLC parallel circuit is an electrical circuit consisting of a resistor $$R$$, an inductor $$L$$, and a capacitor $$C$$ connected in parallel, driven by a voltage source or current source.

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:

\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}=\frac{1}{j{\omega}C}\tag{3}
\end{eqnarray}

, where $${\omega}$$ is the angular frequency, which is equal to $$2{\pi}f$$, and $$X_L\left(={\omega}L\right)$$ is called inductive reactance, which is the resistive component of inductor $$L$$ and $$X_C\left(=\displaystyle\frac{1}{{\omega}C}\right)$$ is called capacitive reactance, which is the resistive component of capacitor $$C$$.

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}{j{\omega}L}+\frac{1}{\displaystyle\frac{1}{j{\omega}C}}\\
\\
&=&\frac{1}{R}+\frac{1}{j{\omega}L}+j{\omega}C\\
\\
&=&\frac{j{\omega}L+R+j{\omega}C×j{\omega}LR}{j{\omega}LR}\\
\\
&=&\frac{j{\omega}L+R+j^2{\omega}^2LCR}{j{\omega}LR}\\
\\
&=&\frac{j{\omega}L+R-{\omega}^2LCR}{j{\omega}LR}\\
\\
&=&\frac{R(1-{\omega}^2LC)+j{\omega}L}{j{\omega}LR}\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}{R}+\displaystyle\frac{1}{j{\omega}L}+j{\omega}C}=\frac{j{\omega}LR}{R(1-{\omega}^2LC)+j{\omega}L}\tag{5}
\end{eqnarray}

From equation (5), there is an imaginary unit $$j$$ in the denominator. To make sure that only the numerator has an imaginary unit $$j$$, multiply the denominator and numerator by "$$\{R(1-{\omega}^2LC)-j{\omega}L\}$$".

\begin{eqnarray}
{\dot{Z}}&=&\frac{j{\omega}LR\{R(1-{\omega}^2LC)-j{\omega}L\}}{\{R(1-{\omega}^2LC)+j{\omega}L\}\{R(1-{\omega}^2LC)-j{\omega}L\}}\\
\\
&=&\frac{j{\omega}LR^2(1-{\omega}^2LC)-j^2{\omega}^2L^2R}{R^2(1-{\omega}^2LC)^2-j{\omega}LR(1-{\omega}^2LC)+j{\omega}LR(1-{\omega}^2LC)-j^2{\omega}^2L^2}\\
\\
&=&\frac{j{\omega}LR^2(1-{\omega}^2LC)+{\omega}^2L^2R}{R^2(1-{\omega}^2LC)^2-j{\omega}LR(1-{\omega}^2LC)+j{\omega}LR(1-{\omega}^2LC)+{\omega}^2L^2}\\
\\
&=&\frac{{\omega}^2L^2R+j{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\\
\\
&=&\frac{{\omega}^2L^2R}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}+j\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\tag{6}
\end{eqnarray}

From the above, the impedance $${\dot{Z}}$$ of the RLC parallel circuit can be expressed as:

The impedance of the RLC parallel circuit

\begin{eqnarray}
{\dot{Z}}&=&\frac{1}{\displaystyle\frac{1}{R}+\displaystyle\frac{1}{j{\omega}L}+j{\omega}C}\\
\\
&=&\frac{{\omega}^2L^2R}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}+j\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\tag{7}
\end{eqnarray}

The magnitude of the inductive reactance $$X_L(={\omega}L)$$ and capacitive reactance $$X_C\left(=\displaystyle\frac{1}{{\omega}C}\right)$$ determine whether the impedance $${\dot{Z}}$$ of the RLC parallel circuit is inductive or capacitive.

• In the case of $$X_L{\;}{\lt}{\;}X_C$$
• In the case of $$X_L{\;}{\gt}{\;}X_C$$
• In the case of $$X_L=X_C$$

In the case of $$X_L{\;}{\lt}{\;}X_C$$

If the inductive reactance $$X_L$$ is smaller than the capacitive reactance $$X_C$$, the following equation holds.

\begin{eqnarray}
&&X_L{\;}{\lt}{\;}X_C\\
\\
{\Leftrightarrow}&&{\omega}L{\;}{\lt}{\;}\displaystyle\frac{1}{{\omega}C}\\
\\
{\Leftrightarrow}&&{\omega}^2LC{\;}{\lt}{\;}1\\
\\
{\Leftrightarrow}&&1-{\omega}^2LC{\;}{\gt}{\;}0\tag{8}
\end{eqnarray}

In this case, the imaginary part $$\displaystyle\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}$$ of the impedance $${\dot{Z}}$$ of the RLC parallel circuit becomes "positive" (in other words, the value multiplied by the imaginary unit "$$j$$" becomes "positive"), so the impedance $${\dot{Z}}$$ is inductive.

In the case of $$X_L{\;}{\gt}{\;}X_C$$

If the inductive reactance $$X_L$$ is bigger than the capacitive reactance $$X_C$$, the following equation holds.

\begin{eqnarray}
&&X_L{\;}{\gt}{\;}X_C\\
\\
{\Leftrightarrow}&&{\omega}L{\;}{\gt}{\;}\displaystyle\frac{1}{{\omega}C}\\
\\
{\Leftrightarrow}&&{\omega}^2LC{\;}{\gt}{\;}1\\
\\
{\Leftrightarrow}&&1-{\omega}^2LC{\;}{\lt}{\;}0\tag{9}
\end{eqnarray}

In this case, the imaginary part $$\displaystyle\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}$$ of the impedance $${\dot{Z}}$$ of the RLC parallel circuit becomes "negative" (in other words, the value multiplied by the imaginary unit "$$j$$" becomes "negative"), so the impedance $${\dot{Z}}$$ is capacitive.

In the case of $$X_L=X_C$$

If the inductive reactance is equal to the capacitive reactance, the following equation holds.

\begin{eqnarray}
&&X_L=X_C\\
\\
{\Leftrightarrow}&&{\omega}L=\displaystyle\frac{1}{{\omega}C}\\
\\
{\Leftrightarrow}&&{\omega}^2LC=1\\
\\
{\Leftrightarrow}&&1-{\omega}^2LC=0\tag{10}
\end{eqnarray}

In this case, the impedance $${\dot{Z}}$$ of the RLC parallel circuit is given by:

\begin{eqnarray}
{\dot{Z}}&=&\frac{{\omega}^2L^2R}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}+j\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\\
\\
&=&\frac{{\omega}^2L^2R}{R^2×0^2+{\omega}^2L^2}+j\frac{{\omega}LR^2×0}{R^2×0^2+{\omega}^2L^2}\\
\\
&=&R\tag{11}
\end{eqnarray}

In this case, the circuit is in parallel resonance. When parallel resonance is established, the part of the parallel circuit between the inductor $$L$$ and the capacitor $$C$$ is open. Therefore, the impedance $${\dot{Z}}$$ of the RLC parallel circuit is "$${\dot{Z}}=R$$". In this case, the angular frequency $${\omega}$$ and frequency $$f$$ are as follows:

\begin{eqnarray}
X_L&=&X_C\\
\\
{\omega}L&=&\frac{1}{{\omega}C}\\
\\
{\Leftrightarrow}{\omega}&=&\frac{1}{\displaystyle\sqrt{LC}}\\
\\
{\Leftrightarrow}f&=&\frac{1}{2{\pi}\displaystyle\sqrt{LC}}\tag{12}
\end{eqnarray}

### Magnitude of the impedance of the RLC parallel circuit

We have just obtained the impedance $${\dot{Z}}$$ expressed by the following equation.

\begin{eqnarray}
{\dot{Z}}=\frac{{\omega}^2L^2R}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}+j\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\tag{13}
\end{eqnarray}

The magnitude $$Z$$ of the impedance of the RLC parallel circuit is the absolute value of the impedance $${\dot{Z}}$$ in equation (13).

In more detail, the magnitude $$Z$$ of the impedance $${\dot{Z}}$$ can be obtained by adding the square of the real part $$\displaystyle\frac{{\omega}^2L^2R}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}$$ and the square of the imaginary part $$\displaystyle\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}$$ and taking the square root, which can be expressed in the following equation:

\begin{eqnarray}
Z&=&|{\dot{Z}}|\\
\\
&=&\displaystyle\sqrt{\left\{\frac{{\omega}^2L^2R}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\right\}^2+\left\{\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\right\}^2}\\
\\
&=&\displaystyle\sqrt{\frac{{\omega}^4L^4R^2+{\omega}^2L^2R^4(1-{\omega}^2LC)^2}{\left\{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2\right\}^2}}\\
\\
&=&\displaystyle\sqrt{\frac{{\omega}^2L^2R^2\left\{{\omega}^2L^2+R^2(1-{\omega}^2LC)^2\right\}}{\left\{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2\right\}^2}}\\
\\
&=&\displaystyle\sqrt{\frac{{\omega}^2L^2R^2\left\{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2\right\}}{\left\{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2\right\}^2}}\\
\\
&=&\displaystyle\sqrt{\frac{{\omega}^2L^2R^2}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}}\\
\\
&=&\frac{{\omega}LR}{\sqrt{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}}\tag{14}
\end{eqnarray}

Next, to express equation (14) in terms of "inductive reactance $$X_L$$" and "capacitive reactance $$X_C$$", the denominator and numerator are divided by $${\omega}L$$.

\begin{eqnarray}
Z&=&\frac{\displaystyle\frac{{\omega}LR}{{\omega}LR}}{\sqrt{\displaystyle\frac{1}{{\omega}^2L^2R^2}\left\{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2\right\}}}\\
\\
&=&\frac{1}{\sqrt{\displaystyle\frac{1}{{\omega}^2L^2}(1-{\omega}^2LC)^2+\displaystyle\frac{1}{R^2}}}\\
\\
&=&\frac{1}{\sqrt{\left(\displaystyle\frac{1}{R}\right)^2+\left(\displaystyle\frac{1}{{\omega}L}-{\omega}C\right)^2}}\\
\\
&=&\frac{1}{\sqrt{\left(\displaystyle\frac{1}{R}\right)^2+\left(\displaystyle\frac{1}{X_L}-\displaystyle\frac{1}{X_C}\right)^2}}\tag{15}
\end{eqnarray}

From the above, the magnitude $$Z$$ of the impedance of the RLC parallel circuit can be expressed as:

The magnitude of the impedance of the RLC parallel circuit

\begin{eqnarray}
Z&=&|{\dot{Z}}|\\
\\
&=&\frac{{\omega}LR}{\sqrt{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}}\\
\\
&=&\frac{1}{\sqrt{\left(\displaystyle\frac{1}{R}\right)^2+\left(\displaystyle\frac{1}{{\omega}L}-{\omega}C\right)^2}}\\
\\
&=&\frac{1}{\sqrt{\left(\displaystyle\frac{1}{R}\right)^2+\left(\displaystyle\frac{1}{X_L}-\displaystyle\frac{1}{X_C}\right)^2}}\tag{16}
\end{eqnarray}

Supplement

Some impedance $$Z$$ symbols have a ". (dot)" above them and are labeled $${\dot{Z}}$$.

$${\dot{Z}}$$ with this dot represents a vector.

If it has a dot (e.g. $${\dot{Z}}$$), it represents a vector (complex number), and if it does not have a dot (e.g. $$Z$$), it represents the absolute value (magnitude, length) of the vector.

## Vector diagram of the RLC parallel circuit

The vector diagram of the impedance $${\dot{Z}}$$ of the RLC parallel circuit can be drawn in the following steps.

How to draw a Vector Diagram

• Draw the vector of impedance $${\dot{Z}_{RE}}$$ of the real part of impedance $${\dot{Z}}$$
• Draw the vector of impedance $${\dot{Z}_{IM}}$$ of the imaginary part of impedance $${\dot{Z}}$$
• Combine the vectors

Let's take a look at each step in turn.

### Draw the vector of impedance $${\dot{Z}_{RE}}$$ of the real part of impedance $${\dot{Z}}$$

The impedance $${\dot{Z}_{RE}}$$ of the real part of impedance $${\dot{Z}}$$ is expressed by:

\begin{eqnarray}
{\dot{Z}_{RE}}=\frac{{\omega}^2L^2R}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\tag{17}
\end{eqnarray}

Therefore, the vector direction of the impedance $${\dot{Z}_{RE}}$$ is the direction of the real axis. How to determine the vector orientation will be explained in more detail later.

The magnitude (length) $$Z_{RE}$$ of the vector of impedances $${\dot{Z}_{RE}}$$ of the real part is given by

\begin{eqnarray}
Z_{RE}=|{\dot{Z}_{RE}}|=\displaystyle\sqrt{\left(\frac{{\omega}^2L^2R}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\right)^2}=\frac{{\omega}^2L^2R}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\tag{18}
\end{eqnarray}

### Draw the vector of impedance $${\dot{Z}_{IM}}$$ of the imaginary part of impedance $${\dot{Z}}$$

The impedance $${\dot{Z}_{IM}}$$ of the imaginary part of impedance $${\dot{Z}}$$ is expressed by:

\begin{eqnarray}
{\dot{Z}_{IM}}=j\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\tag{19}
\end{eqnarray}

The vector direction of the impedance $${\dot{Z}_{IM}}$$ depends on the magnitude of the "inductive reactance $$X_L$$" and "capacitive reactance $$X_C$$" shown below.

• In the case of $$X_L{\;}{\lt}{\;}X_C$$
• In the case of $$X_L{\;}{\gt}{\;}X_C$$
• In the case of $$X_L=X_C$$

We will now discuss the vector direction of the impedance $${\dot{Z}_{IM}}$$ in case "$$X_L{\;}{\lt}{\;}X_C$$"(we will discuss the vector direction in each case later).

If the inductive reactance $$X_L$$ is smaller than the capacitive reactance $$X_C$$, then "$$1-{\omega}^2LC{\;}{\gt}{\;}0$$".

Therefore, since the value $$\displaystyle\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}$$ multiplied by the imaginary unit "$$j$$" of the impedance $${\dot{Z}_{IM}}$$ is positive, the vector direction of the impedance $${\dot{Z}_{IM}}$$ is 90° counterclockwise around the real axis. How to determine the vector orientation will be explained in more detail later.

The magnitude (length) $$Z_{IM}$$ of the vector of impedances $${\dot{Z}_{IM}}$$ of the imaginary part is given by

\begin{eqnarray}
Z_{IM}=|{\dot{Z}_{IM}}|=\displaystyle\sqrt{\left(\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\right)^2}=\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\tag{20}
\end{eqnarray}

### Combine the vectors

Combining the vector of "impedance $${\dot{Z}}_{RE}$$ of the real part" and "impedance $${\dot{Z}}_{IM}$$ of the imaginary part" is the vector diagram of the impedance $${\dot{Z}}$$ of the RLC parallel circuit.

Again, the impedance $${\dot{Z}}$$ of an RLC parallel circuit is expressed by:

\begin{eqnarray}
{\dot{Z}}&=&\frac{{\omega}^2L^2R}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}+j\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\tag{21}
\end{eqnarray}

The vector direction of the impedance $${\dot{Z}}$$ of an RLC parallel circuit depends on the magnitude of the "inductive reactance $$X_L$$" and "capacitive reactance $$X_C$$" shown below.

• In the case of $$X_L{\;}{\lt}{\;}X_C$$
• In the case of $$X_L{\;}{\gt}{\;}X_C$$
• In the case of $$X_L=X_C$$

In the case of $$X_L{\;}{\lt}{\;}X_C$$

If the inductive reactance $$X_L$$ is smaller than the capacitive reactance $$X_C$$, then "$$1-{\omega}^2LC{\;}{\gt}{\;}0$$".

Therefore, since the value $$\displaystyle\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}$$ multiplied by the imaginary unit "$$j$$" of the impedance $${\dot{Z}_{IM}}$$ is positive, the vector direction of the impedance $${\dot{Z}_{IM}}$$ is 90° counterclockwise around the real axis. How to determine the vector orientation will be explained in more detail later.

Hence, the vector direction of the impedance $${\dot{Z}}$$ is upward to the right.

In the case of $$X_L{\;}{\gt}{\;}X_C$$

If the inductive reactance $$X_L$$ is bigger than the capacitive reactance $$X_C$$, then "$$1-{\omega}^2LC{\;}{\lt}{\;}0$$".

Therefore, since the value $$\displaystyle\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}$$ multiplied by the imaginary unit "$$j$$" of the impedance $${\dot{Z}_{IM}}$$ is negative, the vector direction of the impedance $${\dot{Z}_{IM}}$$ is 90° clockwise around the real axis. How to determine the vector orientation will be explained in more detail later.

Hence, the vector direction of the impedance $${\dot{Z}}$$ is downward to the right.

In the case of $$X_L=X_C$$

If the inductive reactance is equal to the capacitive reactance, then "$$1-{\omega}^2LC=0$$".

In this case, the vector direction of the impedance $${\dot{Z}}$$ is to the right because the impedance $${\dot{Z}}$$ of the RLC parallel circuit is "$${\dot{Z}}=R$$".

### Vector orientation

Here is a more detailed explanation of how vector orientation is determined.

Vector orientation

When an imaginary unit "$$j$$" is added to the expression, the direction of the vector is rotated by 90°.

• With "$$+j$$" is attached(or when the value multiplied by the imaginary unit "$$j$$" is "positive")
• The vector rotates 90° counterclockwise.
• With "$$-j$$" is attached(or when the value multiplied by the imaginary unit "$$j$$" is "negative")
• The vector rotates 90° clockwise.

The impedance $${\dot{Z}}$$ of an RLC parallel circuit is expressed by the following equation:

\begin{eqnarray}
{\dot{Z}_{IM}}=j\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}\tag{22}
\end{eqnarray}

In the case of $$X_L{\;}{\lt}{\;}X_C$$, since "$$1-{\omega}^2LC{\;}{\gt}{\;}0$$", the value multiplied by the imaginary unit "$$j$$" of the impedance $${\dot{Z}_{IM}}$$ is "positive". Therefore, the direction of vector $${\dot{Z}_{IM}}$$ is 90° counterclockwise around the real axis.

In the case of $$X_L{\;}{\gt}{\;}X_C$$, since "$$1-{\omega}^2LC{\;}{\lt}{\;}0$$", the value multiplied by the imaginary unit "$$j$$" of the impedance $${\dot{Z}_{IM}}$$ is "negative". Therefore, the direction of vector $${\dot{Z}_{IM}}$$ is 90° clockwise around the real axis.

## Impedance phase angle of the RLC parallel circuit

The impedance phase angle $${\theta}$$ of the RLC parallel circuit can be obtained from the vector diagram.

\begin{eqnarray}
{\tan}{\theta}&=&\displaystyle\frac{\displaystyle\frac{{\omega}LR^2(1-{\omega}^2LC)}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}}{\displaystyle\frac{{\omega}^2L^2R}{R^2(1-{\omega}^2LC)^2+{\omega}^2L^2}}\\
\\
&=&\displaystyle\frac{{\omega}LR^2(1-{\omega}^2LC)}{{\omega}^2L^2R}\tag{23}
\end{eqnarray}

Next, to express equation (23) in terms of "inductive reactance $$X_L$$" and "capacitive reactance $$X_C$$", the denominator and numerator are divided by $${\omega}^2L^2R^2$$.

\begin{eqnarray}
{\tan}{\theta}&=&\displaystyle\frac{\displaystyle\frac{{\omega}LR^2(1-{\omega}^2LC)}{{\omega}^2L^2R^2}}{\displaystyle\frac{{\omega}^2L^2R}{{\omega}^2L^2R^2}}\\
\\
&=&\displaystyle\frac{\displaystyle\frac{1}{{\omega}L}-{\omega}C}{\displaystyle\frac{1}{R}}\\
\\
&=&\displaystyle\frac{\displaystyle\frac{1}{X_L}-\displaystyle\frac{1}{X_C}}{\displaystyle\frac{1}{R}}\tag{24}
\end{eqnarray}

From the above, the impedance phase angle $${\theta}$$ of the RLC parallel circuit is expressed by the following equation:

Impedance phase angle of the RLC parallel circuit

\begin{eqnarray}
\\
\end{eqnarray}

The magnitude of the inductive reactance $$X_L(={\omega}L)$$ and capacitive reactance $$X_C\left(=\displaystyle\frac{1}{{\omega}C}\right)$$ determine whether the impedance phase angle $${\theta}$$ of the RLC parallel circuit is positive or negative.

• In the case of $$X_L{\;}{\lt}{\;}X_C$$
• In the case of $$X_L{\;}{\gt}{\;}X_C$$
• In the case of $$X_L=X_C$$

In the case of $$X_L{\;}{\lt}{\;}X_C$$

If the inductive reactance $$X_L$$ is smaller than the capacitive reactance $$X_C$$, the following equation holds.

\begin{eqnarray}
&&X_L{\;}{\lt}{\;}X_C\\
\\
{\Leftrightarrow}&&\frac{1}{X_L}-\frac{1}{X_C}{\;}{\gt}{\;}0\tag{26}
\end{eqnarray}

Therefore, the impedance phase angle $${\theta}$$ of the RLC parallel circuit is "positive".

In the case of $$X_L{\;}{\gt}{\;}X_C$$

If the inductive reactance $$X_L$$ is bigger than the capacitive reactance $$X_C$$, the following equation holds.

\begin{eqnarray}
&&X_L{\;}{\gt}{\;}X_C\\
\\
{\Leftrightarrow}&&\frac{1}{X_L}-\frac{1}{X_C}{\;}{\lt}{\;}0\tag{27}
\end{eqnarray}

Therefore, the impedance phase angle $${\theta}$$ of the RLC parallel circuit is "negative".

In the case of $$X_L=X_C$$

If the inductive reactance is equal to the capacitive reactance, the following equation holds.

\begin{eqnarray}
&&X_L=X_C\\
\\
{\Leftrightarrow}&&\frac{1}{X_L}-\frac{1}{X_C}=0\tag{28}
\end{eqnarray}

Therefore, the impedance phase angle $${\theta}$$ of the RLC parallel circuit is "$${\theta}=0{\mathrm{[rad]}}$$".

#### Summary

In this article, the following information on "RLC parallel circuit was explained.

1. Equation, magnitude, vector diagram, and impedance phase angle of RLC parallel circuit impedance