# RC Parallel Circuit (Impedance, Phasor Diagram)

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

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

## Iimpedance of the RC parallel circuit

An RC parallel circuit (also known as an RC filter or RC network) is an electrical circuit consisting of a resistor $$R$$ and a capacitor $$C$$ connected in parallel, driven by a voltage source or current source.

The impedance $${\dot{Z}}_R$$ of the resistor $$R$$ and the impedance $${\dot{Z}}_C$$ of capacitor $$C$$ can be expressed by the following equations:

\begin{eqnarray}
{\dot{Z}_R}&=&R\tag{1}\\
\\
{\dot{Z}_C}&=&-jX_C=-j\frac{1}{{\omega}C}=\frac{1}{j{\omega}C}\tag{2}
\end{eqnarray}

,where $${\omega}$$ is the angular frequency, which is equal to $$2{\pi}f$$, 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 RC 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}_C}}\\
\\
&=&\frac{1}{R}+\frac{1}{\displaystyle\frac{1}{j{\omega}C}}\\
\\
&=&\frac{1}{R}+j{\omega}C\\
\\
&=&\frac{1+j{\omega}CR}{R}\tag{3}
\end{eqnarray}

From equation (3), by interchanging the denominator and numerator, the following equation is obtained:

\begin{eqnarray}
{\dot{Z}}=\frac{1}{\displaystyle\frac{1}{R}+j{\omega}C}=\frac{R}{1+j{\omega}CR}\tag{4}
\end{eqnarray}

From equation (4), 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 "$$1-j{\omega}CR$$".

\begin{eqnarray}
{\dot{Z}}&=&\frac{R(1-j{\omega}CR)}{(1+j{\omega}CR)(1-j{\omega}CR)}\\
\\
&=&\frac{R-j{\omega}CR^2}{1-j{\omega}CR+j{\omega}CR-j^2{\omega}^2C^2R^2}\\
\\
&=&\frac{R-j{\omega}CR^2}{1+{\omega}^2C^2R^2}\\
\\
&=&\frac{R}{1+{\omega}^2C^2R^2}-j\frac{{\omega}CR^2}{1+{\omega}^2C^2R^2}\tag{5}
\end{eqnarray}

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

The impedance of the RC parallel circuit

\begin{eqnarray}
{\dot{Z}}=\frac{1}{\displaystyle\frac{1}{R}+j{\omega}C}=\frac{R}{1+{\omega}^2C^2R^2}-j\frac{{\omega}CR^2}{1+{\omega}^2C^2R^2}{\mathrm{[{\Omega}]}}\tag{6}
\end{eqnarray}

### Magnitude of the impedance of the RC parallel circuit

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

\begin{eqnarray}
{\dot{Z}}=\frac{R}{1+{\omega}^2C^2R^2}-j\frac{{\omega}CR^2}{1+{\omega}^2C^2R^2}{\mathrm{[{\Omega}]}}\tag{7}
\end{eqnarray}

The magnitude $$Z$$ of the impedance of the RC parallel circuit is the absolute value of the impedance $${\dot{Z}}$$ in equation (7).

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

\begin{eqnarray}
Z&=&|{\dot{Z}}|\\
\\
&=&\displaystyle\sqrt{\left(\displaystyle\frac{R}{1+{\omega}^2C^2R^2}\right)^2+\left(\displaystyle\frac{{\omega}CR^2}{1+{\omega}^2C^2R^2}\right)^2}\\
\\
&=&\displaystyle\sqrt{\displaystyle\frac{R^2+{\omega}^2C^2R^4}{\left(1+{\omega}^2C^2R^2\right)^2}}\\
\\
&=&\displaystyle\sqrt{\displaystyle\frac{R^2\left(1+{\omega}^2C^2R^2\right)}{\left(1+{\omega}^2C^2R^2\right)^2}}\\
\\
&=&\displaystyle\sqrt{\displaystyle\frac{R^2}{1+{\omega}^2C^2R^2}}\\
\\
&=&\displaystyle\frac{R}{\displaystyle\sqrt{1+{\omega}^2C^2R^2}}\\
\\
&=&\displaystyle\frac{1}{\displaystyle\sqrt{\left(\displaystyle\frac{1}{R}\right)^2+\left({\omega}C\right)^2}}\tag{8}
\end{eqnarray}

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

The magnitude of the impedance of the RC parallel circuit

\begin{eqnarray}
Z=|{\dot{Z}}|=\displaystyle\frac{R}{\displaystyle\sqrt{1+{\omega}^2C^2R^2}}=\displaystyle\frac{1}{\displaystyle\sqrt{\left(\displaystyle\frac{1}{R}\right)^2+\left({\omega}C\right)^2}}{\mathrm{[{\Omega}]}}\tag{9}
\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 RC parallel circuit

The vector diagram of the impedance $${\dot{Z}}$$ of the RC 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{R}{1+{\omega}^2C^2R^2}\tag{10}
\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{R}{1+{\omega}^2C^2R^2}\right)^2}=\frac{R}{1+{\omega}^2C^2R^2}\tag{11}
\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}CR^2}{1+{\omega}^2C^2R^2}\tag{12}
\end{eqnarray}

Therefore, the vector direction of the impedance $${\dot{Z}_{IM}}$$ is 90° clockwise around the real axis (with "$$-j$$", it rotates 90° clockwise). 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}CR^2}{1+{\omega}^2C^2R^2}\right)^2}=\frac{{\omega}CR^2}{1+{\omega}^2C^2R^2}\tag{13}
\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 RC parallel circuit.

The magnitude (length) $$Z$$ of the vector of impedance $${\dot{Z}}$$ of an RC parallel circuit is expressed by:

\begin{eqnarray}
Z=|{\dot{Z}}|=\displaystyle\frac{R}{\displaystyle\sqrt{1+{\omega}^2C^2R^2}}=\displaystyle\frac{1}{\displaystyle\sqrt{\left(\displaystyle\frac{1}{R}\right)^2+\left({\omega}C\right)^2}}{\mathrm{[{\Omega}]}}\tag{14}
\end{eqnarray}

Supplement

The magnitude (length) $$Z$$ of the vector of the impedance $${\dot{Z}}$$ of the RC parallel circuit can also be obtained using the Pythagorean theorem in the vector diagram.

### 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
• The vector rotates 90° counterclockwise.
• With "$$-j$$" is attached
• The vector rotates 90° clockwise.

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

\begin{eqnarray}
{\dot{Z}_{IM}}=-j\frac{{\omega}CR^2}{1+{\omega}^2C^2R^2}\tag{15}
\end{eqnarray}

Since the expression for the impedance $${\dot{Z}}_{IM}$$ in the imaginary part has a "$$-j$$", the direction of the vector $${\dot{Z}}_{IM}$$ is 90° clockwise around the real axis.

## Impedance phase angle of the RC parallel circuit

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

The impedance phase angle $${\theta}$$ of the RC parallel circuit is expressed by the following equation:

\begin{eqnarray}
{\tan}{\theta}&=&\displaystyle\frac{-\displaystyle\frac{{\omega}CR^2}{1+{\omega}^2C^2R^2}}{\displaystyle\frac{R}{1+{\omega}^2C^2R^2}}\\
\\
&=&-{\omega}CR\\
\\
{\Leftrightarrow}{\theta}&=&{\tan}^{-1}\left(-{\omega}CR\right)\\
\\
&=&-{\tan}^{-1}\left({\omega}CR\right)\tag{16}
\end{eqnarray}

#### Summary

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

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