# LC Parallel Circuit (Impedance, Phasor Diagram)

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

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

## Impedance of the LC parallel circuit

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

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}}_L&=&jX_L=j{\omega}L\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 $$X_L\left(={\omega}L\right)$$ 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 LC parallel circuit. Therefore, it can be expressed by the following equation:

\begin{eqnarray}
\frac{1}{{\dot{Z}}}&=&\frac{1}{{\dot{Z}_L}}+\frac{1}{{\dot{Z}_C}}\\
\\
&=&\frac{1}{j{\omega}L}+\frac{1}{\displaystyle\frac{1}{j{\omega}C}}\\
\\
&=&\frac{1}{j{\omega}L}+j{\omega}C\\
\\
&=&\frac{1-{\omega}^2LC}{j{\omega}L}\tag{3}
\end{eqnarray}

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

\begin{eqnarray}
{\dot{Z}}=\frac{j{\omega}L}{1-{\omega}^2LC}=j\frac{{\omega}L}{1-{\omega}^2LC}\tag{4}
\end{eqnarray}

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

The impedance of the LC parallel circuit

\begin{eqnarray}
{\dot{Z}}=j\frac{{\omega}L}{1-{\omega}^2LC}\tag{5}
\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 LC 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{6}
\end{eqnarray}

In this case, the imaginary part $$\displaystyle\frac{{\omega}L}{1-{\omega}^2LC}$$ of the impedance $${\dot{Z}}$$ of the LC 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{7}
\end{eqnarray}

In this case, the imaginary part $$\displaystyle\frac{{\omega}L}{1-{\omega}^2LC}$$ of the impedance $${\dot{Z}}$$ of the LC 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{8}
\end{eqnarray}

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

\begin{eqnarray}
{\dot{Z}}&=&j\frac{{\omega}L}{1-{\omega}^2LC}\\
\\
&=&j\frac{{\omega}L}{0}\\
\\
&=&∞\tag{9}
\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, and 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{10}
\end{eqnarray}

### Magnitude of the impedance of the LC parallel circuit

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

\begin{eqnarray}
{\dot{Z}}=j\frac{{\omega}L}{1-{\omega}^2LC}\tag{11}
\end{eqnarray}

The magnitude $$Z$$ of the impedance of the LC parallel circuit is the absolute value of the impedance $${\dot{Z}}$$ in equation (11).

In more detail, the magnitude $$Z$$ of the impedance $${\dot{Z}}$$ is obtained by taking the square root of the square of the imaginary part $$\displaystyle\frac{{\omega}L}{1-{\omega}^2LC}$$, which can be expressed in the following equation.

\begin{eqnarray}
Z=|{\dot{Z}}|=\sqrt{\left(\frac{{\omega}L}{1-{\omega}^2LC}\right)^2}=\left|\frac{{\omega}L}{1-{\omega}^2LC}\right|\tag{12}
\end{eqnarray}

Next, to express equation (12) in terms of "inductive reactance $$X_L$$" and "capacitive reactance $$X_L$$", the denominator and numerator are divided by $${\omega}L$$.

\begin{eqnarray}
Z&=&\left|\frac{\displaystyle\frac{{\omega}L}{{\omega}L}}{\displaystyle\frac{1-{\omega}^2LC}{{\omega}L}}\right|\\
\\
&=&\left|\frac{1}{\displaystyle\frac{1}{{\omega}L}-{\omega}C}\right|\\
\\
&=&\left|\frac{1}{\displaystyle\frac{1}{X_L}-\displaystyle\frac{1}{X_C}}\right|\tag{13}
\end{eqnarray}

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

The magnitude of the impedance of the LC parallel circuit

\begin{eqnarray}
Z&=&|{\dot{Z}}|\\
\\
&=&\left|\frac{{\omega}L}{1-{\omega}^2LC}\right|\\
\\
&=&\left|\frac{1}{\displaystyle\frac{1}{{\omega}L}-{\omega}C}\right|\\
\\
&=&\left|\frac{1}{\displaystyle\frac{1}{X_L}-\displaystyle\frac{1}{X_C}}\right|\tag{14}
\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 LC parallel circuit

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

\begin{eqnarray}
{\dot{Z}}=j\frac{{\omega}L}{1-{\omega}^2LC}\tag{15}
\end{eqnarray}

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

\begin{eqnarray}
Z&=&|{\dot{Z}}|\\
\\
&=&\left|\frac{{\omega}L}{1-{\omega}^2LC}\right|\tag{16}
\end{eqnarray}

The vector direction of the impedance $${\dot{Z}}$$ of an LC 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 $$X_L{\;}{\lt}{\;}X_C$$

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

Therefore, since the value $$\displaystyle\frac{{\omega}L}{1-{\omega}^2LC}$$ multiplied by the imaginary unit "$$j$$" of the impedance $${\dot{Z}}$$ is positive, the vector direction of the impedance $${\dot{Z}}$$ 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.

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 $$\displaystyle\frac{{\omega}L}{1-{\omega}^2LC}$$ multiplied by the imaginary unit "$$j$$" of the impedance $${\dot{Z}}$$ is negative, the vector direction of the impedance $${\dot{Z}}$$ 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.

### 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 LC parallel circuit is expressed by the following equation:

\begin{eqnarray}
{\dot{Z}}=j\frac{{\omega}L}{1-{\omega}^2LC}\tag{17}
\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 $${\dot{Z}}$$ of the LC parallel circuit is "positive". Therefore, the direction of vector $${\dot{Z}}$$ 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 $${\dot{Z}}$$ of the LC parallel circuit is "negative". Therefore, the direction of vector $${\dot{Z}}$$ is 90° clockwise around the real axis.

## Impedance phase angle of the LC parallel circuit

The impedance angle $${\theta}$$ varies depending on the magnitude of the inductive reactance $$X_L={\omega}L$$ and the capacitive reactance $$X_C=\displaystyle\frac{1}{{\omega}C}$$.

• In the case of $$X_L{\;}{\lt}{\;}X_C$$
• If the inductive reactance $$X_L$$ is smaller than the capacitive reactance $$X_C$$, the impedance angle $${\theta}$$ will be the following value.
\begin{eqnarray}
\end{eqnarray}
• In the case of $$X_L{\;}{\gt}{\;}X_C$$
• If the inductive reactance $$X_L$$ is bigger than the capacitive reactance $$X_C$$, the impedance angle $${\theta}$$ will be the following value.
\begin{eqnarray}