Polar Impedance Calculator

Table of Contents

An online calculator to add, subtract, multiply and divide polar impedances is presented. Operations on polar impedances are needed in order to find equivalent impedances in AC circuits.
\( \) \( \) In what follows \( j \) is the imaginary unit such that \( j^2 = -1 \) or \( j = \sqrt{-1} \).

Impedances in Complex Forms

Impedances are represented by complex numbers in polar form as follows:
\( Z = \rho \: \; \angle \; \: \theta \) , where \( \rho \) is the magnitude of \( Z \) and \( \theta \) its phase in degrees or radians.
\( Z \) in standard complex form is written as
\( Z = \rho \cos \theta + j \; \rho \sin \theta \)

1) A capacitor of capacitance \( C \) has an impedance \( Z_C \) whose magnitude is \( \dfrac{1}{\omega C} \) , where \( \omega = 2 \pi f \) and \( f \) is the frequency of the signal, and a phase equal to \( - \dfrac {\pi}{2} \). Hence \( Z_C \) is written
in standard complex form as
\( Z_C = - \dfrac{j}{\omega C} \)
and in polar form as
\( Z_C = \dfrac{1}{\omega C} \; \angle \; - \dfrac {\pi}{2} \)

2) An inductor of inductance \( L \) has an impedance \( Z_L \) whose magnitude is \( \omega L \) , where \( \omega = 2 \pi f \) and \( f \) is the frequency of the signal, and a phase equal to \( \dfrac {\pi}{2} \). Hence \( Z_L \) is written
in standard complex form as
\( Z_L = j \; \omega L \)
and in polar form as
\( Z_L = \omega L \; \angle \; \dfrac {\pi}{2} \)

3) A resistor of resistance \( R \) has an impedance \( Z_R \) whose magnitude is \( R \) and a phase equal to \( 0 \). Hence \( Z_R \) is written
in standard complex form as
\( Z_R = R + j \; 0 \)
and in polar form as
\( Z_R = R \; \angle \; 0 \)

Formulas to Add, Subtract, Multiply and Divide Polar Impedances

Adding polar impedances

Let \( z_1 = \rho_1 \; \angle \; \theta_1 \) and \( z_2 = \rho_2 \; \angle \; \theta_2 \)
Write \( Z_1 \) and \(Z_2 \) in standard complex forms
\( Z_1 = \rho_1 \cos \theta_1 + j \; \rho_1 \sin \theta_1 \)
\(Z_2 = \rho_2 \cos \theta_2 + j \; \rho_2 \sin \theta_2 \)
\( Z_1 + Z_2 = \rho_1 \cos \theta_1 + \rho_2 \cos \theta_2 + j \; ( \rho_1 \sin \theta_1 + \rho_2 \sin \theta_2) \)
in polar form
\[ Z_1 + Z_2 = \rho \; \; \angle \; \theta \]
where
\( \rho = \sqrt {(\rho_1 \cos \theta_1 + \rho_2 \cos \theta_2)^2 + (\rho_1 \sin \theta_1 + \rho_2 \sin \theta_2)^2} \)
and
\( \theta = \arctan (\dfrac{\rho_1 \sin \theta_1 + \rho_2 \sin \theta_2}{\rho_1 \cos \theta_1 + \rho_2 \cos \theta_2}) \)

Subtracting polar impedances

In standard complex form
\( Z_1 - Z_2 = \rho_1 \cos \theta_1 - \rho_2 \cos \theta_2 + j \; ( \rho_1 \sin \theta_1 - \rho_2 \sin \theta_2) \)
in polar form
\[ Z_1 - Z_2 = \rho \; \; \angle \; \theta \]
where
\( \rho = \sqrt {(\rho_1 \cos \theta_1 - \rho_2 \cos \theta_2)^2 + (\rho_1 \sin \theta_1 - \rho_2 \sin \theta_2)^2} \)
and
\( \theta = \arctan (\dfrac{\rho_1 \sin \theta_1 - \rho_2 \sin \theta_2}{\rho_1 \cos \theta_1 - \rho_2 \cos \theta_2}) \)

It is much easier to multiply and divide polar impedances

Multiplying polar impedances

\[ Z_1 \times Z_2 = \rho \; \; \angle \; \theta \] where
\( \rho = \rho_1 \times \rho_2 \)
and
\( \theta = \theta_1 + \theta_2 \)

Dividing polar impedances

\[ \dfrac{Z_1}{Z_2} = \rho \; \; \angle \; \theta \] where
\( \rho = \dfrac{\rho_1}{\rho_2} \)
and
\( \theta = \theta_1 - \theta_2 \)

Use of Polar Impedance Calculator

1 - Enter the magnitude and phase \( \rho_1 \) and \( \theta_1 \) of impedance \( Z_1 \) and the magnitude and phase \( \rho_2 \) and \( \theta_2 \) of impedance \( Z_2 \) as real numbers with the phases \( \theta_1 \) and \( \theta_2\) in either radians or degrees and then press "Calculate".
The outputs are:
\( Z_1 \) and \( Z_2 \) in complex standard form
and
\( Z_1+Z_2\) , \( Z_1-Z_2\) , \( Z_1 \times Z_2 \) and \( \dfrac{Z_1}{Z_2} \) in polar form with phase in degrees.