Impedance in Series and Parallel

If your circuit has more than one element in it you're going to need a way to combine all their impedances into one equivalent impedance (just like combining resistors in DC circuits). This topic will cover how to simplify impedances when they're in series or parallel.

Two impedances in series are combined by adding together, just like resistances

Simplify the circuit below:

We need the impedances in rectangular form in order to add them so we'll do that first. \(Z_1 = 5\angle 45^\circ = 5\cos(45^\circ) + j5\sin(45^\circ)\) \(Z_1 = 3.54 + j3.54\) \(Z_2 = 3\angle 30^\circ = 3\cos(30^\circ) + j3\sin(30^\circ) \) \(Z_2 = 2.60 + j1.5\) So \(Z_{eq} = Z_1 + Z_2 = (3.54 + 2.60) + j(3.54 + 1.5)\) \(Z_{eq} = 6.14 + j5.04\) Now convert it back into phasor form: \(Z_{eq} = 7.94\angle 39.4^\circ\)

Two impedances in parallel are combined just like two resistances in parallel, using the following formulas: \(Z_1 \parallel Z_2 = \frac{Z_1Z_2}{Z_1+Z_2}\) \(Z_1 \parallel Z_2 = \frac{1}{\frac{1}{Z_1}+\frac{1}{Z_2}}\)

Find the equivalent impedance of the following circuit section:

First we get these in rectangular form. \(Z_1 = 5\angle 45^\circ = 3.54 + j3.54\) \(Z_2 = 3\angle 30^\circ = 2.60 + j1.5\) We need \(Z_1Z_2\) and \(Z_1+Z_2\) \(Z_1Z_2 = 15\angle 75^\circ\) \(Z_1 + Z_2 = 6.14 + j5.04 = 7.94\angle 39.4^\circ\) Now \(Z_{eq} = \frac{Z_1Z_2}{Z_1+Z_2} = \frac{15}{7.94}\angle (75-39.4)^\circ\) \(Z_{eq} = 1.89\angle 35.6^\circ\)