## PHYS345 Electricity and Electronics

Effective Impedance Example

Find the effective impedance of the following reactive network.

The two branches are in parallel; let's label the inductive one (on the left) as Z1 and the capacitive one as Z2. So

 Z1 = (4 + j3) ohm Z2 = (12 - j5) ohm

Now what?

One way to proceed is to evaluate Zeq in the following expression:

Zeq-1 = Z1-1 + Z2-1

Unfortunately this involves too much conversion between standard and polar form (and back). It is easier to do a bit of algebra and find that

Zeq = Z1Z2 / (Z1 + Z2)

Then if we convert Z1, Z2, and Z1 + Z2 to polar form, the evaluation of the effective impedance may be done more readily.

 Z1 = (4 + j3) ohm = 5 ohm \/ 36.9 deg Z2 = (12 - j5) ohm = 13 ohm \/ -22.6 deg Z1 + Z2 = (16 - j2) ohm = 16.1 ohm \/ -7.1 deg

 So Zeq = Z1Z2 / (Z1 + Z2) = (5 ohm \/ 36.9 deg)(13 ohm \/ -22.6 deg) / (16.1 ohm \/ -7.1 deg) = 4.0 ohm \/ 21.4 deg = (3.7 + j1.5) ohm

The equivalent impedance is inductive.

"http://www.physics.udel.edu/~watson/phys345/examples/impedance-example.html"
Last updated Sept. 27, 1999.
Copyright George Watson, Univ. of Delaware, 1999.