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|
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.