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.