Phasor Example

Find $Z_{eq}$ “looking into” the two terminals. We’re dealing with a particular frequency here ($5\text{[rad/s]}$)

First, we convert inductors and capacitors and capacitors into inductances.

We have:

$$\begin{array}{rl}{Z}_{C_1}& =\frac{1}{j{\omega }_0{C}_1}=-\frac{j}{5\cdot 200\times 10^{-3}}=-j(1\Omega )=1\angle -90°[\Omega ]\\ Z_L& =j{\omega }_{0}L=j\cdot 5\cdot 2=j(10\Omega )=10\angle 90°[\Omega ]\\ Z_{C_2}& =\frac{1}{j{\omega }_0{C}_2}=-\frac{j}{5\cdot 500\times 10^{-3}}=-j\cdot 0.4\Omega =0.4\angle -90°[\Omega ]\end{array}$$

Doing some resistor simplification, we have:

$$\begin{array}{rl}{Z}_{eq}& =10\parallel (Z_{C_1}+Z_L+[6\parallel Z_{C_2}])\\ & =4.255+j(4.929)\Omega \\ & =6.511\angle 49.2°[\Omega ]\end{array}$$