Circuit Theory/Phasors/Examples/example11/phasor solution

time domain
Can not substitute and get one differential equation. Must solve all equations simultaneously.

phasor domain
Can solve simultaneous linear equations in the phasor domain ... so must convert them to phasor domain:
 * $${V_s}(t) \rightarrow {\mathbb{V}_s} = 1$$
 * $${I_s}(t) \rightarrow {\mathbb{I}_s} = -2*sin(\frac{\pi}{3}) - j*2*cos(\frac{\pi}{3}) = -\sqrt{3} - j$$
 * $$i_1 \rightarrow \mathbb{I}_1$$
 * $$i_2 \rightarrow \mathbb{I}_2$$
 * $$i_C \rightarrow \mathbb{I}_C$$
 * $$i_L \rightarrow \mathbb{I}_L$$
 * $$v_1 \rightarrow {\mathbb{V}_1}$$
 * $$v_2 \rightarrow {\mathbb{V}_2}$$
 * $$v_C \rightarrow {\mathbb{V}_C}$$
 * $$v_L \rightarrow {\mathbb{V}_L}$$

now transform the calculus operations into the phasor domain ..
 * $${d \over dt}i\rightarrow j\omega\mathbb{I}$$
 * $${d \over dt}v(t)\rightarrow j\omega\mathbb{V}$$

So:
 * $$ \mathbb{V}_1 = R_1 * \mathbb{I}_1$$
 * $$ \mathbb{V}_2 = R_2 * \mathbb{I}_2$$
 * $$\mathbb{I}_C = C * j\omega \mathbb{V}_C$$
 * $$\mathbb{V}_L = L * j\omega \mathbb{I}_L$$
 * $$ \mathbb{I}_1 + \mathbb{I}_s - \mathbb{I}_2 - \mathbb{I}_C = 0$$
 * $$ \mathbb{I}_c - \mathbb{I}_s - \mathbb{I}_L = 0$$
 * $$ \mathbb{V}_1 + \mathbb{V}_2 - \mathbb{V}_S = 0$$
 * $$ \mathbb{V}_c + \mathbb{V}_L - \mathbb{V}_2 = 0$$

The symbolic solution is too complicated to mark up in wiki, but can be seen in a screen shot. Translating this into the time domain symbolically doubles the complexity (and mistakes).

The numeric solution is: