User:1sfoerster/sandbox

Given that the voltage source is defined by $$ V_s(t) = cos(t)$$ and the current source is defined by $$ I_s(t) = 2sin(t-\frac{pi}{3})$$, find all other voltages, currents and check power.

Label Loops Junctions


The values of capacitor, inductor, and $$\omega$$ are all odd. Obviously the goal is to make the phasor math easy. Not intending this problem to be worked symbolically. But we shall anyway .. in the phasor domain.

Knowns, Unknowns and Equations

 * Knowns: $$ V_s, R_1, R_2, L, C, I_s $$
 * Unknowns: $$ v_1, v_2, v_C, v_L, i_1, i_2, i_C, i_L $$
 * Equations:
 * $$ v_1 = R_1 * i_1$$
 * $$ v_2 = R_2 * i_2$$
 * $$ i_C = C * {d \over dt}v_C$$
 * $$ v_L = L * {d \over dt}i_L$$
 * $$ i_1 + I_s - i_2 -i_C = 0$$
 * $$ i_c - I_s - i_L = 0$$
 * $$ v_1 + v_2 - V_s = 0$$
 * $$ v_C + v_L - v_2 = 0$$

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:
 * $${V_s}(t) \rightarrow {\mathbb{V}_s} = 1$$
 * $${I_s}(t) \rightarrow {\mathbb{I}_s} = -2*sin(\frac{\pi}{3}) - j*2*cos(\frac{\pi}{3}) = - 1.7321 - j$$
 * $$i_1 \rightarrow \mathbb{I}_1$$
 * $$i_2 \rightarrow \mathbb{I}_2$$
 * $$i_C \rightarrow \mathbb{I}_C$$
 * $$i_L \rightarrow \mathbb{I}_L$$
 * $$v_1 \rightarrow {\mathbb{V}_1}$$
 * $$v_2 \rightarrow {\mathbb{V}_2}$$
 * $$v_C \rightarrow {\mathbb{V}_C}$$
 * $$v_L \rightarrow {\mathbb{V}_L}$$

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 is doubles the complexity (and mistakes).

The numeric solution is: