User:1sfoerster/sandbox

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Complex 2R, L, C, voltage circuit

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[edit]

Circuit marked up for analysis

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[edit]

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

Phasor Symbolic and Numeric[edit]

time domain[edit]

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

phasor domain[edit]

mupad symbolic solution .. code
mupad numeric solution .. code

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:

variable real imaginary magnitude angle
v1 2 -1.73j 2.644 ∠ -0.7131
v2 -1 1.73j 1,9982 ∠ 2.0949
vC -3.598 - 0.76795j 3.6789 ∠ -2.9313
VL 2.598 2.5j 3.6055 ∠ 0.7662
i1 2 - 1.73j 2.644 ∠ -0.7131
i2 -0.5 0.866j 1 ∠ 2.0944
iC 0.76785 - 3.598j 3.679 ∠ -1.3605
iL 2.5 - 2.598j 3.6055 ∠ - 0.8046