User:1sfoerster/sandbox

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

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

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

time domain

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

phasor domain

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