User:1sfoerster/sandbox

Complex 2R, L, C, voltage circuit

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

Label Loops Junctions

Circuit marked up for analysis

The values of capacitor, inductor, and ${\displaystyle \omega }$ 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: ${\displaystyle V_{s},R_{1},R_{2},L,C,I_{s}}$
Unknowns: ${\displaystyle v_{1},v_{2},v_{C},v_{L},i_{1},i_{2},i_{C},i_{L}}$
Equations:
${\displaystyle v_{1}=R_{1}*i_{1}}$
${\displaystyle v_{2}=R_{2}*i_{2}}$
${\displaystyle i_{C}=C*{d \over dt}v_{C}}$
${\displaystyle v_{L}=L*{d \over dt}i_{L}}$
${\displaystyle i_{1}+I_{s}-i_{2}-i_{C}=0}$
${\displaystyle i_{c}-I_{s}-i_{L}=0}$
${\displaystyle v_{1}+v_{2}-V_{s}=0}$
${\displaystyle v_{C}+v_{L}-v_{2}=0}$

Phasor Symbolic and Numeric

time domain

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

phasor domain

Can solve simultaneous linear equations in the phasor domain ... so must convert them to phasor domain:

${\displaystyle {V_{s}}(t)\rightarrow {\mathbb {V} _{s}}=1}$
${\displaystyle {I_{s}}(t)\rightarrow {\mathbb {I} _{s}}=-2*sin({\frac {\pi }{3}})-j*2*cos({\frac {\pi }{3}})=-1.7321-j}$
${\displaystyle i_{1}\rightarrow \mathbb {I} _{1}}$
${\displaystyle i_{2}\rightarrow \mathbb {I} _{2}}$
${\displaystyle i_{C}\rightarrow \mathbb {I} _{C}}$
${\displaystyle i_{L}\rightarrow \mathbb {I} _{L}}$
${\displaystyle v_{1}\rightarrow {\mathbb {V} _{1}}}$
${\displaystyle v_{2}\rightarrow {\mathbb {V} _{2}}}$
${\displaystyle v_{C}\rightarrow {\mathbb {V} _{C}}}$
${\displaystyle v_{L}\rightarrow {\mathbb {V} _{L}}}$

now transform the calculus operations into the phasor domain ..

${\displaystyle {d \over dt}i\rightarrow j\omega \mathbb {I} }$
${\displaystyle {d \over dt}v(t)\rightarrow j\omega \mathbb {V} }$

So:

${\displaystyle \mathbb {V} _{1}=R_{1}*\mathbb {I} _{1}}$
${\displaystyle \mathbb {V} _{2}=R_{2}*\mathbb {I} _{2}}$
${\displaystyle \mathbb {I} _{C}=C*j\omega \mathbb {V} _{C}}$
${\displaystyle \mathbb {V} _{L}=L*j\omega \mathbb {I} _{L}}$
${\displaystyle \mathbb {I} _{1}+\mathbb {I} _{s}-\mathbb {I} _{2}-\mathbb {I} _{C}=0}$
${\displaystyle \mathbb {I} _{c}-\mathbb {I} _{s}-\mathbb {I} _{L}=0}$
${\displaystyle \mathbb {V} _{1}+\mathbb {V} _{2}-\mathbb {V} _{S}=0}$
${\displaystyle \mathbb {V} _{c}+\mathbb {V} _{L}-\mathbb {V} _{2}=0}$

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