Circuit Theory/Phasors/Examples/example11/phasor solution

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 .. done with kirchhoff and node analysis and something inbetween code

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

$V_s(t)\to {𝕍}_s=1$

$I_s(t)\to {𝕀}_s=-2*sin(\frac{\pi }{3})-j*2*cos(\frac{\pi }{3})=-\sqrt{3}-j$

$i_1\to {𝕀}_1$

$i_2\to {𝕀}_2$

$i_C\to {𝕀}_C$

$i_L\to {𝕀}_L$

$v_1\to {𝕍}_1$

$v_2\to {𝕍}_2$

$v_C\to {𝕍}_C$

$v_L\to {𝕍}_L$

now transform the calculus operations into the phasor domain ..

$\frac{d}{dt}i\to j\omega 𝕀$

$\frac{d}{dt}v(t)\to j\omega 𝕍$

So:

${𝕍}_1=R_1*{𝕀}_1$

${𝕍}_2=R_2*{𝕀}_2$

${𝕀}_C=C*j\omega {𝕍}_C$

${𝕍}_L=L*j\omega {𝕀}_L$

${𝕀}_1+{𝕀}_s-{𝕀}_2-{𝕀}_C=0$

${𝕀}_c-{𝕀}_s-{𝕀}_L=0$

${𝕍}_1+{𝕍}_2-{𝕍}_S=0$

${𝕍}_c+{𝕍}_L-{𝕍}_2=0$

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:

variable	real	imaginary	magnitude	radians	degrees	time domain
i1	3.70	-2.39i	4.41	-0.574	-32.9	4.41 cos(t-0.574)
i2	0.648	1.20i	1.36	1.07	61.6	1.36 cos(t + 1.07)
ic	1.32	-4.59i	4.78	-1.29	-73.9	4.78 cos(t-1.29)
iL	3.06	-3.59i	4.72	-0.866	-49.6	4.72 cos(t-0.866)
v1	3.70	-2.39i	4.41	-0.574	-32.9	4.41 cos(t-0.574)
v2	1.30	2.39i	2.72	1.07	61.6	2.72 cos(t+1.07)
vc	-2.30	-0.662i	2.39	-2.86	-164	2.39 cos(t-2.86)
vL	3.60	3.06i	4.72	0.705	40.4	4.72 cos(t+0.705)