Circuit analysis -- Find active and reactive power in the branch...

  • #1
208
1

Homework Statement


Given the circuit of sinusoidal current (attachment 1) with given data:
[tex]\underline{E}=100V,\underline{E_1}=40V,\underline{Z}=(10+j10)\Omega,\omega=10^5rad/s,L=1mH,
C=0.1uF.[/tex] Find [itex]\underline{I_L},\underline{U_{16}}[/itex], active and reactive power in the branch [itex]2-5[/itex].

im1.PNG

im2.PNG


2. The attempt at a solution
Using the loop current analysis we can find four loops (attachment 2) that correspond to linear system of four complex equations:

[tex]C_1: (2\underline{Z}+jX_L)\underline{I_{C1}}-\underline{Z}\underline{I_{C2}}-\underline{Z}\underline{I_{C3}}+\underline{Z}\underline{I_{C4}}=\underline{E_1}-\underline{E}[/tex]
[tex]C_2: 2\underline{Z}\underline{I_{C2}}-\underline{Z}\underline{I_{C1}}+\underline{Z}\underline{I_{C3}}+\underline{Z}\underline{I_{C4}}=\underline{E_1}+\underline{E}[/tex]
[tex]C_3: 2\underline{Z}\underline{I_{C3}}-\underline{Z}\underline{I_{C1}}+\underline{Z}\underline{I_{C2}}-\underline{Z}\underline{I_{C4}}=\underline{E}[/tex]
[tex]C_4: (2\underline{Z}-jX_C)\underline{I_{C4}}+2\underline{Z}\underline{I_{C1}}+\underline{Z}\underline{I_{C2}}-\underline{Z}\underline{I_{C3}}=\underline{E_1}-\underline{E}[/tex]

This gives:
[tex](20+j120)\underline{I_{C1}}-(10+j10)\underline{I_{C2}}-(10+j10)\underline{I_{C3}}+(20+j20)\underline{I_{C4}}=-60[/tex]

[tex](-10-j10)\underline{I_{C1}}+(20+j20)\underline{I_{C2}}+(10+j10)\underline{I_{C3}}+(10+j10)\underline{I_{C4}}=140[/tex]

[tex](-10-j10)\underline{I_{C1}}+(10+j10)\underline{I_{C2}}+(20+j20)\underline{I_{C3}}+(-10-j10)\underline{I_{C4}}=100[/tex]

[tex](20+j20)\underline{I_{C1}}+(10+j10)\underline{I_{C2}}-(10+j10)\underline{I_{C3}}+(20-j80)\underline{I_{C4}}=-60[/tex]

After reducing to 3x3 system:

[tex](30+j230)\underline{I_{C1}}+(-10-j10)\underline{I_{C3}}+(50+j50)\underline{I_{C4}}=20[/tex]

[tex](10+j110)\underline{I_{C1}}+(10+j10)\underline{I_{C3}}+(10+j10)\underline{I_{C4}}=20[/tex]

[tex](40+j140)\underline{I_{C1}}+(-20-j20)\underline{I_{C3}}+(40-j60)\underline{I_{C4}}=-120[/tex]


After reducing to 2x2 system:

[tex](40+j340)\underline{I_{C1}}+(60+j60)\underline{I_{C4}}=60[/tex]

[tex](-20-j320)\underline{I_{C1}}+(-60-j160)\underline{I_{C4}}=-160[/tex]

[tex]
\begin{bmatrix}
40+j340 & 60+j60 \\
-20-j320 & -60-j160 \\
\end{bmatrix} \begin{bmatrix}
\underline{I_{C1}} \\
\underline{I_{C4}} \\
\end{bmatrix}=\begin{bmatrix}
60 \\
-160 \\
\end{bmatrix}\Rightarrow
[/tex]

[tex]\begin{bmatrix}
40+j340 & 60+j60 & 60+j0 \\
-20-j320 & -60-j160 & -160+j0 \\
\end{bmatrix}=[/tex]

[tex]\begin{bmatrix}
40 & -340 & 60 & -60 & 60 & 0 \\
340 & 40 & 60 & 60 & 0 & 60 \\
-20 & 320 & -60 & 160 & -160 & 0 \\
-320 & -20 & -160 & -60 & 0 & -160 \\
\end{bmatrix}
[/tex]

Reduced row echelon form of this matrix is:
[tex]\begin{bmatrix}
1 & 0 & 0 & 0 & 1275/7481 & -240/7481 \\
0 & 1 & 0 & 0 & 240/7481 & 1275/7481 \\
0 & 0 & 1 & 0 & 303/7481 & 7688/7481\\
0 & 0 & 0 & 1 & -7688/7481 & 303/7481 \\
\end{bmatrix}[/tex]

Now:

[tex]\underline{I_{C1}}=\frac{1275}{7481}+j\frac{240}{7481},\underline{I_{C4}}=\frac{303}{7481}-j\frac{7688}{7481}\Rightarrow \underline{I_{C3}}=\frac{8209}{7481}-j\frac{15089}{7481},$$$$\underline{I_{C2}}=\frac{22565}{7481}-j\frac{14675}{7481}[/tex]

[tex]\underline{I_L}=\underline{I_{C1}},\underline{U_{16}}=-jX_C \underline{I_{16}},\underline{I_{16}}=\underline{I_{C2}}\Rightarrow \underline{U_{16}}=-\frac{1467500}{7481}-j\frac{2256500}{7481}[/tex]

Active and reactive power in the branch 2-5 can be found by complex apparent power, [itex]\underline{S_{25}}=\underline{U_{25}}\underline{{I_{52}}^{*}}[/itex]

[tex]\underline{I_{52}}=\underline{I_{C1}}+\underline{I_{C2}}+\underline{I_{C3}}=\frac{32049}{7481}-j\frac{29524}{7481}[/tex]
[tex]\underline{U_{25}}=\underline{E_1}-\underline{I_{52}}\underline{Z}=-\frac{316490}{7481}-j\frac{25250}{7481}\Rightarrow \underline{S_{25}}=-\frac{9397707010}{55965361}-j\frac{10153288010}{55965361}[/tex]

[tex]\Rightarrow P=-\frac{9397707010}{55965361} W,Q=-\frac{10153288010}{55965361} var[/tex]

Question: Could someone check if the results are correct?

UPDATE:

Question: What type of simulation in OrCAD Capture CIS Lite 16.6 can be used for checking these results?
 
Last edited:

Answers and Replies

  • #2
The Electrician
Gold Member
1,285
169
Near the end of all your analysis you have: I52 = IC1+IC2+IC3 but I think it should be I52 = IC1+IC2+IC4

You sure used a cumbersome method of solving the system. Modern symbolic algebra software can solve the system in one go:

CmplxPwr1.png
 
  • #3
208
1
Near the end of all your analysis you have: I52 = IC1+IC2+IC3 but I think it should be I52 = IC1+IC2+IC4
You sure used a cumbersome method of solving the system.

You are correct about the current [itex]\underline{I_{52}}[/itex]. It should be [itex]\underline{I_{52}}=\underline{I_{C1}}+\underline{I_{C2}}+\underline{I_{C4}}[/itex].

Do you know the easier method for solving the system of linear complex equations without any software?

Also, do you know if it is possible to check the results in OrCAD?
 
  • #4
The Electrician
Gold Member
1,285
169
Why do you want to solve the system without any software? How about using a calculator that can do complex arithmetic? That would at least relieve some of the burden of the massive amount of number crunching.

I don't use OrCad, so I can't help you there.
 
  • #5
gneill
Mentor
20,925
2,866
@gruba: How did you arrive at 120 Ω for XL and 20 Ω for XC?

I'm not familiar with OrCad, but it would be straightforward to set up a simulation using a Spice package such as LTSpice (which is free).
 
  • #6
gneill
Mentor
20,925
2,866
@gruba: How did you arrive at 120 Ω for XL and 20 Ω for XC?
The reason I ask is that with a source angular frequency of 105 rad/sec and part values for the inductor and capacitor being 1 mH and 0.1 μF, impedance values of 120 and 20 Ohms are not possible.

You might also find that their actual impedance values have a particularly fortuitous relationship that can greatly simplify your circuit analysis approach... :wink:
 

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