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

[gruba]

Homework Statement

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

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:

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}
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}
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}
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}

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

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

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

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

After reducing to 3x3 system:

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

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

(40+j140)\underline{I_{C1}}+(-20-j20)\underline{I_{C3}}+(40-j60)\underline{I_{C4}}=-120After reducing to 2x2 system:

(40+j340)\underline{I_{C1}}+(60+j60)\underline{I_{C4}}=60

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

<br /> \begin{bmatrix}<br /> 40+j340 &amp; 60+j60 \\<br /> -20-j320 &amp; -60-j160 \\<br /> \end{bmatrix} \begin{bmatrix}<br /> \underline{I_{C1}} \\<br /> \underline{I_{C4}} \\<br /> \end{bmatrix}=\begin{bmatrix}<br /> 60 \\<br /> -160 \\<br /> \end{bmatrix}\Rightarrow<br />

\begin{bmatrix}<br /> 40+j340 &amp; 60+j60 &amp; 60+j0 \\<br /> -20-j320 &amp; -60-j160 &amp; -160+j0 \\<br /> \end{bmatrix}=

\begin{bmatrix}<br /> 40 &amp; -340 &amp; 60 &amp; -60 &amp; 60 &amp; 0 \\<br /> 340 &amp; 40 &amp; 60 &amp; 60 &amp; 0 &amp; 60 \\<br /> -20 &amp; 320 &amp; -60 &amp; 160 &amp; -160 &amp; 0 \\<br /> -320 &amp; -20 &amp; -160 &amp; -60 &amp; 0 &amp; -160 \\<br /> \end{bmatrix}<br />

Reduced row echelon form of this matrix is:
\begin{bmatrix}<br /> 1 &amp; 0 &amp; 0 &amp; 0 &amp; 1275/7481 &amp; -240/7481 \\<br /> 0 &amp; 1 &amp; 0 &amp; 0 &amp; 240/7481 &amp; 1275/7481 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 &amp; 303/7481 &amp; 7688/7481\\<br /> 0 &amp; 0 &amp; 0 &amp; 1 &amp; -7688/7481 &amp; 303/7481 \\<br /> \end{bmatrix}

Now:

\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}

\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}

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

\underline{I_{52}}=\underline{I_{C1}}+\underline{I_{C2}}+\underline{I_{C3}}=\frac{32049}{7481}-j\frac{29524}{7481}
\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}

\Rightarrow P=-\frac{9397707010}{55965361} W,Q=-\frac{10153288010}{55965361} var

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?

 

[The Electrician]

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:

 

[gruba]

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 \underline{I_{52}}. It should be \underline{I_{52}}=\underline{I_{C1}}+\underline{I_{C2}}+\underline{I_{C4}}.

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?

 

[The Electrician]

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.

 

[gneill]

@gruba: How did you arrive at 120 Ω for X_L and 20 Ω for X_C?

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).

 

[gneill]

@gruba: How did you arrive at 120 Ω for X_L and 20 Ω for X_C?

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...