AC circuit problem

Question

I need help with the following problem:

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

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:

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

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

$$(40+j140)\underline{I_{C1}}+(-20-j20)\underline{I_{C3}}+(40-j60)\underline{I_{C4}}=-120$$

After reducing to 2x2 system:

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

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

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

$$\begin{bmatrix} 40+j340 & 60+j60 & 60+j0 \\ -20-j320 & -60-j160 & -160+j0 \\ \end{bmatrix}=$$

$$\begin{bmatrix} 40 & -340 & 60 & -60 & 60 & 0 \\ 340 & 40 & 60 & 60 & 0 & 60 \\ -20 & 320 & -60 & 160 & -160 & 0 \\ -320 & -20 & -160 & -60 & 0 & -160 \\ \end{bmatrix} $$

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

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?

Answer 1

Am I missing something? Where is the AC source? Lots of reactance and some batteries is all I see. In steady state, it looks like a DC problem to me.

Answer 2

Using Orcad Capture CS Lite it is fairly easy to perform an AC analysis and see the behavior of this circuit. Below are the steps that I would take if I were in your shoes:

1. Build the circuit in PSpice
2. Click on the "Create new simulation profile" button
3. Select AC sweep, and enter the desired sweep range, points per decade, etc.
4. Run the simulation
5. Plot the traces of interest
6. Use the cursor to take measurements as needed.

There are some good tutorials for this type of simulation here and here.

Answer 3

You can rebuild the circuit as it is in OrCAD, with the following modifications:

1. substitute the \$\underline Z\$ with an inductive impedance made of a series RL, with R=10\$\Omega\$ and L=10/\$\omega\$=0.1mH;

2. modify the \$\underline E\$ and \$\underline E_1\$ sources to have these values: AC 100 and AC 40, respectively;

3. only simulate for a single frequency, which is \$10^5\$/(2\$\pi\$). There should be an option in the simulation control panel for a list of frequencies, only choose one. Unfortunately, since I don't have OrCAD, I can only give an example of how it would look like in LTspice: .ac list {10k/(2*pi)}.

You can also name the nodes in your schematic and, not lastly, don't forget to add a ground somewhere, since every calculation is done with respect to zero potential. Then, all the results should be referenced to that point.