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

• Engineering

## Homework Statement

Given the circuit of sinusoidal current (attachment 1) with given data:
$$\underline{E}=100V,\underline{E_1}=40V,\underline{Z}=(10+j10)\Omega,\omega=10^5rad/s,L=1mH, C=0.1uF.$$ 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:

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

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

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

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?

Last edited:
Delta2

## Answers and Replies

The Electrician
Gold Member
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:

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
Gold Member
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
Mentor
@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).

gneill
Mentor
@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...