Question 5.14: Phasor Mesh-Current Analysis with MATLAB Determine the value...
Phasor Mesh-Current Analysis with MATLAB
Determine the values for the mesh currents, the real power supplied by V1, and the reactive power supplied by V1 in the circuit of Figure 5.47.
First, we apply KVL to each loop obtaining the mesh-current equations:
\left(5+j3\right)\mathrm{I}_1+\left(50\angle -10^\circ\right) \left(\mathrm{I}_1-\mathrm{I}_2\right) =2200\sqrt{2}
\left(50\angle -10^\circ \right) \left(\mathrm{I}_1-\mathrm{I}_2\right)+\left(4+j\right)\mathrm{I}_2+2000\sqrt{2}\angle30 =0
In matrix form, these equations become
\left [ \begin{matrix} \left(5+j3+50\angle-10^\circ\right) & -50\angle-10^\circ \\ -50\angle10^\circ & \left(4+j+50\angle-10^\circ\right) \end{matrix} \right ] \left [ \begin{matrix} \mathrm{I}_{1} \\ \mathrm{I}_{2} \end{matrix} \right ] =\left [ \begin{matrix} 2200\sqrt{2} \\ -2000\sqrt{2}\angle -10^\circ \end{matrix} \right ]
We will solve these equations for I1 and I2. Then, we will compute the complex power delivered by V1
\mathrm{S}_1=\frac{1}{2}\mathrm{V}_1\mathrm{I}_1^*
Finally, the power is the real part of S1 and the reactive power is the imaginary part.
We enter the coefficient matrix Z and the voltage matrix V into MATLAB, making use of our new pin function to enter polar values. Then, we calculate the current matrix.
>> Z = [(5 + i*3 + pin(50,-10)) (-pin(50,-10));...
(-pin(50,-10)) (4 + i + pin(50,-10))];
>> V = [2200*sqrt(2); -pin(2000*sqrt(2),-10)];
>> I = inv(Z)*V
I =
74.1634 + 29.0852i
17.1906 + 26.5112i
This has given us the values of the mesh currents in rectangular form. Next, we obtain the polar form for the mesh currents, making use of our new pout function:
>> pout(I(1))
ans =
79.6628 21.4140
>> pout(I(2))
ans =
31.5968 57.0394
Thus, the currents are I1 = 79.66 ∠21.41° A and I2 = 31.60 ∠57.04° A, rounded to two decimal places. Next, we compute the complex power, real power, and reactive power for the first source.
\mathrm{S}_1=\frac{1}{2}\mathrm{V}_1\mathrm{I}_1^*
>> S1 = (1/2)*(2200*sqrt(2))*conj(I(1));
>> P1 = real(S1)
P1 =
1.1537e + 005
>> Q1 = imag(S1)
Q1 =
-4.5246e + 004
Thus, the power supplied by V1 is 115.37 kW and the reactive power is -45.25 kVAR. The commands for this example appear in the m- file named Example_5_14.