Question E.5: In the unbalanced three-phase system shown in Fig. E.11, fin......
In the unbalanced three-phase system shown in Fig. E.11, find currents I_{1} , I_{2} , I_{3} , and I_{Bb} . Let
Z_{A} = 12 + j10 Ω , Z_{B} = 10 – j 8 Ω , Z_{C} = 15 + j6 Ω
For mesh 1,
120 \underline{/ – 120°} – 120 \underline{/0°} + I_{1} (2 + 1 + 12 + j10) – I_{2} – I_{3} (12 + j10) = 0
or
I_{1} (15 + j10) – I_{2} – I_{3} (12 + j10) = 120\underline{/0°} – 120 \underline{/ – 120°} (E.5.1)
For mesh 2,
120 \underline{/ 120°} – 120 \underline{/ – 120°} + I_{2} (2 + 1 + 10 – j8) – I_{1} – I_{3} (10 – j8) = 0
or
– I_{1} + I_{2} (13 – j8) – I_{3} (10 – j8) = 120 \underline{/- 120°} – 120 \underline{/ 120° } (E.5.2)
For mesh 3,
I_{3} (12 + j10 + 10 – j8 + 15 + j6) – I_{1} (12 + j10) – I_{2} (10 – j8) = 0
or
– I_{1} (12 + j10) – I_{2} (10 – j8) – I_{3} (37 + j8) = 0 (E.5.3)
In matrix form, we can express Eqs. (E.5.1) to (E.5.3) as
\begin{bmatrix} 15 + j10 & -1 & -12 – j10 \\ -1 & 13 – j8 & -10 + j8 \\ -12 – j10 & -10 + j8 & 37 + j8 \end{bmatrix} \begin{bmatrix}I_{1}\\ I_{2} \\ I_{3} \end{bmatrix}
= \begin{bmatrix}120 \underline{/0°} – 120\underline{/ – 120°} \\ 120\underline{/ – 120°} – 120\underline{/ 120°} \\ 0 \end{bmatrix}
or
ZI = V
Script File
We input matrices Z and V into MATLAB to get I .
> z = [(15 + 10j) -1 (-12 - 10j);
-1 (13 - 8j) (-10 + 8j);
(-12 - 10j) (-10 + 8j) (37 + 8j)];
>> c1=120*exp(j*pi*(-120)/180);
>> c2=120*exp(j*pi*(-120)/180);
>> a1=120 - c1; a2=c1 - c2;
>> V = [a1; a2; 0]
>> I = inv(z)*V
I=
16.9910 - 6.5953i
12.4023 - 16.9993i
5.6621 - 6.0471i
>> IbB = I(2) - I(1)
IbB =
-4.5887 - 10.4039i
>> abs(I(1))
ans =
18.2261
>> angle(I(1))*180/pi
ans =
-21.2146
>> abs (I(2))
ans =
21.0426
>> angle(I(2))*180/pi
ans =
-53.8864
>> abs(I(3))
ans =
8.2841
>> angle(I(3))*180/pi
ans =
-46.8833
>> abs(IbB)
ans =
11.3709
>> angle(IbB)*180/pi
ans =
-113.8001
Thus, I_{1} = 18.23 \underline{/ - 21 .21°} , I_{2} = 21.04 \underline{/ -58.89°}
I_{3} = 8.284 \underline{/ - 46.88° } , and I_{bB} = 11.37 \underline{/ - 113. 8°} A