Question 3.1: Figure 3.3a shows a simple network of three buses with serie......
Figure 3.3a shows a simple network of three buses with series and shunt elements. Numerical values of circuit elements are impedances. The shunt resistors may represent unity power factor loads. Write the bus admittance matrix by examination and by use of Equations 3.14 and 3.15.
Υ_{22} = \frac{I_{2}}{V_{2}}(V_{1} = V_{3} = ⋅ ⋅ ⋅ \ V_{n} = 0) (3.14)
Υ_{21} = \frac{I_{2}}{V_{1}}(V_{2} = V_{3} = ⋅ ⋅ ⋅ \ V_{n} = 0) (3.15)
The bus admittance matrix is formed by inspection. At node 1, the self-admittance Y_{11} is 1 + 1/j0.2 = 1 – j5, and the transfer admittance between node 1 and 2, Y_{12} = -(1/j0.2) = j5.
Similarly, the other admittance elements are easily calculated:
= \left|\begin{matrix} 1-j5 & j5 & 0 \\ j5 & 0.5 – j8.33 & j3.33 \\ 0 & j3.33 & 0.33-j3.33 \end{matrix} \right| (3.16)
Alternatively, the bus admittance matrix can be constructed by use of Equations 3.14 and 3.15. Apply unit voltages, one at a time, to each bus while short-circuiting the other bus voltage sources to ground. Figure 3.3b shows unit voltage applied to bus 1, while buses 2 and 3 are short-circuited to ground. The input current I to bus 1 gives the driving point admittance. This current is given by
\frac{V}{1.0} + \frac{V}{j0.2}=1- j5 = Y_{11}Bus 2 is short-circuited to ground; the current flowing to bus 2 is
\frac{-V}{j0.2} = Y_{12} = Y_{21}Only buses directly connected to bus 1 have current contributions. The current flowing to bus 3 is zero:
Y_{13}=Y_{31}=0The other elements of the matrix can be similarly found and the result obtained is the same as in Equation 3.16.