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A three-bus network is shown in the figure below, indicating the p.u. impedances of each element. The corresponding bus admittance matrix is

1. j [0.3 -0.2 0 -0.2 0.12 0.08 0 0.08 0.02]
2. j [-15 5 0 5 7.5 -12.5 0 -12.5 2.5]
3. j [0.1 0.2 0 0.2 0.12 -0.08 0 -0.08 0.10]
4. j [-10 5 0 5 7.5 12.5 0 12.5 -10]

Correct answer: j [-15 5 0 5 7.5 -12.5 0 -12.5 2.5]

Solution

The bus admittance matrix is formed from the network branch admittances. For a line with reactance jX, the admittance is 1/(jX) = -j/X, so the matrix entries are purely imaginary. Using the given network connections and summing the connected admittances at each bus gives the option with j[-15 5 0; 5 7.5 -12.5; 0 -12.5 2.5].

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