How is a generator with series reactance and transformer treated in Ybus?

As a shunt admittance from its bus to reference using the series sum of reactances

As a series line between two buses

Source branch -j10 plus lines 1-2 and 1-3 admittances

Only the line admittances at bus 1

2080 Bhadra · 4 marks

Ybus for network with three sources

Short answer

Combine each generator and its transformer series reactance into a shunt admittance to ground. Compute line admittances from impedances. Diagonal entries: sum of all admittances connected to the bus; off-diagonals: negative of line admittance. Ybus = [[11.8966–j42.2414, –6.8966+j17.2414, –5+j15]; [–6.8966+j17.2414, 16.8966–j51.2414, –10+j30]; [–5+j15, –10+j30, 15–j50]] pu.

For the network shown below obtain a bus admittance matrix. [Figure: three buses; line 1-2 = 0.02 + j0.05, line 1-3 = 0.02 + j0.06, line 2-3 = 0.01 + j0.03; source E1 with x = j0.02 connected to bus 1 through transformer T1 of j0.08; source E2 with x = j0.13 connected to bus 2 through transformer T2 of j0.12; source E3 with x = j0.1 connected to bus 3 through transformer T3 of j0.1]

Each source reactance is in series with its transformer, so each source branch is a shunt admittance from its bus to the reference (generator emfs become current injections and do not enter $Y_{bus}$ ).

Admittances

Branch	Impedance (pu)	Admittance (pu)
Bus 1 to ground	$j0.02 + j0.08 = j0.10$	$-j10$
Bus 2 to ground	$j0.13 + j0.12 = j0.25$	$-j4$
Bus 3 to ground	$j0.10 + j0.10 = j0.20$	$-j5$
Line 1-2	$0.02 + j0.05$	$6.8966 - j17.2414$
Line 1-3	$0.02 + j0.06$	$5.0000 - j15.0000$
Line 2-3	$0.01 + j0.03$	$10.0000 - j30.0000$

Elements

\begin{aligned} Y_{11} &= -j10 + (6.8966 - j17.2414) + (5 - j15) = 11.8966 - j42.2414 \\ Y_{22} &= -j4 + (6.8966 - j17.2414) + (10 - j30) = 16.8966 - j51.2414 \\ Y_{33} &= -j5 + (5 - j15) + (10 - j30) = 15.0000 - j50.0000 \\ Y_{12} &= -y_{12} = -6.8966 + j17.2414 \\ Y_{13} &= -y_{13} = -5.0000 + j15.0000 \\ Y_{23} &= -y_{23} = -10.0000 + j30.0000 \end{aligned}

Answer:

Y_{bus} = \begin{bmatrix} 11.8966 - j42.2414 & -6.8966 + j17.2414 & -5 + j15 \\ -6.8966 + j17.2414 & 16.8966 - j51.2414 & -10 + j30 \\ -5 + j15 & -10 + j30 & 15 - j50 \end{bmatrix}\ \text{pu}

✓ Answer checked by Claude Sonnet 5.5

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Ybus for network with three sources

Admittances

Elements

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