Why does Y33 contain no shunt term?

Bus 3 has no generator or shunt element

2076 Chaitra · 4 marks

Ybus with generator and transformer impedances

Short answer

Treat generator plus transformer reactances as shunt admittances to ground. Compute line admittances from impedances. Diagonal Y_ii sum of connected admittances; off-diagonal Y_ik = –y_ik. Ybus = [[11.0061–j33.2003, –4.1096+j10.9589, –6.8966+j17.2414]; [–4.1096+j10.9589, 10.0267–j29.3268, –5.9172+j14.2012]; [–6.8966+j17.2414, –5.9172+j14.2012, 12.8137–j31.4426]] pu.

Form the Ybus matrix for the following network. All impedance values are in per unit system. [Figure: G1 through T1 to bus 1; G2 through T2 to bus 2; line L1 between buses 1 and 2, L2 between buses 1 and 3, L3 between buses 2 and 3]

Component	G1	G2	T1	T2	L1	L2	L3
Z, pu	j0.1	j0.12	j0.1	j0.12	0.03+j0.08	0.02+j0.05	0.025+j0.06

Each generator reactance is in series with its transformer, so it forms a shunt branch from its bus to the reference (the generator emf is a current injection). Bus 3 has no generator.

Admittances

Branch	Impedance (pu)	Admittance (pu)
G1 + T1 at bus 1	$j0.1 + j0.1 = j0.2$	$-j5$
G2 + T2 at bus 2	$j0.12 + j0.12 = j0.24$	$-j4.1667$
L1 (1-2)	$0.03 + j0.08$	$4.1096 - j10.9589$
L2 (1-3)	$0.02 + j0.05$	$6.8966 - j17.2414$
L3 (2-3)	$0.025 + j0.06$	$5.9172 - j14.2012$

Example: $y_{12} = \dfrac{0.03 - j0.08}{0.03^2 + 0.08^2} = \dfrac{0.03 - j0.08}{0.0073} = 4.1096 - j10.9589$ .

Elements

\begin{aligned} Y_{11} &= -j5 + y_{12} + y_{13} = 11.0061 - j33.2003 \\ Y_{22} &= -j4.1667 + y_{12} + y_{23} = 10.0267 - j29.3268 \\ Y_{33} &= y_{13} + y_{23} = 12.8137 - j31.4426 \end{aligned}

Answer:

Y_{bus} = \begin{bmatrix} 11.0061 - j33.2003 & -4.1096 + j10.9589 & -6.8966 + j17.2414 \\ -4.1096 + j10.9589 & 10.0267 - j29.3268 & -5.9172 + j14.2012 \\ -6.8966 + j17.2414 & -5.9172 + j14.2012 & 12.8137 - j31.4426 \end{bmatrix}\ \text{pu}

✓ Answer checked by Claude Sonnet 5.5

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Ybus with generator and transformer impedances

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