Skip to main content

Node equations and Ybus, three-source network

Short answer

Convert each source with its series reactance to a Norton equivalent (current injection and shunt admittance). Compute line admittances. Ybus (pu) = [[20–j55, –10+j20, –10+j30], [–10+j20, 26–j56, –16+j32], [–10+j30, –16+j32, 26–j67]].

For the network given below, obtain node equations and then form admittance matrix (YBus). All impedances and voltages are marked in pu. [Figure: three buses; line 1-2 = 0.02 + j0.04, line 1-3 = 0.01 + j0.03, line 2-3 = 0.0125 + j0.025; source E1 = 1∠0° with X = j0.12 connected to bus 1 through transformer T1 of j0.08; source E2 = 1∠0° with X = j0.13 connected to bus 2 through transformer T2 of j0.12; source E3 = 1∠−36.87° with X = j0.1 connected to bus 3 through transformer T3 of j0.1]

Convert each voltage source with its series impedance into an equivalent current source (Norton form) at its bus, then write KCL at each bus.

Step 1: Source branches

Each source reactance is in series with its transformer, so the total series reactance to each bus is:

BusSource + transformerShunt admittance yi0y_{i0}
1j0.12+j0.08=j0.20j0.12 + j0.08 = j0.201/j0.2=−j51/j0.2 = -j5
2j0.13+j0.12=j0.25j0.13 + j0.12 = j0.251/j0.25=−j41/j0.25 = -j4
3j0.10+j0.10=j0.20j0.10 + j0.10 = j0.201/j0.2=−j51/j0.2 = -j5

Equivalent injected currents Ii=Ei yi0I_i = E_i\,y_{i0}:

I1=1∠0∘j0.2=−j5 puI2=1∠0∘j0.25=−j4 puI3=1∠−36.87∘j0.2=5∠−126.87∘=−3−j4 pu\begin{aligned} I_1 &= \frac{1\angle 0^\circ}{j0.2} = -j5\ \text{pu} \\ I_2 &= \frac{1\angle 0^\circ}{j0.25} = -j4\ \text{pu} \\ I_3 &= \frac{1\angle -36.87^\circ}{j0.2} = 5\angle -126.87^\circ = -3 - j4\ \text{pu} \end{aligned}

Step 2: Line admittances

y12=10.02+j0.04=10−j20 puy13=10.01+j0.03=10−j30 puy23=10.0125+j0.025=16−j32 pu\begin{aligned} y_{12} &= \frac{1}{0.02 + j0.04} = 10 - j20\ \text{pu} \\ y_{13} &= \frac{1}{0.01 + j0.03} = 10 - j30\ \text{pu} \\ y_{23} &= \frac{1}{0.0125 + j0.025} = 16 - j32\ \text{pu} \end{aligned}

Step 3: Node equations (KCL)

I1=y10V1+y12(V1−V2)+y13(V1−V3)I2=y20V2+y12(V2−V1)+y23(V2−V3)I3=y30V3+y13(V3−V1)+y23(V3−V2)\begin{aligned} I_1 &= y_{10}V_1 + y_{12}(V_1 - V_2) + y_{13}(V_1 - V_3) \\ I_2 &= y_{20}V_2 + y_{12}(V_2 - V_1) + y_{23}(V_2 - V_3) \\ I_3 &= y_{30}V_3 + y_{13}(V_3 - V_1) + y_{23}(V_3 - V_2) \end{aligned}

Collecting terms:

−j5=(y10+y12+y13)V1−y12V2−y13V3−j4=−y12V1+(y20+y12+y23)V2−y23V3−3−j4=−y13V1−y23V2+(y30+y13+y23)V3\begin{aligned} -j5 &= (y_{10} + y_{12} + y_{13})V_1 - y_{12}V_2 - y_{13}V_3 \\ -j4 &= -y_{12}V_1 + (y_{20} + y_{12} + y_{23})V_2 - y_{23}V_3 \\ -3 - j4 &= -y_{13}V_1 - y_{23}V_2 + (y_{30} + y_{13} + y_{23})V_3 \end{aligned}

Step 4: Ybus elements

Y11=−j5+(10−j20)+(10−j30)=20−j55Y22=−j4+(10−j20)+(16−j32)=26−j56Y33=−j5+(10−j30)+(16−j32)=26−j67Y12=Y21=−10+j20Y13=Y31=−10+j30Y23=Y32=−16+j32\begin{aligned} Y_{11} &= -j5 + (10 - j20) + (10 - j30) = 20 - j55 \\ Y_{22} &= -j4 + (10 - j20) + (16 - j32) = 26 - j56 \\ Y_{33} &= -j5 + (10 - j30) + (16 - j32) = 26 - j67 \\ Y_{12} &= Y_{21} = -10 + j20 \\ Y_{13} &= Y_{31} = -10 + j30 \\ Y_{23} &= Y_{32} = -16 + j32 \end{aligned}

Result

[−j5−j4−3−j4]=[20−j55−10+j20−10+j30−10+j2026−j56−16+j32−10+j30−16+j3226−j67][V1V2V3]\begin{bmatrix} -j5 \\ -j4 \\ -3 - j4 \end{bmatrix} = \begin{bmatrix} 20 - j55 & -10 + j20 & -10 + j30 \\ -10 + j20 & 26 - j56 & -16 + j32 \\ -10 + j30 & -16 + j32 & 26 - j67 \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ V_3 \end{bmatrix}

Answer: YbusY_{bus} is the 3×3 matrix above (pu); it is symmetric, and its diagonal terms include the source-branch admittances −j5-j5, −j4-j4, −j5-j5.

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Power System Analysis II chapter 1 questions →