Test: Z Bus Matrix - Electrical Engineering (EE) MCQ

Fault current at bus 3 is given by If₃  

= -j 2.1762 p.u (V_p = voltage of the bus prior to fault).
Due to fault at bus-3 its voltage becomes 0 V and the voltage of Bus-1 and Bus-2 will also reduce. 
Changes in voltage can be found with the help of the Z-bus matrix as.

ΔV₁ = (j0.3243)(j2.1762)  = - 0.7057
ΔV₂ = (j0.2297)(j2.1762) = - 0.4998
Final voltage at Bus-1: V_f1 = V_p1 + ΔV₁ = 1 - 0.7057 = 0.2943 p.u
Final voltage at Bus-2: V_f2 = V_p2 + ΔV₂ = 1 - 0.4998 = 0.5002 p.u

Z bus matrix:

• The Z bus matrix or Bus impedance matrix is an important tool for the fault analysis of the power system.
• Z bus matrix can be formed by either inverting the Y bus matrix or by the Z bus building algorithm.
• The diagonal elements of the Z bus are referred to as driving point impedances of the buses and the off-diagonal elements are called transfer impedances.
•  The impedance between two far-away buses becomes very large, so there are no zero elements.
• As most of the elements in the Z bus matrix are non zero elements, hence Z bus matrix usually considered as a full matrix or dense matrix.
• The Z bus matrix is also a symmetric matrix.
• Z bus matrix is not preferred for load flow analysis since it requires more time to compute when the number of buses are more than three, and also more memory is required.

Y bus matrix:

• The Y bus or admittance matrix is the most preferred tool for the load flow analysis of the power systems.
• Y bus represents the nodal admittance of the buses in a power system, so it is also called a nodal admittance matrix.
• The admittance between two far-away buses becomes negligible, so most of the elements are zero.
• More than 80% of elements of the Y bus matrix are zero, hence it can be considered as a sparse matrix.
• The Y bus matrix is also a symmetric matrix.
• As most of the elements are zero, the computational time and memory required for load flow analysis are low.

New element Z_b = j0.2Ω is connected in i^th bus and reference bus i = 2, n = 4 so


Given that we are required to find only Z₂₂, Z₂₃

= j 0.1260

= j 0.0956

Power system having n nodes.
Z₃₃ = j 0.5 PU
V₃ = 1.3∠ -10° PU
A capacitor having a reactance of -j 3.5 PU is now added to the network

I_C = 0.433∠ 80° 

By considering line - 1,

Here Z_BUS is formed between new bus & reference bus, so type - 1 formation
∴  Z_Bus = [j 0.2]
By adding line - 2 to the above Bus

Here Z_BUS is formed between new bus & old bus, so type - 2 formation