ABCD parameters of medium transmission line : Nominal
π method
The total capacitance of the transmission line is divided into two parts, half at the sending end side and another half at the receiving end side in the nominal π method. All the parameters are given considering three phase transmission line.
Let
V_{S} =
Sending end voltage to neutral
V_{R} =
Receiving end voltage to neutral
I_{S} = Sending
end current
I_{R} =
Receiving end current
I_{C1} =
Capacitance current at the receiving end side
I_{C2} =
Capacitance current at the sending end side
R = Resistance of transmission
line, resistance to the neutral in the threephase line
X = Reactance of
transmission line, reactance to the neutral in the threephase line
Z = R + j X = Total
impedance
Y = jX_{C}
= Admittance
A, B, C and D =
Constant
Generalized
Constant of the transmission line is given by sending end voltage and sending
end current equation
V_{S} = AV_{R}
+ BI_{R}……. ( 1 )
I_{S }= CV_{R}
+ DI_{R}……....( 2 )
Important Points to be noted
Apply KCL at point
Q,
I_{1} = I_{R }+ I_{C1}
I_{1} = I_{R
}+ ( Y/2 ) V_{R} ……( 3 )
Apply KCL at point
P,
I_{s} = I_{1 }+ I_{C2}
I_{s} = I_{1
}+ ( Y/2 ) V_{s }………( 4 )
Substitute value
of I_{1} from equation ( 3 )
I_{s} = I_{R
}+ ( Y/2 ) V_{R} + ( Y/2 ) V_{s }………( 5 )
Sending end
voltage V_{S} = V_{R} + I_{1}( Z )
Substitute value
of I_{1} from equation ( 3 )
V_{S} = V_{R}
+ { I_{R }+ ( Y/2 ) V_{R} } ( Z )
= V_{R} + Z I_{R }+ ( ZY/2
) V_{R}
= V_{R} + ( ZY/2 ) V_{R }+ Z I_{R}
= ( 1 + ZY/2 ) V_{R }+ Z I_{R}
V_{S} = ( 1 + ZY/2 ) V_{R }+ Z I_{R} ….. ( 6 )
Compare equation (
6 ) and ( 1 )
V_{S} = AV_{R} + BI_{R} 
V_{S} = ( 1 + ZY/2 ) V_{R }_{} _{ }+ Z I_{R}

A = ( 1 + YZ/2 ) and B = Z 
From
equation ( 5 )
I_{s} = I_{R
}+ ( Y/2 ) V_{R} + ( Y/2 ) V_{s }
Substitute value
of V_{s} from equation ( 6 )
I_{s} = I_{R
}+ ( Y/2 ) V_{R} + ( Y/2 ) {( 1 + ZY/2 ) V_{R }+ Z I_{R }}
I_{s} = I_{R }+ ( Y/2 ) V_{R} + ( Y/2 + Y^{2}Z/4 ) V_{R }+ ( YZ/2 ) I_{R}
I_{s} = ( Y/2 ) V_{R} + ( Y/2 + Y^{2}Z/4 ) V_{R }+ I_{R }+ ( YZ / 2 ) I_{R }+ _{ }
I_{s} = { Y/2 + Y/2 + Y^{2}Z/4 ) V_{R }+ { 1 + ( YZ / 2 ) } I_{R }
I_{s} = { Y + Y^{2}Z/4 ) V_{R }+ { 1 + ( YZ / 2 ) } I_{R }
I_{s} = { 1 + ( YZ / 2 ) } I_{R }+ Y ( 1 + Y^{2}Z/4 ) V_{R }….. ( 7 )_{ }_{}
Compare equation (
7 ) and ( 2 )
I_{S }= CV_{R} + DI_{R} 
I_{s} = Y ( 1 + YZ/4 ) V_{R} + { 1 + (
YZ / 2 ) } I_{R } 
C = Y ( 1 + YZ/4
) and D = 1 + ( YZ / 2 ) 
From the above
value of A, B, C and D
A = ( 1 + YZ/2 )
B = Z
C = Y ( 1 + YZ/4 )
D = 1 + ( YZ / 2 )
Therefore
AD – BC = ( 1 + YZ/2 ) × ( 1 + YZ/2
) – Z × Y ( 1 + YZ/4 )
= ( 1 + YZ + Y^{2}Z^{2}/4 ) – YZ – Y^{2}Z^{2}/4
= 1
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