## 11/04/2022

### ABCD Parameters of Nominal T method

ABCD parameters of medium transmission line

The total capacitance of the transmission line is considered at the middle of the line in the nominal T method. Half the impedance at the sending end side and another half at the receiving end side. All the parameters are given considering three phase transmission line.

Let

VS = Sending end voltage to neutral

VR = Receiving end voltage to neutral

IS = Sending end current

IR = Receiving end current

IC = Capacitance current

R = Resistance of transmission line, resistance to the neutral in the three-phase line

X = Reactance of transmission line, reactance to the neutral in the three-phase line

Z = R + j X = Total impedance

Y = jXC = Admittance

A, B, C and D = Constant

Generalized Constant of the transmission line is given by sending end voltage and sending end current equation

VS = AVR + BIR……. ( 1 )

IS = CVR + DIR……....( 2 )

Important Points to be noted

Here, the total impedance is divided into two parts, half on the sending end and another half on receiving side but both have same value.

Z/2 = R/2 + jXL / 2

Voltage across capacitor V1 = VR + IR ( Z/2 )…. ( 3 )

Sending end current is sum of capacitance current and receiving end current.

IS = IR + IC

IS = IR + YV1

IS = IR + Y { VR + IR ( Z/2 ) } { from equation ( 3 ) }

IS = IR + YVR + YIR ( Z/2 )

IS = YVR + IR + YIR ( Z/2 )

IS = YVR + ( 1 + YZ/2 ) IR ……( 4 )

Compare equation ( 2 ) and ( 4 )

 IS = CVR + DIR IS = YVR + ( 1 + YZ/2 ) IR C = Y and D = 1 + YZ/2

Now, sending end voltage

VS = V1 + IS ( Z/2 )

= VR + IR ( Z/2 ) + IS ( Z/2 ) { from equation ( 3 ) }

Putting value of Is from equation ( 4 )

= VR + IR ( Z/2 ) + ( Z/2 )( YVR + ( 1 + YZ/2 ) IR

= VR + IR ( Z/2 ) + ( YZ/2 )VR + ( Z/2 )( 1 + YZ/2 ) IR

= ( 1 + YZ/2 )VR + IR ( Z/2 ) + ( Z/2 )( 1 + YZ/2 ) IR

= ( 1 + YZ/2 )VR + ( Z/2 + Z/2 + YZ2/4 ) IR

= ( 1 + YZ/2 )VR + ( Z + YZ2/4 ) IR

VS = ( 1 + YZ/2 )VR + Z( 1 + YZ/4 ) IR ……( 5 )

Compare equation ( 1 ) and ( 5 )

 VS = AVR + BIR VS = ( 1 + YZ/2 )VR +         Z( 1 + YZ/4 ) IR A = ( 1 + YZ/2 ) and B = Z( 1 + YZ/4 )

From the above value of A, B, C and D

A = ( 1 + YZ/2 )

B = Z( 1 + YZ/4 )

C = Y

D = 1 + YZ/2

Therefore

AD – BC = ( 1 + YZ/2 )( 1 + YZ/2 ) – Z( 1 + YZ/4 ) Y

= 1 + YZ/2 + YZ/2 + Y2Z2 /4 – YZ – Y2Z2 / 4

= 1 + YZ + Y2Z2 /4 – YZ – Y2Z2 / 4

= 1

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