# Nominal Pi Model of a Medium Transmission Line

In the nominal pi model of a medium transmission line, the series impedance of the line is concentrated at the centre and half of each capacitance is placed at the centre of the line. The nominal Pi model of the line is shown in the diagram below.

By ohm’s law

Sending-end current is found by applying KCL at node c

or

Equations can be written in matrix form as

Also,

Hence, the ABCD constants for nominal pi-circuit model of a medium line are

### Phasor diagram of nominal pi model

The phasor diagram of a nominal pi-circuit is shown in the figure below.

It is also drawn for a lagging power factor of the load. In the phasor diagram the quantities shown are as follows;

OA = V_{r} – receiving end voltage. It is taken as reference phasor.

OB = I_{r} – load current lagging V_{r} by an angle ∅_{r}.

BE = I_{ab }– current in receiving-end capacitance. It leads V_{r} by 90°.

The line current I is the phasor sum of I_{r} and I_{ab}. It is shown by OE in the diagram.

AC = IR – voltage drop in the resistance of the line. It is parallel to I.

CD = IX -inductive voltage drop in the line. It is perpendicular to I.

AD = IZ – voltage drop in the line impedance.

OD = V_{s} – sending–end voltage to neutral. It is phasor sum of V_{r} and IZ.

The current taken by the capacitance at the sending end is I_{cd}. It leads the sending–end voltage V_{s} by 90

OF = I_{s} – the sending–end current. It is the phasor sum of I and I_{cd}.

∅s – phase angle between V_{s} and I_{s} at the sending end, and cos∅s will give the sending-end power factor.

## 1 Comment

Explanation super