A detailed, beginner-friendly explanation of the Nominal-π equivalent circuit and the step-by-step derivation of its four transmission-line constants.
1. Introduction
A transmission line is used to transfer electrical power from a generating station to a receiving station. When electrical power travels through a transmission line, some voltage drop and power losses occur because of the electrical properties of the conductor.
Every transmission line has four important electrical parameters:
- Resistance (R): The conductor offers resistance to the flow of electric current. This resistance causes power loss in the form of heat.
- Inductance (L): Current flowing through a conductor produces a magnetic field around it. This magnetic field gives rise to inductance.
- Capacitance (C): There is a potential difference between transmission-line conductors. Due to this potential difference, the conductors behave like the plates of a capacitor.
- Conductance (G): A small leakage current may flow through the insulation between conductors and earth. Conductance represents this leakage path.
In a short transmission line, the effect of capacitance is generally neglected. However, in a medium transmission line, capacitance becomes important and must be considered in the calculations.
For analysing a medium transmission line, two commonly used equivalent circuit models are:
- Nominal-π (Pi) Model
- Nominal-T Model
In this section, we will understand the Nominal-π Model and derive its four important transmission-line constants: A, B, C and D.
2. What Is the Nominal-π Model?
The Nominal-π model is an approximate equivalent circuit used to represent a medium transmission line.
In this model, the complete series impedance of the transmission line is concentrated at the centre, while the total shunt admittance is divided into two equal parts.
The arrangement is as follows:
- Half of the total shunt admittance (Y/2) is connected at the sending end.
- The complete series impedance (Z) is connected at the centre.
- The remaining half of the shunt admittance (Y/2) is connected at the receiving end.
Because the equivalent circuit resembles the Greek letter π (Pi), it is called the Nominal-π Model.
Why is the shunt admittance divided into two equal parts?
The capacitance and leakage conductance of a transmission line exist throughout its length. Instead of representing them at every point, the Nominal-π model simplifies the circuit by placing half of the total shunt admittance at each end.
This arrangement provides a convenient approximate representation of the medium transmission line.
3. Equivalent Circuit and Important Symbols
The equivalent circuit consists of three main sections:
Section 1: Sending-end shunt admittance
A shunt admittance of Y/2 is connected between the sending-end conductor and neutral.
Section 2: Series impedance
The complete series impedance Z is connected between the sending and receiving ends.
Section 3: Receiving-end shunt admittance
The remaining shunt admittance Y/2 is connected between the receiving-end conductor and neutral.
The electrical quantities used in the circuit are:
| Symbol | Meaning |
|---|---|
| Vs | Sending-end voltage |
| Is | Sending-end current |
| Vr | Receiving-end voltage |
| Ir | Receiving-end load current |
| Z | Total series impedance |
| Y | Total shunt admittance |
| Y/2 | Shunt admittance at each end |
Series impedance (Z)
The series impedance consists of conductor resistance and inductive reactance.
Where:
- R = Total resistance of the line in ohms (Ω).
- XL = Total inductive reactance of the line in ohms (Ω).
- j = Imaginary operator, equal to √−1.
Shunt admittance (Y)
The total shunt admittance consists of conductance and capacitive susceptance.
Where:
- G = Total shunt conductance in siemens (S).
- C = Total line capacitance in farads (F).
- ω = Angular frequency in radians per second.
- f = Supply frequency in hertz (Hz).
Angular frequency is calculated as:
Important: The quantities Z and Y are complex quantities in AC circuit analysis. The imaginary part represents the effect of inductance or capacitance.
4. What Are ABCD Constants?
Before starting the derivation, let us understand the meaning of ABCD constants.
A transmission line can be considered a two-port electrical network:
- The sending end is the input port.
- The receiving end is the output port.
The voltage and current at the sending end depend on the voltage and current at the receiving end.
This relationship is expressed using four constants called ABCD constants, also known as generalised circuit constants.
The general equations are:
Here:
- A = Voltage ratio constant.
- B = Transfer impedance constant.
- C = Transfer admittance constant.
- D = Current ratio constant.
The units of these constants are:
| Constant | Unit |
|---|---|
| A | Dimensionless |
| B | Ohm (Ω) |
| C | Siemens (S) |
| D | Dimensionless |
Our objective is to derive the values of A, B, C and D for the Nominal-π model.
5. Derivation of ABCD Constants
To derive the ABCD constants, we will use two basic electrical laws:
Ohm's Law: The current flowing through an admittance is equal to the product of voltage and admittance.
Similarly, for an impedance:
Kirchhoff's Current Law (KCL): The total current entering a junction is equal to the total current leaving that junction.
We will apply these laws step by step to the Nominal-π equivalent circuit.
Step 1: Calculate the current through the receiving-end shunt admittance
First, consider the receiving-end shunt admittance.
In the Nominal-π model, half of the total shunt admittance is connected at the receiving end.
Therefore:
This admittance is connected across the receiving-end voltage VR.
According to Ohm's law, current is equal to voltage multiplied by admittance.
Therefore, the current through the receiving-end shunt branch is:
Hence,
What does this equation mean?
It means that a part of the current supplied by the transmission line is used to supply the receiving-end shunt branch.
This current is associated mainly with the charging effect of the line capacitance and any leakage represented by conductance.
Therefore, the current entering the series impedance is not only the load current IR. It also includes the current flowing through the receiving-end shunt admittance.
Step 2: Calculate the current through the series impedance
Now, consider the junction between the receiving-end shunt branch and the series impedance.
At this junction, the current entering the series impedance divides into two parts:
- Current supplied to the load (IR).
- Current flowing through the receiving-end shunt admittance (Iab).
Applying Kirchhoff's Current Law (KCL):
From Step 1:
Substituting this value:
Therefore,
Why do we add these two currents?
Because the series impedance must carry the total current required by both the load and the receiving-end shunt branch.
Thus, the current through the series impedance is greater than the load current by the amount of current flowing through the receiving-end shunt admittance.
Step 3: Calculate the sending-end voltage (VS)
Now, consider the complete series impedance Z connected at the centre of the circuit.
The voltage at the sending end is equal to:
- Receiving-end voltage (VR).
- Voltage drop across the series impedance (ZI).
According to Ohm's law:
Therefore:
From Step 2:
Substitute this value into the sending-end voltage equation:
Now, multiply Z by the terms inside the bracket:
Rearranging the terms:
Taking VR common from the first two terms:
Therefore, the sending-end voltage equation becomes:
Comparison with the general ABCD equation
The general sending-end voltage equation is:
Our derived equation is:
Compare the coefficients of VR and IR:
Thus, we have obtained the first two ABCD constants.
Step 4: Calculate the sending-end current (IS)
Next, we calculate the sending-end current.
At the sending end, the current entering the transmission line divides into two parts:
- Current flowing through the sending-end shunt admittance (Iab or Ish, depending on notation).
- Current flowing through the series impedance (I).
Applying Kirchhoff's Current Law:
The sending-end shunt admittance is Y/2 and is connected across the sending-end voltage VS.
Therefore, the current through this branch is:
Substituting this value:
From Step 2:
Therefore:
We already derived the sending-end voltage in Step 3:
Substitute this expression into the sending-end current equation:
Now, multiply Y/2 by each term inside the bracket:
Now, arrange the terms containing VR and IR separately:
Let us simplify the coefficient of VR.
First, expand the expression:
Adding the first two terms:
Taking Y common:
Therefore, the sending-end current equation becomes:
Comparison with the general ABCD equation
The general sending-end current equation is:
Our derived equation is:
Comparing the coefficients of VR and IR:
Thus, we have obtained all four ABCD constants.
6. Final Expressions of ABCD Constants
Therefore, the ABCD constants of a medium transmission line using the Nominal-π model are:
A constant:
B constant:
C constant:
D constant:
Notice that:
This is because the Nominal-π equivalent circuit is symmetrical: both ends have equal shunt admittances of Y/2.
7. General Transmission Equations in Matrix Form
The sending-end voltage and current equations can be combined into a single matrix equation.
We know:
Substituting the derived ABCD constants:
This matrix equation represents the complete relationship between the sending-end and receiving-end voltage and current of a medium transmission line using the Nominal-π model.
8. Important Properties of the Nominal-π Model
1. Symmetrical Network
The sending-end and receiving-end shunt admittances are equal.
2. Reciprocal Network
A reciprocal transmission-line network satisfies the following condition:
For the Nominal-π model, this condition is also satisfied.
3. Capacitance Is Considered
Unlike the short transmission-line model, the Nominal-π model considers the effect of line capacitance.
4. Lumped Parameter Model
The total line impedance and admittance are represented by concentrated circuit elements rather than being distributed continuously along the entire line.
5. Application
The Nominal-π model is used to analyse medium-length transmission lines and calculate sending-end quantities, voltage regulation, efficiency and other performance parameters.
9. Remember Points
| Parameter | Nominal-π Model |
|---|---|
| Series impedance | Complete Z at the centre |
| Sending-end shunt admittance | Y/2 |
| Receiving-end shunt admittance | Y/2 |
| A constant | \(1+YZ/2\) |
| B constant | \(Z\) |
| C constant | \(Y(1+YZ/4)\) |
| D constant | \(1+YZ/2\) |
| Symmetry condition | A = D |
| Reciprocity condition | AD − BC = 1 |
Important examination point: In the Nominal-π model, the total shunt admittance is divided equally between the sending and receiving ends, while the complete series impedance is placed at the centre. This arrangement gives the equivalent circuit its π-shaped appearance.
10. A Simple Way to Understand the Entire Derivation
Remember the sequence below. It is the easiest way to reproduce the derivation in an examination:
- Receiving-end shunt current\[I_{ab}=\frac{Y}{2}V_R\]
- Current through series impedance\[I=I_R+\frac{Y}{2}V_R\]
- Sending-end voltage\[V_S=V_R+ZI\]This gives A and B.
- Sending-end current\[I_S=I+\frac{Y}{2}V_S\]This gives C and D.