ElectricalPower System

The calculations performed using short-line approximate model instead of nomial-π model for a medium length transmission line delivering lagging load at a given receiving-end voltage always result in higher.

sending-end current.
sending-end power.
regulation.
Efficiency.

Which of these statements are correct ?

1 and 2

3 and 4

1, 2 and 3

1, 2 and 4

Correct Answer

Concept & PrincipleElectricalPower System

Option A

1 and 2

Quick Summary: The short-line model neglects the shunt capacitance ($C$) of the transmission line, whereas the nominal-$\pi$ model accounts for it by placing half the capacitance at each end. For a lagging load, the charging current of the shunt capacitance partially compensates for the lagging inductive current of the load; neglecting this effect in the short-line approximation leads to overestimations of the required source current, power, and voltage regulation.

💡 Explanation

The short-line model neglects the shunt capacitance ( $C$ ) of the transmission line, whereas the nominal- $\pi$ model accounts for it by placing half the capacitance at each end. For a lagging load, the charging current of the shunt capacitance partially compensates for the lagging inductive current of the load; neglecting this effect in the short-line approximation leads to overestimations of the required source current, power, and voltage regulation.

🔢 Key Formulas

$V_s = A V_r + B I_r$ — Transmission line ABCD parameters equation

$\% \text{Regulation} = \frac{|V_{s, no-load}| - |V_r|}{|V_r|} \times 100$ — Voltage regulation definition

⚙️ Working Principle

In the nominal- $\pi$ model, the shunt capacitive current $I_{ch} = V \cdot j\omega C$ flows, which reduces the total current drawn from the source compared to a model that ignores this path. Because the short-line model assumes $I_s = I_r$ , the calculated sending-end parameters are inherently higher because the charging current (which provides some reactive compensation) is omitted from the line current calculations.

📌 Key Points

▸
Short-line model assumes $A=1, B=Z, C=0, D=1$ .
▸
Nominal- $\pi$ model correctly accounts for shunt admittance ( $Y$ ).
▸
Neglecting shunt admittance results in higher calculated sending-end current for lagging loads.
▸
Higher current leads to higher calculated $I^2R$ losses, thus higher sending-end power.
▸
Voltage regulation increases as the simplified model overestimates the voltage drop across the series impedance $Z$ .

✅ Advantages

▸
Simplicity in manual calculation for short lines.
▸
Reduced complexity for quick transmission line approximations.

❌ Disadvantages / Limitations

▸
Inaccurate for lines where shunt capacitance is significant.
▸
Overestimates losses and voltage drops for lagging loads.

🛠️ Applications / Uses

▸
Distribution lines less than 80 km.
▸
Quick verification of power flow models.

🔄 Comparison Table

Feature	Short-Line	Nominal-$\pi$
Shunt Admittance ( $Y$ )	Neglected	Included (as $Y/2$ at both ends)

📄 Additional Information

▸
The efficiency is defined as $\frac{P_{out}}{P_{in}}$ . Since the short-line model overestimates input power ( $P_{in}$ ) while the output power ( $P_{out}$ ) remains constant, the calculated efficiency will actually be lower, not higher.
▸
Statement 4 is incorrect because efficiency is lower, whereas statements 1, 2, and 3 are higher.

📊 Diagram / Illustration

✅

C is correct — The short-line model neglects shunt capacitance, causing an overestimation of the sending-end current, power, and voltage regulation for lagging loads.

Core Concepts Used

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Transmission Line Models Nominal-$\pi$ Model Voltage Regulation

💡 EXAM TIP

Always remember that short-line approximation is mathematically equivalent to setting the shunt admittance $Y=0$ in the ABCD parameter matrix.

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