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ElectricalPower System
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The voltages at the two ends of a line are 132 kV and its reactance is 40 Ω. The capacity of the line is

A

435.6 MW

B

217.5 MW

C

251.5 MW

D

500 MW

Correct Answer

⚙️ TE • Technical Direct FormulaElectricalPower System
Option A

435.6 MW

Quick Summary:

Given: Voltage V = 132 kV, Reactance X = 40 Ω.

📐MAMath SolutionDirect Formula
📋 Given

Voltage V = 132 kV, Reactance X = 40 Ω.

🔢 Formula Used

P=V2XP = \frac{V²}{X}P=XV2​

🔢 Step-by-Step Solution
1

Identify given parameters

The line voltage VVV is given as 132 kV132 \text{ kV}132 kV and the line reactance XXX is 40 Ω40 \text{ } \Omega40 Ω.

V=132×103 V,X=40 ΩV = 132 \times 10³ \text{ V}, X = 40 \text{ } \OmegaV=132×103 V,X=40 Ω

2

Apply Steady State Stability Limit formula

The capacity of the transmission line (maximum power transfer) is determined by the formula P=V2XP = \frac{V²}{X}P=XV2​ where δ=90°\delta = 90°δ=90° for maximum power.

P=(132×103)240P = \frac{(132 \times 10³)^2}{40}P=40(132×103)2​

3

Calculate the power

Calculating the value: P=17424×10640=435.6×106 WP = \frac{17424 \times 10⁶}{40} = 435.6 \times 10⁶ \text{ W}P=4017424×106​=435.6×106 W. Converting to MW gives 435.6 MW435.6 \text{ MW}435.6 MW.

P=435.6 MWP = 435.6 \text{ MW}P=435.6 MW

✅

A is correct because the maximum power capacity of the line calculated using the steady-state stability formula is 435.6 MW.

Core Concepts Used
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Power System Stability Steady State Power Limit Transmission Line Reactance
💡 EXAM TIP

This formula is a fundamental concept in Power System Stability analysis; it assumes a lossless line where the maximum power transfer occurs at a load angle of 90°90°90°.

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