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Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
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ElectricalPower System
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The per unit value of a 2 Ω resister at 100 MVA base and 10 kV base is

A

4 pu

B

2 pu

C

0.5 pu

D

0.2 pu

Correct Answer

⚙️ TE • Technical Direct FormulaElectricalPower System
Option B

2 pu

Quick Summary:

Given: Resistance R = 2 Ω,\Omega,Ω,Base Power S_{base} = 100 MVA, Base Voltage V_{base} = 10 kV

📐MAMath SolutionDirect Formula
📋 Given

Resistance R = 2 Ω,\Omega,Ω,Base Power S_{base} = 100 MVA, Base Voltage V_{base} = 10 kV

🔢 Formula Used

Zpu=RactualZbase=Ractual×Sbase(Vbase)2Z_{pu} = \frac{R_{actual}}{Z_{base}} = \frac{R_{actual} \times S_{base}}{(V_{base})^2}Zpu​=Zbase​Ractual​​=(Vbase​)2Ractual​×Sbase​​

🔢 Step-by-Step Solution
1

Calculate Base Impedance

The base impedance ZbaseZ_{base}Zbase​ is calculated using the base voltage and base power as Zbase=(Vbase)2SbaseZ_{base} = \frac{(V_{base})^2}{S_{base}}Zbase​=Sbase​(Vbase​)2​. Given Vbase=10kVV_{base} = 10 kVVbase​=10kV and Sbase=100MVAS_{base} = 100 MVASbase​=100MVA.

Zbase=(10×103)2100×106=100×106100×106=1 ΩZ_{base} = \frac{(10 \times 10³)^2}{100 \times 10⁶} = \frac{100 \times 10⁶}{100 \times 10⁶} = 1 \, \OmegaZbase​=100×106(10×103)2​=100×106100×106​=1Ω

2

Calculate Per Unit Resistance

Substitute the actual resistance Ractual=2 ΩR_{actual} = 2 \, \OmegaRactual​=2Ω and the base impedance Zbase=1 ΩZ_{base} = 1 \, \OmegaZbase​=1Ω into the per-unit formula Zpu=RactualZbaseZ_{pu} = \frac{R_{actual}}{Z_{base}}Zpu​=Zbase​Ractual​​.

Rpu=21=2 puR_{pu} = \frac{2}{1} = 2 \, \text{pu}Rpu​=12​=2pu

✅

B is correct because the calculated per unit resistance for the given base values is 2 pu.

Core Concepts Used
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Per Unit System Base Impedance Power System Normalization
💡 EXAM TIP

This concept is fundamental for load flow studies and fault analysis; always remember that Zbase=(Vbase,kV)2/Sbase,MVAZ_{base} = (V_{base, kV})^2 / S_{base, MVA}Zbase​=(Vbase,kV​)2/Sbase,MVA​.

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