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ElectricalPower System
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The impedance per phase of a 3-phase transmission line on the base of 100 MVA, 100 kV is 2 pu ┬╖ The value of this impedance on a base of 400 MVA and 400 kV would be

A

5 pu

B

1 pu

C

0.5 pu

D

0.25 pu

Correct Answer

тЪЩя╕П TE тАв Technical Direct FormulaElectricalPower System
Option C

0.5 pu

Quick Summary:

Given: Base MVA (old) = 100 MVA, Base kV (old) = 100 kV, Impedance (old) = 2 pu, Base MVA (new) = 400 MVA, Base kV (new) = 400 kV

ЁЯУРMAMath SolutionDirect Formula
ЁЯУЛ Given

Base MVA (old) = 100 MVA, Base kV (old) = 100 kV, Impedance (old) = 2 pu, Base MVA (new) = 400 MVA, Base kV (new) = 400 kV

ЁЯФв Formula Used

Znew=Zold├Ч(MVAnewMVAold)├Ч(kVoldkVnew)2Z_{new} = Z_{old} \times \left( \frac{MVA_{new}}{MVA_{old}} \right) \times \left( \frac{kV_{old}}{kV_{new}} \right)^2ZnewтАЛ=ZoldтАЛ├Ч(MVAoldтАЛMVAnewтАЛтАЛ)├Ч(kVnewтАЛkVoldтАЛтАЛ)2

ЁЯФв Step-by-Step Solution
1

Identify given variables

Identify the old base values (MVAold=100MVA_{old} = 100MVAoldтАЛ=100, kVold=100kV_{old} = 100kVoldтАЛ=100) and the new base values (MVAnew=400MVA_{new} = 400MVAnewтАЛ=400, kVnew=400kV_{new} = 400kVnewтАЛ=400) along with the existing impedance Zold=2Z_{old} = 2ZoldтАЛ=2 pu.

Zold=2┬аpu,MVAold=100,kVold=100,MVAnew=400,kVnew=400Z_{old} = 2 \text{ pu}, MVA_{old} = 100, kV_{old} = 100, MVA_{new} = 400, kV_{new} = 400ZoldтАЛ=2┬аpu,MVAoldтАЛ=100,kVoldтАЛ=100,MVAnewтАЛ=400,kVnewтАЛ=400

2

Apply change of base formula

Use the per-unit impedance transformation formula to calculate the new impedance value based on the new power and voltage bases.

Znew=Zold├Ч(MVAnewMVAold)├Ч(kVoldkVnew)2Z_{new} = Z_{old} \times \left( \frac{MVA_{new}}{MVA_{old}} \right) \times \left( \frac{kV_{old}}{kV_{new}} \right)^2ZnewтАЛ=ZoldтАЛ├Ч(MVAoldтАЛMVAnewтАЛтАЛ)├Ч(kVnewтАЛkVoldтАЛтАЛ)2

3

Substitute and compute

Substitute the given values into the formula: Znew=2├Ч(400/100)├Ч(100/400)2Z_{new} = 2 \times (400/100) \times (100/400)^2ZnewтАЛ=2├Ч(400/100)├Ч(100/400)2. Simplify the expression: 2├Ч4├Ч(1/4)2=8├Ч(1/16)=0.52 \times 4 \times (1/4)^2 = 8 \times (1/16) = 0.52├Ч4├Ч(1/4)2=8├Ч(1/16)=0.5.

Znew=2├Ч(400100)├Ч(100400)2=2├Ч4├Ч116=0.5┬аpuZ_{new} = 2 \times \left( \frac{400}{100} \right) \times \left( \frac{100}{400} \right)^2 = 2 \times 4 \times \frac{1}{16} = 0.5 \text{ pu}ZnewтАЛ=2├Ч(100400тАЛ)├Ч(400100тАЛ)2=2├Ч4├Ч161тАЛ=0.5┬аpu

тЬЕ

C is correct because the new per-unit impedance is calculated to be 0.5 pu after adjusting for the changes in base MVA and base kV.

Core Concepts Used
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Per-unit system Change of Base Transmission Line Impedance
ЁЯТб EXAM TIP

Always ensure the voltage base is squared in the conversion formula, as impedance is directly proportional to voltage squared (ZтИЭV2Z \propto V^2ZтИЭV2) when power is constant.

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