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For a given base voltage and base volt-amperes, the per unit impedance value of an element is x. The per unit impedance value of this element when the voltage and volt-amperes bases are both doubled is
1. 0.5x
2. 2x
3. 4x
4. x

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Correct Answer - Option 1 : 0.5x
Concept:
The relation between new per-unit value & old per unit value impedance

\({({Z_{pu}})_{new}}\; = {({Z_{pu}})_{old}} \times {\left( {\frac{{k{V_{base}}}}{{k{V_{new}}}}} \right)^2} \times \;\left( {\frac{{MV{A_{new}}}}{{MV{A_{old}}}}} \right)\)

Also \({Z_{pu}} = \frac{{{Z_{Actual}}}}{{{Z_{base}}}}\)

 \({Z_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)

Where,

(Zpu)new = New per unit value of impedance

(Zpu)old = Old per unit value of impedance

kVbase = Old base value of voltage

kVnew = New base value of voltage

MVAnew = New base value of power

MVAold = Old base value of power

Calculation:

Given that,

(Zpu)old = x

MVA(old) = P (let)

MVA(new) = 2P

KV(old) = V (let)

KV(new) = 2V

\({X_{d\left( {new} \right)}} = X \times \frac{{2P}}{{P}} \times {\left( {\frac{{V}}{{2V}}} \right)^2} = 0.5x~P.U.\)

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