# The per unit impedance of a synchronous machine is 0.242. If base voltage is increased by 1.1 times, then per unit value will be:

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The per unit impedance of a synchronous machine is 0.242. If base voltage is increased by 1.1 times, then per unit value will be:
1. 0.266
2. 0.242
3. 0.220
4. 0.200

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Correct Answer - Option 4 : 0.200

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:

When the voltage bases is increased by 1.1 times  (i.e. kVnew = 1.1kVbase), the new per unit impedance is

$(Z_{pu})_{new}=0.242\times \frac{1}{1.1^2}=0.2\ \ pu$