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A 500 MVA, 11KV synchronous generator has 0.2 p.u. synchronous reactance. The p.u. synchronous reactance on the base values of 100 MVA and 22 KV is

1. 0.16
2. 0.01
3. 4.0
4. 0.25

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Correct Answer - Option 2 : 0.01


The relation between new per-unit value & old per unit value of reactance

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

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

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


(Xpu)new = New per unit value of reactance

(Xpu)old = Old per unit value of reactance

kVbase = Old base value of voltage

kVnew = New base value of voltage

MVAnew = New base value of power

MVAold = Old base value of power


Given that,

Xd(old) = 0.2 pu

MVA(new) = 100

MVA(old) = 500

KV(old) = 11

KV(new) = 22

∴ \({X_{d\left( {new} \right)}} = 0.2 \times \frac{{100}}{{500}} \times {\left( {\frac{{11}}{{22}}} \right)^2} = 0.01\;pu\)

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