Correct Answer - Option 1 : 2 pu
Concept:
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 \({X_{pu}} = \frac{{{X_{Actual}}}}{{{X_{base}}}}\)
\({X_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)
Where,
(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
Calculation:
Given that,
Xd(old) = 1 pu
MVA(new) = 200
MVA(old) = 100
kV(old) = 20
kV(new) = 20
∴ \({X_{d\left( {new} \right)}} = 1 \times \frac{{200}}{{100}} \times {\left( {\frac{{20}}{{20}}} \right)^2} = 2\;pu\)