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An alternator has a per unit impedance of 0.9 p.u. to a base of 20 MVA, 33 kV, Then the p.u. impedance to the base of 50 MVA and 11 kV.

Base 1: S = 20 MVA, V = 33 kV, Z = 0.9 pu Base 2: S = 50 MVA, V = 11 kV

We can calculate the per unit impedance on Base 2 as follows:

Z2 (pu) = Z1 (pu) x (S1 / S2) x (V2 / V1)2

Z2 (pu) = 0.9 x (20 x 106 / 50 x 10^6) x (11 x 103 / 33 x 103)2

Z2 (pu) = 20.25

Therefore, the per unit impedance to the base of 50 MVA and 11 kV is 20.25 pu.

\(\begin{array}{l}{Z_{pu\;new}} = {Z_{pu\;old}}\left( {\dfrac{{MV{A_{base\;new}}}}{{MV{A_{base\;old}}}}} \right){\left( {\dfrac{{K{V_{base\;old}}}}{{K{V_{base\;new}}}}} \right)^2}\\\\ = 0.9 \times \dfrac{{50}}{{20}} \times {\left( {\dfrac{{33}}{{11}}} \right)^2=20.25}\end{array}\)