SSC JE Electrical Engineering 2011 Paper

Question : 139 of 150

Marks: +1, -0

An equipment has an impedance of 0.9 p.u. to a base of 20 MVA, 33 kV. To the base of 50 MVA, 11 kV, the p.u. impedance will be

Solution:

(P.U. Impedence)

_\New

= (P.U. Inpedence)

_\old×{(\kVA)_\New}/{(\kVA)_\old}×{(\kV)^2}_\old/{(\kV)^2}_\New

9×50/20×(33/11)^2

= 6.75 × 3

= 20.25

P.U.

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