SSE JE Electrical Engineering 10 Dec 2020 Papers

Question : 161 of 200

Marks: +1, -0

In a 1-phase 250 / 3000 V, 50 Hz transformer, if the EMF per turn is 8 V and

\I_\LV = 400 \A

, determine the LV and HV side turns and power rating of the transformer.

Solution:

\E_\LV = 250 \V, \E_\HV = 3000, \I_\LV = 400 \A,

Emf per turn = 8 V

\E_\LV = \N_\LV × \text "emf per turn"

∴ \N_\LV= \E_\LV/ \text "emf per turn "

Similarly

\N_\HV = \E_\HV / \text "emf per turn"

\N_\LV= 250 / 4 = 32

(aprox)

\N_\HV = 3000 / 8 = 375

Power rating P

= \E_\LV× \I_\LV

= 250 × 400 = 100 kVA

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