The expected answers were:
(a) (i) Any two correct applications
- step up and step down voltages; impedance matching
- protect instruments
- isolate
(ii) It cools the transformer.
It provides electrical installation between interval live parts.
(b) Iron losses = 400W
Copper losses = 400W
Total losses = 800W
Efficiency = Output x 100
Output + Losses
= 40,000 x 100
40,000 + 800
= 98.04%
This question was on electrical machines. Many candidates attempted it.
Most of the candidates stated the two applications and only one function in part (a).
In part (b), candidates mistook the input for the output thus missing the calculation of efficiency.
