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Applied Electricity Paper 2 , Nov/Dec. 2008  
Questions:   1 2 3 4 5 6   Main
General Comments
Weakness/Remedies
Strength











































question 5

(a) State:   
    1. two applications of a transformer;
    2. two functions of oil in a transformer.

                241

(b) A 40kVA three-phase transformer has a core loss of 400W and a full

load copper loss of 400W.  If the power factor is unity, calculate the efficiency of the transformer
_____________________________________________________________________________________________________
OBSERVATION

The expected answers were:
            (a) (i)    Any two correct applications

    1. step up and step down voltages; impedance matching
    2. protect instruments
    3. 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
                            ii 
                    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.

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