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Applied Electricity Paper 2, May/June2007  
Questions: 1 2 3 4 5 6 Main
General Comments
Weakness/Remedies
Strength

 

 

 

 

 

 

 

 

 

 

 

 

 

Question 6

(a)Define conductivity of a material.

(b)a piece of copper wire of length 100m has a coss-sectional area of 0.6 mm2. If the resistivity of copper is 1.73 x 10-8Ωm, calculate the resistance of the wire.

(c)A current of 15 A flows for 3 hours through a 5Ω resistor. Calculte the:
(i)Power dissipated in the resistor;
(ii)Energy consumed by the resistor.

_____________________________________________________________________________________________________
Observation

The expected answers were:

 (a) Conductivity is the ability of a material to allow the passage of electric current. It is the inverse of resistivity.
    σ = 1
         p
(b)        Using R = pl 
                                A
Where p = resistivity l = length
      A = area
       R = 1.73 X 10-8 x 100 = 2.88
            = 0.6 x 10-6

(c)(i) Power = 12R = 152 x 5 = 1125W or 1.125kW

(ii) E = P x T = 1125 x 3 = 3375Wh or 3.375kWh
This question on direct current circuit theory was attempted by many c8no;J<: (cs. The performance was fairly good.

Most of the candidates gave correct formulae for resistance but missed the correct
answer because of wrong conversion from mm2 to m2. Many of the candidates got the power dissipated in R but could not go beyond this to compute the energy consumed by R.

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