Applied Electricity Paper 2 , May/June 2010

Questions:	1	2	3	4	5	6	7	8	Main

General Comments
Weakness/Remedies
Strength

Question 1
(a) (i) Define energy. (ii) State the unit of energy (b) In figure 3, the supply current Is is 8 mA. Calculate the: (i) value of resistance R1; (ii) total resistance; (iii) power in the circuit.
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Observation
The expected answers were: (a) (i) Energy is the ability to do work. (ii) Joule or kilowatt-hour, watt-sec. (b) Is= 8 mA IR2= 6 mA :.IR1= Is-IR2 =(8-6) mA = 2 mA (i) R1=V/IR1 R1= 24/2 mA = 12 kn (ii) Total resistance, RT =R1RJRl + R2 =4 X 103 x 12 X 103/(4 x103 + 12 X 103) =48 x 103/16 =3kn (iii) Power = Is 2RT = (8 X 10-3)2 X 3 X 103= 192 mW OR = (ld2R2 + (IR1)2R1 = (6 X 10-3)2 x 4 X 103 + {(2 X 10-3)2 X 12 X 103} = 0.192 watts = 192 mW The question was on Direct Current Circuit Theory. It was quite popular with the candidates. Many of the candidates who attempted the question got it right. However, only a few could not parallel the resistors correctly.