Applied Electricity Paper 2 , May/June 2010
 Questions: 1 2 3 4 5 6 7 8 Main
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

(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.