Applied Electricity 1 , May/June 2009

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Questions:		1	2		Main

General Comments
Weakness/Remedies
Strength

Question 2

(a) Connect the circuit as shown in fig 2.

(b)Ask the supervisor to check the circuit.

(c)Copy Table 3.

Table 3

Quantity	Value
V₁ V₂ V₃ V₄ I

(d) Adjust the power supply to 9 V and close switch S.

     (e)       Use the voltmeter to measure voltages V₁, V₂, V₃,
                 and record in Table 3.
     (f)       Take the reading of the ammeter and record in Table 3.
     (g)      Evaluate V ₁ + V₂.
     (h)      Evaluate V₃ + V₄
     (i)         Compare the results obtained in step (g) with that obtained in step (h).
     (j)         Evaluate     α   = V₃
                                           R₃
    (k)        Evaluate    β    = V₁
                                          R₁
    (l)        Evaluate   α + β.
    (m)      Compare the results obtained in step (l) with the value of I in Table 3.

_____________________________________________________________________________________________________

Observation

The candidates were required to:

(i) connect the circuit as shown and obtain voltage and current readings;

(ii) calculate the voltage and current in each branch.

The expected answers were:
Tables 3

Quantity	Value
V₁	5.625V
V₂	3.375V
V₃	6.907V
V₄	2.093V
I	490.6mA

(g)        V₁ + V₂ = (5.625 + 3.375)V = 9v
(h)        V₃ + V₄ = (6.907 + 2.093)V = 9v
(i)         Results obtained from steps (g) and (h) were found to be the same.
(j)         I₃ = V₃ =     6.907       = 0.2093A or 209.3mA
                    R₃              33

(k)          I₁ = V₁   =   5.625       = 0.2813A or 281.3m A
                         R₁             20
(l)         I₃ + I₁    = (0.2093 + 0.2813) A = 0.4906A or 490.6mA
(m)      On comparing step (l) with the value of I, they were the same.

The candidates understood the task, they connected the resistors as required and obtained readings of the current and voltage drop. However, some of the candidates’ readings were not correct or not expressed in the correct units.