Part (a). Poor responses were given to the expected explanation of the observed spreading out of
the candle flame.

Part (b). Correct responses were quite high

Part (c). Generally, candidates did not show enough knowledge of what was required and the
attendant mathematical computation due to poor interpretations of the given circuit diagram.

The expected answers are:

14. (a)The candle flame ionizes the air around it. .

The positively charged conductor attracts the negative charges
in the air and repels the positive charges.

(b)F = qvB Sin θ .

= 1.6 X 10-19 X 5.0 X 105 X 0.2 X Sin 30°

= 0.8 X 10^{-1}

(c)(i)E = I (Rc + r)

9 = 4(Rc + 0.5)

R: = *9/4-0.5 *

= 1.75Ω

(ii) Lost volt = Ir = 4.0 X 0.5 =2.0 V

V oltage across 2Ωand RΩ

V = 9.0-2.0 =7.0V

Current in A_{2} = I_{2} = *V/R*_{2}= *7/2 *= 3.5A

Current in A_{3} =I_{3} = V/R = 7/14 = 0.5A

OR

A_{3} = A_{1} - A_{2}

= 4.0 - 3.5 = 0.5A

(iii) 1/R_{C} = *1/R *+ *1/R2 *

1/1.75 = *1/R *+ 1/2

*1/R *= 1/1.75- 1/2

R = 14Ω