SECTION 01

(a) How does the increase of length and cross-section area of a conductor affect its resistance?

Increase of length of conductor increases its resistance while increase of cross-sectional area of the conductor decreases its resistance

Resistance of conductor α (length / cross-sectional area)

SECTION 02

(b)(i) State the function of a circuit breaker in a wiring system.

The main function of circuit breaker is to disconnect electricity when electric current exceeds the rated value of electric current.

(ii) Determine the ratio of resistance of wire A to that of wire B which are made up of the same material such that wire A has half the length and twice the diameter

of wire B.

Resistance,   R α l/A  , R = (kl)/A ... (i) hence R_A= kl_A/A_A = kl_A/d²_A ... (ii) and R_B = kl_B/d²_B …(iii)

(assume π/4 from cross-sectional area included within the constant …)

Resistance ratio = R_A/ R_B ... (iv), but l_A = l_B /2 = 0.5l_B and d_A = 2d_B

Substitute equation (ii) and (iii) to equation (iv) we obtain ...

Resistance ratio = R_A/ R_B = (kl_A/d²_A)/(kl_B/d²_B) = (kl_A x d²_B)/ (kl_B x d²_A) … (v)   

Substitute the equivalent value of l_A and d_A to equation (v)

Resistance ratio = R_A/ R_B = [k0.5l_B x d²_B]/ [kl_B x (2d)²_B] 

Resistance ratio = [k0.5l_B x d²_B ]/[ kl_B x 4d²_B ]= 0.5/4 = 1/8

Resistance ratio = 1/8

SECTION 03

(c) An electric kettle contains 720 W heating units:

(i) What current does it take from 240 V mains?

P=VI

V = 240 v and P= 720W

I=P/V=(720W)/(240V)

I=3A

(ii) How long will the kettle take to raise the temperature of 2 kg of water at 30°C to its boiling point?

Data and Solution:

Temperature change = 30℃ to 100℃

Mass of water = 2Kg

Power equation:

P = E/t …. (i)

E=Pt 

E = Specific Heat Capacity x Mass x Change in Temperature

E = SHC x M x ∆℃ …. (ii)

Comparing equation (i) and equation (ii)

SHC x M x ∆℃ = Pt

t = [SHC x M x ∆℃]/P

t= [4200 x 2 x (100 - 30)]/720

t = 816.67sec = 13.6 min