(a)(i) How does the increase of length and cross-section area of a conductor affect its resistance?
(b)(i) State the function of a circuit breaker in a wiring system.
(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.
(c) An electric kettle contains 720 W heating units:
(i) What current does it take from 240 V mains?
(ii) How long will the kettle take to raise the temperature of 2 kg of water at 30Â°C to its boiling point?
(a)(i) How does the increase of length and cross-section area of a conductor affect its resistance?
(b)(i) State the function of a circuit breaker in a wiring system.
(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.
(c) An electric kettle contains 720 W heating units:
(i) What current does it take from 240 V mains?
(ii) How long will the kettle take to raise the temperature of 2 kg of water at 30°C to its boiling point?
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}/d^{2}_{A} ... (ii) and R_{B }= kl_{B}/d^{2}_{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}/d^{2}_{A})/(kl_{B}/d^{2}_{B}) = (kl_{A} x d^{2}_{B})/ (kl_{B} x d^{2}_{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 d^{2}_{B}]/ [kl_{B} x (2d)^{2}_{B}]
Resistance ratio = [k0.5l_{B} x d^{2}_{B} ]/[ kl_{B} x 4d^{2}_{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
Therefor it will take 13.6 min for the heater to raise temperature of water to boiling point.
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