Page 149 - Physics Coursebook 2015 (A level)

Chapter 9: Electric current, potential difference and resistance
Fuses
A fuse is a device which is fitted in an electric circuit; it is usually there to protect the wiring from excessive currents. For example, the fuses in a domestic fuse box will ‘blow’
if the current is too large. High currents cause wires to
get hot, and this can lead to damaged wires, fumes from melting insulation, and even fires.
Fuses (Figure 9.15) are usually marked with their current rating; that is, the maximum current which they will permit. Inside the fuse cartridge is a thin wire which gets hot and melts if the current exceeds this value. This breaks the circuit and stops any hazardous current. Worked example 7 shows how an appropriate fuse is chosen.
Figure 9.15 Fuses of different current ratings. QUESTION
19 An electric cooker is usually connected to the mains supply in a separate circuit from other appliances, because it draws a high current. A particular cooker is rated at 10 kW, 230 V.
a Calculate the current in the cooker when it is fully switched on.
b Suggest a suitable current rating for the fuse for this cooker.
WORKED EXAMPLES
6 a
A power station produces 20 MW of power at a voltage of 200 kV. Calculate the current supplied to the grid cables.
Step1 HerewehavePandVandwehaveto findI,sowecanuseP=VI.
Step2 Rearrangingtheequationand substituting the values we know gives:
currentI = P = 20×106 =100A V 200 × 103
Hint: Remember to convert megawatts into watts and kilovolts into volts.
So the power station supplies a current of 100 A.
b The grid cables are 15 km long, with a resistance per unit length of 0.20 Ω km−1. How much power is wasted as heat in these cables?
Step1 Firstwemustcalculatetheresistanceof the cables:
resistance R = 15km × 0.20Ωkm−1 = 3.0Ω Step2 NowweknowIandRandwewantto
findP.WecanuseP=I2R:
power wasted as heat, P = I 2R = (100)2 × 3.0
=3.0×104W
=30kW
Hence, of the 20 MW of power produced by the power station, 30 kW is wasted – just 0.15%.
7 An electric kettle is rated at 2.5 kW, 230 V. Determine a suitable current rating of the fuse to put in the three-pin plug. Choose from 1 A, 5 A, 13 A, 30 A.
Step1 Calculatethecurrentinthekettleinnormal operation. Rearranging P = VI to make I the subject gives:
I = PV
So:
I = 2500 = 10.9 A 230
Step2 Nowweknowthatthenormalcurrentinthe kettle is 10.9 A. We must choose a fuse with a slightly higher rating than this. Therefore the value of the fuse rating is 13 A.
Hint: A 5 A fuse would not be suitable because it would melt as soon as the kettle is switched on. A 30 A fuse would allow more than twice the normal current before blowing, which would not provide suitable protection.
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