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Domestic power Higher tier question

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1 Domestic power Higher tier question
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2 Question (part 1) A mains electric fire has two settings. On the high setting, it has a current of 10 A. (a) Show that on the high setting the power is 2.3 kilowatts. [2] (b) (i) Calculate the energy in joules that is transferred when the fire is used on the high setting for five minutes. energy = __________J

3 Question (part 2) (ii) Each kilowatt hour costs 10 p. Calculate the weekly cost of using the fire on the high setting, assuming that it is used on average for 90 minutes each day. cost = £__________ [2] On the low setting, the fire has a power of 1.1 kW. Calculate the current taken by the fire on the low setting.

4 Explanation of question (part 1)
In (a), you need to recall that the mains voltage is 230 V. As you are being asked to ‘show that’ the power is 2.3 kilowatts, you should carry out the calculation in full, showing all the steps as you go. Don’t for get to show that 1 kW is 1000 W. In (b), you need to carry out calculations with multiple steps. In (i) you must remember to convert from minutes to seconds and use power in watts rather than kilowatts before you calculate the energy. Part (ii) is more straightforward, even though it is a three-stage process. The sequence of calculations to follow is daily energy  weekly energy  cost.

5 Explanation of question (part 2)
In (c), you need to rearrange the equation power = voltage × current. You are required to rearrange equations in Higher tier examinations.

6 power = voltage × current so voltage = power ÷ current
D–C answer power = voltage × current so voltage = power ÷ current = 2300 ÷ 10 = 230 volts which is right. comment This response gets both marks. The watts to kilowatts conversion has been done correctly. Although it seems that the candidate has forgotten the mains voltage, which would explain why the equation was rearranged, the marks would definitely be given. Next Previous .

7 D–C answer (continued) (i) 2300 × 5 = 11,500 energy = 11,500 J
(ii) 11,500 J for 5 minutes = ,000 J for 90 minutes. cost = ____ comment This answer gets one of the two marks as the candidate has forgotten to convert the time to seconds. The candidate has made a start on the question but lost confidence. If you get stuck like this, you may wish to leave the question and come back to it at the end. Next Previous .

8 D–C answer (continued) (c)
(c) current = power ÷ voltage = 1.1 kW ÷ 230 V = A comment The candidate has used a formula triangle to rearrange the equation. This is a popular method but it is too limiting to recommend. Unfortunately, the candidate has been careless and has not converted power into watts. This was done correctly in (a). This suggests that the candidate has not checked their work carefully enough. Next Previous .

9 B–A* answer (a) Voltage = 230 V (mains) So power = voltage × current = 230 V × 10 A = 2300 W (b) (i) 5 minutes = 5 × 60 s = 300 s Energy = power × time = 2300 W × 300 s energy =690,000 J comment This answer to (a) has all the steps in calculation laid out clearly, including changing the subject of the power equation. It is a good idea in a question such as this to write out the equation being used in words. In (b) (i), the candidate has used the correct values for power and time and set out the calculation carefully. This isn’t just for the examiners – it will help you think through the question if you write it out as you go. Next Previous .

10 B–A* answer (continued) (ii) 90 minutes = 1½ hours per day
Hours per week = 7 × 1½ h = h So energy in 1 week = 2.3 kW × 10.5 h = kWh cost = £2.42 comment The answer has been rounded off to the nearest penny, although both £ and £2.415 would have been accepted. The arithmetic in working out the cost from the energy was straightforward so the candidate did not write it down. This can be risky in an examination but no mistake is made here. Next Previous .

11 B–A* answer (continued) power = voltage × current
so current = power ÷ voltage = 1100 W ÷ 230 V = 4.78 A comment The candidate has again written out the relevant equation and the re-arranged equation, and has recalled that current is measured in amps. They have given units in all the parts of the calculations which is good exam technique. Next Previous .


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