QUESTIONSHEET 1 (a) (i) 0.4 A 1 (ii) 0.4 A 1 (b) (i) potential difference = current resistance 1 2.4 V 1 (ii) 1.6 V 1 (c) showing all working 1 correct answer with units for total resistance: 16 Ω 1 calculate third resistance: 6 Ω 1 QUESTIONSHEET 2 (a) (i) 144 kj / 144,000 J 1 (ii) Lost in heating the surroundings, eg the kettle itself. 1 (b) (i) 10 A 1 (ii) iron / cooker / covector heater / any appliance whose main purpose is producing heat 2 (c) (i) 2.4 20/10 = 4.80 1 (ii) 0.1 20 10 = 20p 1 (iii) 20 100 = 4.17% 1 480
QUESTIONSHEET 3 (a) time = 5 3 minutes = 15 minutes = 1/4 hour 1 units = 2.4 kw 1/4 hour = 0.6 kwh 1 cost = 0.6 8p = 4.8p 1 (b) (i) Power = current x voltage 1 (ii) current = power/voltage = 2400 W/240 V 1 10 A 1 (iii) 13 A 1 (c) If the cable/cord was damaged could be a large current 1 QUESTIONSHEET 4 (a) (i) 12V - 10V = 2 V 1 (ii) Voltage = current resistance 1 (iii) 2 V/1000 Ω 1 0.002 A or 2 ma 1 (b) (i) 0.008 A 1000 Ω 1 8V 1 12-8 = 4 V 1 (ii) 4 V/0.008 A 1 = 500 Ω 1 QUESTIONSHEET 5 TOTAL / 9 (a) dangerous 1 too much current may flow/overloading the circuit 1 causing much heat and possibly a fire 1 (b) If fault develops circuit breaker acts more quickly than a fuse 1 reducing the chance of electrocution. 1 TOTAL / 5
QUESTIONSHEET 6 (a) (i) High heat parallel connection (A to B & A to D) 1 no connection C to D 1 Medium heat Single connection A to B 1 No connection C to D 1 Low heat Connection A to B 1 And C to D 1 (ii) Parallel circuits each element 1 has same current 1 medium heat has only one element 1 therefore half the heat 1 series circuit shares current 1 between elements 1 QUESTIONSHEET 7 TOTAL / 12 (a) (i) 3000 / 240 1 = 12.5 amps 1 (ii) 13 amps 1 (b) 240 / 12.5 1 = 19.2 ohms 1 (c) earth connected to case of fire 1 TOTAL / 6
QUESTIONSHEET 8 (a) 5A 240 V 1 = 1200 W 1200 / 60 1 = 20 1 (b) mains (c) 6 15 1.2 = total units 1 = 108 kwh Price = 108 12 p 1 = 12.96 1 QUESTIONSHEET 9 3 TOTAL / 10 (a) voltage across coil 1 = number of turns in coil 1 voltage across coil 2 number of turns in coil 2 1 working 1 correct answer with units: 12 V 1 (b) (i) current / electric current / current flow 1 (ii) A2 1 (c) (i) low current 1 (ii) 1 mark for answers which include the idea that: flow of current converts some electrical energy to heat energy which is lost by convection 1 1 mark for answers which include the idea that: transmitting at high voltage is more efficient than at high current. 1
QUESTIONSHEET 10 (a) (i) a resistor whose resistance changes with temperature. 1 (ii) Resistance (kω) 400 350 300 250 200 150 100 50 0 0 10 20 30 40 50 60 70 80 90 100 Temperature ( o C) Deduct 1 mark for each incorrectly plotted point to a maximum of 2. 2 (iii) decreases / goes down 1 (b) (i) decreases / goes down 1 (ii) potential difference = current resistance 1 0.04 ma / 0.00004 A 1 (c) Light dependent resistor (LDR) 1
QUESTIONSHEET 11 A (a) all components present: battery or cell, variable resistor, bulb and ammeter all in series, 1 voltmeter connected in parallel with bulb only. 1 (b) (i) Potential Difference (V) 4 3 2 1 0 0 0.1 0.2 0.3 0.4 Current (A) Deduct 1 mark for each incorrectly plotted point to a maximum of 2. 2 (ii) 1 V + 0.1 V 1 (iii) resistance = potential difference current 1 5 Ω + 0.5 Ω 1 (iv) increases 1 V
QUESTIONSHEET 12 (a) (i) 787, 748, 604, 527: all correct 2 3 correct. 1 (ii) December 1998 1 (iii) Weather colder so more energy required for heating or cooking/ nights longer so more lighting required. 1 (b) (i) times (h): 0.25, 0.25, 2, 0.5, 4 1 kwh: 0.5, 0.175, 0.4, 0.4, 0.41 1 Deduct 1 mark per incorrect answer to a total of 3 marks. (ii) 18.75p 1 QUESTIONSHEET 13 (a) low energy bulb 1 (b) units = 0.15 kw 1000 = 150 units (note the change of watt to kilowatt) 1 150 8p 1 1200 p or 12 1 (c) 12 5 = 2.40 1 QUESTIONSHEET 14 TOTAL / 5 (a) 1000 W (power is measured in watt, W) 1 (b) (i) Power = current voltage 1 (ii) current = power/voltage = 1000 W/230 V 1 4.3(5) A 1 (c)(i) safety device 1 if too much current flows, it breaks and prevents further current flow 1 (ii) 5 A (use the lowest value which will allow the device to operate normally.) 1 TOTAL / 7
QUESTIONSHEET 15 (a) 1100 W 1 (b) (i) Power = current voltage 1 (ii) current = power/voltage = 1100 W/220 V 1 5 A 1 (c) (i) Voltage = current resistance 1 (ii) resistance = voltage/current = 220 V/5 A 1 44 Ω 1 (d) Assume the heating element has the same resistance current = voltage/resistance = 110 V/44 Ω = 2.5 A 1 power = current voltage = 2.5 A 110 V = 275 W 1 new power = 275/1100 = 1/4 of original power 1 QUESTIONSHEET 16 TOTAL / 10 (a) if a short circuit occurs 1 current will flow to earth 1 case will not become live 1 (b) short circuit causes fuse to heat up 1 and melt 1 cutting off power 1 TOTAL / 6
QUESTIONSHEET 17 (a) current = power/voltage = 3000 W/240 V 1 12.5 A 1 (b) wire current /A earth 0 live 12.5 neutral 12.5 (c) Very large current flows 1 in the live and earth wires 1 fuse melts and breaks, preventing further current flow. 1 QUESTIONSHEET 18 (a) Wire Name Colour A NEUTRAL BLUE B EARTH GREEN AND YELLOW C LIVE BROWN 3 (b) (i) earth wire 1 (ii) since the case of the hairdryer is plastic, there are no exposed metal parts 1 no danger of electrocution if the live wire touches the plastic case. 1 (c) Fuse protects wires 1 high current flows & melts fuse wire 1 circuit breaks 1 3 TOTAL / 9
QUESTIONSHEET 19 (a) kilowatt 1 one hour 1 unit 1 (b) 3 600 000 1 3.6 mv (c) (i) 3 amp 13 amp 5 amp (ii) A 3 2 = 6 units 1 B 0.1 12 = 1.2 units 1 Immersion heater 1 QUESTIONSHEET 20 One mark for each of current fire damaged metals melting 5 amps melt live TOTAL / 11