QUESTIONSHEET 1

(a) (i) 0.4 A 1 1

(ii) 0.4 A 1 1

(b) (i) potential difference = current ´ resistance 1 1

2.4 V 1 1

(ii) 1.6 V 1 1

(c) showing all working 1 1

correct answer with units for total resistance: 16 W 1 1

calculate third resistance: 6 W 1 1

TOTAL / 8

QUESTIONSHEET 2

(a) (i) 144 kJ / 144,000 J 1 1

(ii) Lost in heating the surroundings, eg the kettle itself. 1 1

(b) (i) 10 A 1 1

(ii) iron / cooker / covector heater /

any appliance whose main purpose is producing heat 2

(c) (i) 2.4 ´ 20/10 = 4.80 1 1

(ii) 0.1 ´ 20 ´ 10 = 20p 1 1

(iii) 20 ´ 100 = 4.17% 1 1

480

TOTAL / 8

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

TOTAL / 8

QUESTIONSHEET 4

(a) (i) 12V - 10V = 2 V 1

(ii) Voltage = current ´ resistance 1

(iii) 2 V/1000 W 1
0.002 A or 2 mA 1

(b) (i) 0.008 A ´ 1000 W 1
8V 1

12 - 8 = 4 V 1

(ii) 4 V/0.008 A 1

= 500 W 1

TOTAL / 9

QUESTIONSHEET 5

(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 1

no connection C to D 1 1

Medium heat

Single connection A to B 1 1

No connection C to D 1 1

Low heat

Connection A to B 1 1

And C to D 1 1

(ii) Parallel circuits each element 1 1

has same current 1 1

medium heat has only one element 1 1

therefore half the heat 1 1

series circuit shares current 1 1

between elements 1 1

TOTAL / 12

QUESTIONSHEET 7

(a) (i) 3000 / 240 1 1

= 12.5 amps 1 1

(ii) 13 amps 1 1

(b) 240 / 12.5 1 1

= 19.2 ohms 1 1

(c) earth connected to case of fire 1 1

TOTAL / 6

QUESTIONSHEET 8

(a) 5A ´ 240 V 1 1

= 1200 W 1

1200 / 60 1 1

= 20 1 1

(b)

3

(c) 6 ´ 15 ´ 1.2 = total units 1 1

= 108 kWh 1

Price = 108 ´ 12 p 1 1

= £12.96 1

TOTAL / 10

QUESTIONSHEET 9

(a) voltage across coil 1 = number of turns in coil 1

voltage across coil 2 number of turns in coil 2 1

working 1 1

correct answer with units: 12 V 1 1

(b) (i) current / electric current / current flow 1 1

(ii) A2 1 1

(c) (i) low current 1 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

1 mark for answers which include the idea that:

transmitting at high voltage is more efficient than at high current. 1

TOTAL / 8

QUESTIONSHEET 10

(a) (i) a resistor whose resistance changes with temperature. 1

(ii)

Resistance (kW)
	Temperature (^oC)

Deduct 1 mark for each incorrectly plotted point to a maximum of 2. 2 2

(iii) decreases / goes down 1 1

(b) (i) decreases / goes down 1 1

(ii) potential difference = current ´ resistance 1 1

0.04 mA / 0.00004 A 1 1

(c) Light dependent resistor (LDR) 1 1

TOTAL / 8

QUESTIONSHEET 11

(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)

Current (A)

Deduct 1 mark for each incorrectly plotted point to a

maximum of 2. 2 2

(ii) 1 V + 0.1 V 1 1

(iii) resistance = potential difference

current 1 1

5 W + 0.5 W 1 1

(iv) increases 1 1

TOTAL / 8

QUESTIONSHEET 12

(a) (i) 787, 748, 604, 527:

all correct 2 2

3 correct. 1 1

(ii) December 1998 1 1

(iii) Weather colder so more energy required for heating or cooking/

nights longer so more lighting required. 1 1

(b) (i) times (h): 0.25, 0.25, 2, 0.5, 4 1 1

kWh: 0.5, 0.175, 0.4, 0.4, 0.41 1 1

Deduct 1 mark per incorrect answer to a total of 3 marks.

(ii) 18.75p 1 1

TOTAL / 8

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

TOTAL / 5

QUESTIONSHEET 14

(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 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 W 1

(d) Assume the heating element has the same resistance
current = voltage/resistance = 110 V/44 W = 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

TOTAL / 10

QUESTIONSHEET 16

(a) if a short circuit occurs 1 1

current will flow to earth 1 1

case will not become live 1 1

(b) short circuit causes fuse to heat up 1 1

and melt 1 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

3

(c) Very large current flows 1
in the live and earth wires 1
fuse melts and breaks, preventing further current flow. 1

TOTAL / 8

QUESTIONSHEET 18

(a)

Wire	Name	Colour
A	NEUTRAL	BLUE
B	EARTH	GREEN AND YELLOW
C	LIVE	BROWN

3 3

(b) (i) earth wire 1 1

(ii) since the case of the hairdryer is plastic, there are no exposed metal parts 1 1
no danger of electrocution if the live wire touches the plastic case. 1

1

(c) Fuse protects wires 1 1

high current flows & melts fuse wire 1 1

circuit breaks 1 1

TOTAL / 9

QUESTIONSHEET 19

(a) kilowatt 1 1

one hour 1 1

unit 1 1

(b) 3 600 000 1 1

3.6 mV 1

(c) (i) 3 amp 1

13 amp 1

5 amp 1

(ii) A 3 ´ 2 = 6 units 1 1

B 0.1 ´ 12 = 1.2 units 1 1

Immersion heater 1 1

TOTAL / 11

QUESTIONSHEET 20

One mark for each of

current

fire

damaged

metals

melting

5 amps

melt

live

TOTAL / 8