International GCSE Physics · Pearson Edexcel 4PH1 · Topic 2
Electricity
Original exam-style questions with mark schemes. Questions marked “Paper 2 only” use content assessed only on Paper 2. Notes 2(b) →Notes 2(c) →Notes 2(d) →
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Freefull mark schemes
Original questions in the style of each exam boardEvery answer re-solved independently before publishingCreated by Evan Kotronis with AI assistanceUpdated 13 October 2026 · new questions added regularlyHow every question is checked →
1IGCSE-PHY-2-001The volt as a joule per coulombPapers 1 & 2Grade 4–51 mark
A student is learning the units used in electricity. Which of these is equivalent to one volt (1 V)?
Show mark scheme
B (1 joule per coulomb);
A is the inverse — confuses which quantity is divided by which in voltage = energy ÷ charge
C is the watt (power = energy ÷ time) — confuses voltage with power
D is the ampere (current = charge ÷ time) — confuses voltage with current
2IGCSE-PHY-2-002Current at a junction in a parallel circuitPapers 1 & 2Grade 4–51 mark
The diagram shows a circuit with a lamp and a resistor connected in parallel. The reading on ammeter A₁ is 0.80 A. The reading on ammeter A₂ is 0.30 A. What is the reading on ammeter A₃?Show mark scheme
D (0.50 A);
A adds the two readings — thinks the branch currents add up to more than the supply current
B thinks the current is the same everywhere in the circuit, as in a series circuit
C thinks the current always splits equally between the two branches
3IGCSE-PHY-2-003a.c. and d.c., indicator LEDs and charge carriersPapers 1 & 2Grade 4–55 marks
A rechargeable LED torch is charged from the mains using a charger. When the torch is in use, it is powered by a battery inside the torch. A small red LED on the charger lights up while the torch is charging.
(a)
State the type of current supplied by the mains.
(1)
(b)
Describe how the current supplied by the battery is different from the current supplied by the mains.
(2)
(c)
Suggest why the charger has a small LED that lights up while the torch is charging.
(1)
(d)
Name the charged particles that move in the copper wires of the charger cable.
(1)
Show mark scheme
(a)
alternating current / a.c.;
Allow
AC
Do not accept
direct current
mains current
(b)
(battery supplies) direct current / d.c. which flows in one direction only;
mains / a.c. (repeatedly) changes / reverses direction;
Allow
d.c. has a constant direction
a.c. goes backwards and forwards
d.c. voltage is constant / a.c. voltage changes from positive to negative
Do not accept
a.c. is stronger / more powerful than d.c.
d.c. is stored and a.c. is not
(c)
(to show) there is a current (in the circuit) / the torch is charging;
4IGCSE-PHY-2-004Electric blanket: power, current and energyPapers 1 & 2Grade 4–57 marks
An electric blanket is used to warm a bed. It contains a long heating wire inside the fabric. The blanket has a power of 60 W and is connected to the 230 V mains supply.
(a)
Explain why the heating wire gets warm when there is a current in it.
(2)
(b)
State the formula linking power, current and voltage.
(1)
(c)
Calculate the current in the heating wire.
(2)
(d)
The blanket is switched on for 1800 s. Calculate the energy transferred by the blanket in this time. Use the equation energy transferred = current × voltage × time.
(2)
Show mark scheme
(a)
(moving) electrons collide with ions / atoms (in the wire);
energy transferred (electrically) to thermal store (of wire) / ions vibrate more (so temperature rises);
Allow
particles for ions/atoms
lattice for ions
energy is transferred by heating to the wire
Do not accept
friction between electrons and wire
electrons get hot
electricity is converted into heat
(b)
power = current × voltage;
Allow
P = I × V
P = IV
rearranged forms
Do not accept
P = V/I
P = I/V
(c)
substitution into P = I × V;
e.g. 60 = I × 230 or I = 60 ÷ 230
evaluation;
e.g. (I =) 0.26 (A)
Allow
0.261 (A)
0.2608… (A)
Do not accept
3.8 (A) (inverted 230/60)
(d)
substitution into E = I × V × t;
e.g. E = 0.26 × 230 × 1800
evaluation;
e.g. (E =) 108 000 (J)
Allow
107 640 (J) using I = 0.26 A
values in range 107 000 to 108 000 (J)
1.1 × 10⁵ (J)
108 (kJ)
ECF from (c)
60 × 1800 = 108 000 (J) (E = P × t)
Do not accept
1800 (J) (time missed)
6480 (J) / 6.48 (kJ) with t = 30 (min) not converted
5IGCSE-PHY-2-005Thermistor resistance and current when heatedPapers 1 & 2Grade 4–51 mark
The diagram shows a thermistor connected in series with an ammeter and a battery. The thermistor is placed in a beaker of water. The water is heated so that the temperature of the thermistor increases. The voltage of the battery does not change. Which row describes what happens to the resistance of the thermistor and to the reading on the ammeter?Show mark scheme
C (resistance decreases, ammeter reading increases);
A treats the thermistor like a metal wire / filament, whose resistance rises with temperature
B thinks a larger resistance gives a larger current
D thinks every component has a fixed resistance that does not depend on temperature
6IGCSE-PHY-2-021What moves when there is a current in a copper wirePapers 1 & 2Grade 4–51 mark
A battery is connected to a lamp with copper wires. There is a current in the wires and the lamp lights. Which statement describes the current in the copper wires?
Show mark scheme
B (electrons moving from the negative terminal towards the positive terminal);
A protons are fixed in the nuclei of the copper atoms and do not move through the wire
C electrons are not used up; the same number leave the lamp as enter it (charge is conserved), it is energy that is transferred in the lamp
D the positive ions in a metal are held in fixed positions; they vibrate but do not flow along the wire
7IGCSE-PHY-2-022Recognising alternating current on a voltage–time graphPapers 1 & 2Grade 4–51 mark
The graphs A, B, C and D show how a voltage changes with time. The time axis of each graph is drawn at zero voltage. Which graph shows the voltage of the mains electricity supply?Show mark scheme
C (graph C);
A shows a constant voltage in one direction: this is the direct current (d.c.) supplied by a cell or battery
B the voltage changes in size but never becomes negative, so the current is always in the same direction: this is still d.c., not a.c.
D shows a voltage that decreases steadily, e.g. a battery running down: it never reverses direction
8IGCSE-PHY-2-023Power of a USB desk fanPapers 1 & 2Grade 4–57 marks
A small desk fan is plugged into the USB port of a laptop. The USB port supplies a voltage of 5.0 V. On its low setting, the current in the fan is 0.40 A.
(a)
Calculate the power of the fan on its low setting.
(2)
(b)
On its high setting, the current in the fan is 0.60 A. The voltage is still 5.0 V. Determine the power of the fan on its high setting.
(1)
(c)
State the unit of each of these quantities. charge energy resistance
(3)
(d)
The fan's motor contains coils of wire. The coils become warm when the fan is used. State why the coils become warm.
(1)
Show mark scheme
(a)
substitution;
e.g. P = 0.40 × 5.0
evaluation;
e.g. (P =) 2.0 (W)
Allow
2 (W)
award full marks for correct answer with no working
Do not accept
12.5 (W) (5.0 ÷ 0.40)
0.08 (W) (0.40 ÷ 5.0)
(b)
3.0 (W);
Allow
3 (W)
ECF: 1.5 × their (a)
Do not accept
8.3 (W) (5.0 ÷ 0.60)
(c)
(charge) coulomb / C;
(energy) joule / J;
(resistance) ohm / Ω;
Allow
kWh for energy
Do not accept
ampere / amp for charge
watt for energy
volt for energy
volt / ampere for resistance
(d)
(electrical) current in the wire / resistance of the wire transfers energy to the thermal (energy) store (of the coils);
Allow
electrons collide with (ions / atoms in) the wire, increasing its temperature
heating effect of a current
Do not accept
friction of the air / friction in the motor (only)
9IGCSE-PHY-2-024Light-beam security alarm using an LDRPapers 1 & 2Grade 4–56 marks
A jewellery shop uses a light-beam alarm across its doorway. A narrow beam of light shines across the doorway onto a light-dependent resistor (LDR). The LDR is part of a circuit with a battery and an alarm unit. The alarm unit sounds when the current in the circuit falls below a set value. A small green LED in the circuit is lit while the alarm system is switched on.
(a)
A person walks through the doorway and blocks the beam of light. State what happens to the resistance of the LDR.
(1)
(b)
Explain why the alarm sounds when the beam is blocked.
(2)
(c)
Suggest why the circuit includes the green LED.
(1)
(d)
In the afternoon, bright sunlight shines through the doorway onto the LDR. Suggest why the alarm might not sound when a person walks through the beam.
(2)
Show mark scheme
(a)
(resistance) increases;
Allow
goes up / becomes large
Do not accept
decreases
becomes zero
(b)
(resistance of LDR increases so) total resistance of circuit increases;
current decreases (below the set value);
Allow
less light, so higher resistance, so smaller current for 2 marks
Do not accept
current increases
voltage of the battery decreases
(c)
to show / indicate that there is a current (in the circuit);
Allow
to show that the alarm system is switched on / working
Do not accept
to light up the doorway
to provide the light beam
(d)
(sunlight / other light) still falls on the LDR;
(so) resistance (of LDR) stays low / current stays above the set value;
10IGCSE-PHY-2-006Fuse selection, a faulty earth connection and double insulationPapers 1 & 2Grade 6–79 marks
A portable electric heater has a metal case. It is connected to the 230 V mains with a three-core cable and a plug. The power of the heater is 2.8 kW. The plug can be fitted with a 3 A, 5 A or 13 A fuse.
(a)
Calculate the current in the heater when it is working normally.
(2)
(b)
State which fuse should be fitted in the plug. Give a reason for your answer.
(2)
(c)
A fault causes the live wire to touch the metal case of the heater. The earth wire has also come loose, so the case is not connected to earth. A person standing on the floor touches the case. Explain why the person gets an electric shock and why the 13 A fuse does not protect them.
(3)
(d)
A hairdryer has a plastic case and is double insulated. Explain why the hairdryer does not need an earth wire.
(2)
Show mark scheme
(a)
substitution and rearrangement;
e.g. I = 2800 ÷ 230
evaluation;
e.g. (I =) 12 (A)
Allow
12.2 (A)
12.17 (A)
0.012 (A) / 0.0122 (A) for 1 mark only (kW not converted)
Do not accept
82 (A) (inverted)
(b)
13 A;
(13 A is) just above / higher than normal current (and 3 A and 5 A would melt / blow in normal use);
Allow
ECF for nearest rating above answer to (a)
smallest fuse that is bigger than the current
Do not accept
13 A because it is the biggest fuse
5 A
(c)
case is live / at (mains) voltage / 230 V (because it is not earthed);
current flows through the person (to earth / the ground) so they get a shock;
current through the person is (much) less than 13 A / too small to melt the fuse (so the circuit is not broken);
Allow
person's body has a (much) higher resistance than an earth wire so current is small for MP3
fuse only melts when the current is larger than its rating for MP3
electrons / charge flow through the person for MP2
Do not accept
the fuse melts / blows
earth wire absorbs the electricity
the person absorbs the electricity / current is used up in the person
(d)
(plastic) case is an insulator / does not conduct;
so case cannot become live / no current can pass through a person touching it;
Allow
live parts are surrounded by two layers of insulation
11IGCSE-PHY-2-007Practical: I–V characteristic of a filament lampPapers 1 & 2Grade 6–79 marks
A student uses the circuit shown to investigate how the current in a filament lamp varies with the voltage across it. The table shows her results. Voltage / V: 0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0 Current / A: 0, 0.13, 0.21, 0.27, 0.32, 0.36, 0.40
(a)
Describe how the student uses the circuit to obtain her set of results.
(2)
(b)
The student says 'the current in the lamp is directly proportional to the voltage'. Use the data to explain why she is wrong.
(2)
(c)
Calculate the resistance of the lamp when the voltage is 6.0 V.
(2)
(d)
Explain why the resistance of the lamp changes as the current increases.
(2)
(e)
Describe how the student could extend the investigation to obtain readings for negative values of voltage.
(1)
Show mark scheme
(a)
adjust / change variable resistor (to change voltage across lamp);
record readings of ammeter and voltmeter (for each setting);
Allow
change resistance of variable resistor
measure current and voltage
Do not accept
change the battery to change voltage (no mention of variable resistor) for MP1
(b)
(when voltage doubles) current does not double / e.g. from 1.0 V to 2.0 V current rises from 0.13 A to only 0.21 A;
(so) ratio V ÷ I / resistance is not constant / graph would not be a straight line (through origin);
Allow
any correct pair of values from the table used to show current does not double
current increases by smaller amounts for each 1.0 V increase
Do not accept
because the current increases
current is not proportional because it starts at zero
(c)
substitution and rearrangement;
e.g. R = 6.0 ÷ 0.40
evaluation;
e.g. (R =) 15 (Ω)
Allow
15.0 (Ω)
Do not accept
0.067 (Ω) (inverted)
2.4 (Ω) (V × I)
(d)
(larger current makes) filament / lamp hotter / temperature increases;
ions / atoms vibrate more so electrons collide more (so resistance increases);
Allow
resistance increases with temperature for MP1 if hotter not stated
Do not accept
electrons get tired
the lamp uses up the current
(e)
reverse the connections to the battery / power supply (and repeat the measurements);
Allow
turn the battery / power supply round
swap the terminals of the power supply
Do not accept
turn the lamp round / reverse the connections to the lamp
decrease the voltage to below zero with the variable resistor
12IGCSE-PHY-2-008Two heating elements in series in a car-seat heaterPapers 1 & 2Grade 6–77 marks
A car-seat heater has two heating elements, A and B, connected in series to a 12 V car battery. The resistance of element A is 16 Ω. A voltmeter connected across element A reads 4.8 V.
(a)
Calculate the current in element A.
(2)
(b)
Determine the voltage across element B.
(1)
(c)
Calculate the resistance of element B.
(2)
(d)
Element B is replaced by an element with a larger resistance. Explain what happens to the current in element A.
(2)
Show mark scheme
(a)
substitution and rearrangement;
e.g. I = 4.8 ÷ 16
evaluation;
e.g. (I =) 0.30 (A)
Allow
0.3 (A)
Do not accept
3.3 (A) (inverted)
0.75 (A) (12 ÷ 16)
(b)
7.2 (V);
Allow
12 − 4.8 seen
Do not accept
4.8 (V)
(c)
substitution;
e.g. R = 7.2 ÷ 0.30
evaluation;
e.g. (R =) 24 (Ω)
Allow
ECF from (a) and (b)
R(total) = 12 ÷ 0.30 = 40 Ω then 40 − 16 = 24 (Ω)
Do not accept
40 (Ω) (total resistance)
(d)
total resistance (of circuit) increases;
(battery voltage same so) current (in A) decreases;
Allow
current is the same everywhere in a series circuit so current in A also decreases
Do not accept
current in A stays the same because A has not changed
13IGCSE-PHY-2-009Series decorative lights versus parallel domestic lightingPapers 1 & 2Grade 6–78 marks
A string of decorative lights has 20 identical lamps connected in series. The string is connected to the 230 V mains.
(a)
Determine the voltage across each lamp.
(1)
(b)
The filament in one of the lamps breaks. Explain why all of the lamps go out.
(2)
(c)
The lights in the rooms of a house are connected in parallel, not in series. Explain the advantages of connecting the lights in parallel.
(3)
(d)
Two lamps, X and Y, in a house are connected in parallel and are both switched on. The current in lamp X is 0.17 A and the current in lamp Y is 0.26 A. Determine the current in the cable that supplies both lamps. Explain your answer.
(2)
Show mark scheme
(a)
11.5 (V);
Allow
12 (V)
Do not accept
230 (V)
4600 (V)
(b)
(series so) only one path / circuit is broken / incomplete (at that lamp);
so no current (in any lamp);
Allow
gap in the circuit
Do not accept
the current is used up
the broken lamp takes all the voltage so others get none (without mention of no current)
(c)
any three from:
each lamp has the full (mains) voltage / 230 V (across it);
(so) each lamp is at full / normal brightness;
each lamp can be switched on and off independently / separately;
if one lamp fails / blows the others stay on;
Allow
each lamp has its own circuit / path for current
Do not accept
parallel circuits use less current
parallel is cheaper
(d)
0.43 (A);
current is conserved at a junction / supply current = sum of currents in branches;
14IGCSE-PHY-2-010Charge flow and recharging a cordless drill batteryPapers 1 & 2Grade 6–78 marks
A cordless drill is powered by an 18 V battery. When the drill is used to make some holes, the current in the motor is 1.5 A for a total time of 4.0 minutes.
(a)
State what is meant by an electric current.
(1)
(b)
The battery has a voltage of 18 V. State what this tells you about the energy transferred by the battery, in terms of charge.
(2)
(c)
Calculate the charge that flows through the motor in 4.0 minutes.
(3)
(d)
After the holes are made, the battery is recharged with a constant current of 0.60 A. Calculate the time needed for 360 C of charge to flow back into the battery. Give your answer in minutes.
15IGCSE-PHY-2-011Practical data: resistance of an LDR and distance from a lampPapers 1 & 2Grade 6–79 marks
A student investigates how the resistance of a light-dependent resistor (LDR) changes as a lamp is moved further away from it. The voltage across the LDR is 6.0 V for every reading. The table shows his results. Distance d / cm: 10, 20, 30, 40, 50 Current / mA: 12.0, 5.0, 2.6, 1.6, 1.1
(a)
State the dependent variable in this investigation.
(1)
(b)
Calculate the resistance of the LDR when the distance is 20 cm.
(3)
(c)
Describe how the resistance of the LDR changes as the distance increases. Use data from the table in your answer.
(2)
(d)
Explain why the student does the investigation in a darkened room.
(2)
(e)
State one variable, other than the voltage, that the student should keep the same.
(1)
Show mark scheme
(a)
current (in the LDR);
Allow
resistance (of LDR)
ammeter reading
Do not accept
distance
voltage
(b)
conversion of mA to A;
e.g. 5.0 mA = 0.0050 A
substitution and rearrangement;
e.g. R = 6.0 ÷ 0.0050
evaluation;
e.g. (R =) 1200 (Ω)
Allow
1.2 kΩ
1.2 (Ω) for 2 marks (mA not converted)
Do not accept
0.00083 (Ω) (inverted)
(c)
resistance increases (as distance increases);
not proportional / increases by greater amounts each time, supported by data;
Allow
e.g. distance doubles from 10 cm to 20 cm but resistance more than doubles from 500 Ω to 1200 Ω
resistance values calculated from 6.0 ÷ current
current decreases so resistance increases for MP1
Do not accept
resistance is proportional to distance
resistance decreases
(d)
light from other sources / room / windows would reach LDR;
(which would) change / lower the resistance (so only the lamp affects results);
Allow
so that only the lamp's light changes
to make it a fair test for MP2 only if linked to other light
16IGCSE-PHY-2-012Diode I–V graph: resistance and series resistorPapers 1 & 2Grade 6–710 marks
A student obtains the graph shown for a diode connected so that it conducts (forward direction).
(a)
Use the graph to calculate the resistance of the diode when the voltage across it is 0.70 V.
(3)
(b)
Use the graph to determine the resistance of the diode when the voltage across it is 0.60 V.
(2)
(c)
Another student says 'Once a diode starts to conduct, its resistance stays constant.' Use your answers to (a) and (b) to comment on this statement.
(2)
(d)
The diode is connected in series with a 270 Ω resistor and a battery, in the direction that conducts. The current in the circuit is 10 mA. Use the graph to determine the voltage of the battery.
(3)
Show mark scheme
(a)
current read from graph as 20 mA / 0.020 A;
substitution and rearrangement;
e.g. R = 0.70 ÷ 0.020
evaluation;
e.g. (R =) 35 (Ω)
Allow
0.035 (Ω) for 2 marks (mA not converted)
currents in range 19 to 21 mA giving 33 to 37 (Ω)
Do not accept
0.029 (Ω) (inverted)
(b)
current read as 5 mA / 0.005 A;
(R = 0.60 ÷ 0.005 =) 120 (Ω);
Allow
currents in range 4.5 to 5.5 mA giving 109 to 133 (Ω)
(c)
statement is wrong / resistance is not constant;
resistance decreases (as voltage / current increases), e.g. from 120 Ω to 35 Ω;
Allow
ECF from (a) and (b)
Do not accept
resistance increases
(d)
voltage across diode read from graph as 0.65 V;
voltage across resistor = 0.010 × 270 = 2.7 (V);
(battery voltage = 0.65 + 2.7 =) 3.35 (V);
Allow
3.3 / 3.4 (V)
diode readings 0.64 to 0.66 V giving 3.34 to 3.36 (V)
17IGCSE-PHY-2-014Parallel heating strips in a rear-window demisterPapers 1 & 2Grade 6–711 marks
The rear window of a car has a demister made from 10 identical heating strips connected in parallel to the 12 V car battery. Each strip has a resistance of 9.6 Ω.
(a)
State the voltage across each strip. Give a reason for your answer.
(2)
(b)
Calculate the current in one strip.
(2)
(c)
Determine the total current supplied by the battery to the demister.
(1)
(d)
Determine the total resistance of the demister.
(2)
(e)
One strip breaks. Explain the effect on the other strips and calculate the new total power of the demister.
(4)
Show mark scheme
(a)
12 (V);
(strips are in) parallel so each has the same voltage (as the battery);
Allow
each strip connected directly across the battery
Do not accept
1.2 (V) (voltage shared between 10 strips)
(b)
substitution and rearrangement;
e.g. I = 12 ÷ 9.6
evaluation;
e.g. (I =) 1.25 (A)
Allow
1.3 (A)
1.2 (A)
Do not accept
0.80 (A) (inverted)
0.125 (A)
(c)
(10 × 1.25 =) 12.5 (A);
Allow
ECF from (b)
13 (A)
12 (A) from 10 × 1.2
(d)
substitution of total current;
e.g. R = 12 ÷ 12.5
evaluation;
e.g. (R =) 0.96 (Ω)
Allow
ECF from (c)
0.92 to 1.0 (Ω) from rounded currents
Do not accept
96 (Ω) (resistances added as if in series)
(e)
other strips still have 12 V / 1.25 A so (still heat) normally / unaffected;
new total current = 9 × 1.25 = 11.25 (A);
substitution into P = I × V;
e.g. P = 11.25 × 12
evaluation;
e.g. (P =) 135 (W)
Allow
ECF from (b)
P per strip = 12 × 1.25 = 15 W, 9 × 15 = 135 (W) for last three marks
18IGCSE-PHY-2-025Effect of adding a cell and a resistor to a series circuitPapers 1 & 2Grade 6–71 mark
A circuit contains one cell, one resistor and an ammeter connected in series. The ammeter reads 0.30 A. A second cell, identical to the first, and a second resistor, identical to the first, are both added in series in the same circuit. The cells are connected the same way round. What is the new reading on the ammeter?
Show mark scheme
A (0.30 A);
B considers only the extra resistor (doubled resistance halves the current) and ignores the doubled voltage
C considers only the extra cell (doubled voltage doubles the current) and ignores the doubled resistance
D treats the extra resistor as if it increased the current (as an extra parallel path would), as well as the extra cell
19IGCSE-PHY-2-026Fuse in an electric lawnmower plugPapers 1 & 2Grade 6–77 marks
An electric lawnmower is connected to the 230 V mains supply by a long cable. The plug of the lawnmower contains a 5 A fuse. The fuse is connected in the live wire.
(a)
Explain why the fuse is connected in the live wire and not in the neutral wire.
(2)
(b)
While the grass is being cut, the blade cuts through the cable. The live wire and the neutral wire touch each other. Explain how the fuse prevents the cable from overheating.
(3)
(c)
After the fuse has melted, the owner replaces it with a piece of thick copper wire. Explain why this is dangerous.
(2)
Show mark scheme
(a)
(when fuse melts) the appliance is disconnected from the live / high voltage;
fuse in neutral would leave the appliance (still) live / at high voltage (so it could give a shock);
Allow
live wire is the dangerous / high-voltage wire
Do not accept
current flows only in the live wire
neutral wire has no current
(b)
(very) large current (in cable / fuse);
fuse (wire) heats up and melts;
circuit is broken / current stops (before cable overheats);
Allow
very low resistance / short circuit for MP1 if linked to large current
blows for melts
current larger than 5 A for MP1
Do not accept
fuse absorbs / stops the extra current without melting
fuse trips (circuit-breaker idea) without melting / breaking
(c)
copper wire will not melt (at a fault current / at 5 A);
(so) large current continues / cable overheats / fire / risk of shock;
Allow
copper wire has a much higher current rating than 5 A
circuit not broken when there is a fault
Do not accept
copper wire is a better conductor so the lawnmower gets more current in normal use (only)
20IGCSE-PHY-2-027Practical: current–voltage data for a length of thin metal wirePapers 1 & 2Grade 6–78 marks
A student uses the circuit shown to investigate how the current in a length of thin metal wire changes with the voltage across it. The table shows the student's results. Voltage / V: 0.50, 1.00, 1.50, 2.00, 2.50, 3.00 Current / A: 0.10, 0.20, 0.30, 0.46, 0.50, 0.60
(a)
State the purpose of the variable resistor in the circuit.
(1)
(b)
Identify the anomalous result in the table.
(1)
(c)
Describe the relationship between current and voltage for this wire. Use data from the table in your answer.
(2)
(d)
Determine the current in the wire when the voltage across it is 4.00 V.
(1)
(e)
The student leaves the current switched on for several minutes between readings. Explain how this could affect the results and what the student should do instead.
(3)
Show mark scheme
(a)
to change / control the current (in the wire) / the voltage across the wire;
21IGCSE-PHY-2-028Practical data: current and the number of identical resistors in seriesPapers 1 & 2Grade 6–79 marks
A student investigates how the current in a series circuit depends on the number of identical resistors in the circuit. She uses a 6.0 V battery and an ammeter. She adds the resistors one at a time. The table shows her results. Number of resistors / –: 1, 2, 3, 4, 5 Current / mA: 120, 60, 40, 30, 24
(a)
State the independent variable and one control variable in this investigation.
(2)
(b)
Describe the relationship between the number of resistors and the current. Use data from the table in your answer.
(2)
(c)
Calculate the resistance of one resistor.
(3)
(d)
Explain, in terms of resistance, why the current decreases as more resistors are added.
(2)
Show mark scheme
(a)
(independent variable) number of resistors;
(control) battery voltage / same type (or value) of resistor / same ammeter / temperature;
Allow
voltage of the supply for control
Do not accept
current as independent variable
number of resistors as control variable
(b)
current decreases as number of resistors increases;
inversely proportional / doubling the number halves the current, with data e.g. 2 → 4 resistors, 60 mA → 30 mA;
Allow
number × current is constant (120) for MP2
Do not accept
(directly) proportional
(c)
conversion of mA to A;
e.g. 120 mA = 0.120 A
substitution and rearrangement;
e.g. R = 6.0 ÷ 0.120
evaluation;
e.g. (R =) 50 (Ω)
Allow
any correct row, e.g. 6.0 ÷ 0.060 = 100 Ω then ÷ 2 = 50 Ω
POT error (e.g. 0.05 Ω or 50 000 Ω) scores 2
Do not accept
0.05 (Ω) (no conversion) as final answer for full marks
100 / 150 / 200 / 250 (Ω) (total resistance not divided by number)
(d)
total resistance (of circuit) increases;
voltage (of battery) is the same (so current is smaller);
22IGCSE-PHY-2-029Lamp and resistor in series in a model lighthousePapers 1 & 2Grade 6–78 marks
A student builds a model lighthouse. A lamp and a fixed resistor are connected in series with a 6.0 V battery, as shown. When the lamp is lit normally, its resistance is 12 Ω. The resistance of the fixed resistor is 18 Ω.
(a)
Determine the total resistance of the circuit.
(1)
(b)
Calculate the current in the circuit.
(2)
(c)
Calculate the voltage across the lamp.
(2)
(d)
Determine the voltage across the fixed resistor.
(1)
(e)
The student replaces the 6.0 V battery with a 9.0 V battery. He predicts that the current will now be 0.30 A. Explain why the actual current will be less than 0.30 A.
(2)
Show mark scheme
(a)
30 (Ω);
Do not accept
7.2 (Ω) (parallel formula)
216 (Ω) (multiplied)
(b)
substitution and rearrangement;
e.g. I = 6.0 ÷ 30
evaluation;
e.g. (I =) 0.20 (A)
Allow
ECF from (a)
0.2 (A)
Do not accept
0.50 (A) (6.0 ÷ 12)
0.33 (A) (6.0 ÷ 18)
(c)
substitution;
e.g. V = 0.20 × 12
evaluation;
e.g. (V =) 2.4 (V)
Allow
ECF from (b)
Do not accept
3.6 (V) (voltage across the resistor)
6.0 (V)
(d)
3.6 (V);
Allow
6.0 − their (c)
0.20 × 18
(e)
(larger current so) lamp / filament gets hotter;
resistance of lamp increases (so total resistance more than 30 Ω);
Allow
current is not proportional to voltage for a filament lamp, if linked to resistance increasing
23IGCSE-PHY-2-030Kitchen appliances on one extension leadPapers 1 & 2Grade 6–78 marks
An extension lead has four sockets connected in parallel. The plug of the extension lead contains a 13 A fuse and is connected to the 230 V mains. Three kitchen appliances are plugged into the extension lead. Appliance / –: kettle, toaster, coffee machine Power / W: 2600, 900, 1400
(a)
Calculate the current in the kettle.
(2)
(b)
Explain why the current in the cable of the extension lead is equal to the sum of the currents in the three appliances.
(2)
(c)
Evaluate whether all three appliances can be used at the same time without the fuse melting. Your answer should include calculations.
(4)
Show mark scheme
(a)
substitution and rearrangement;
e.g. I = 2600 ÷ 230
evaluation;
e.g. (I =) 11.3 (A)
Allow
11 (A)
11.30 (A)
Do not accept
0.088 (A) (230 ÷ 2600)
598 000 (A) (multiplied)
(b)
current is conserved at a junction / current in = current out at a junction;
charge is not used up / lost (so charge flowing in the cable = total charge flowing in the appliances);
Allow
the appliances are in parallel so the current splits between them, for MP1
Do not accept
current is shared equally between the appliances
current is the same in each appliance
(c)
toaster current = 3.9 (A) and coffee machine current = 6.1 (A);
total current = 21.3 (A);
compares their total with 13 (A);
conclusion: total is greater than 13 A so fuse melts / they cannot all be used together;
Allow
alternative route: total power = 2600 + 900 + 1400 = 4900 W (MP1); maximum power for a 13 A fuse = 13 × 230 = 2990 W (MP2); comparison of 4900 W with 2990 W (MP3); conclusion (MP4)
ECF from (a)
21 (A)
toaster 3.91 A, coffee machine 6.09 A
Do not accept
conclusion without any calculation
fuse does not melt because each appliance draws less than 13 A
24IGCSE-PHY-2-032Resistance and wiring of an electric showerPapers 1 & 2Grade 6–79 marks
An electric shower heats water as it flows over a heating element. The shower is connected to the 230 V mains and has a power of 9.2 kW. It is connected by its own separate cable, which is much thicker than the cable used for a lamp. A thicker cable has a lower resistance than a thinner cable of the same length.
(a)
Calculate the resistance of the heating element.
(3)
(b)
Explain why the shower needs a thick cable.
(3)
(c)
At times of high demand, the mains voltage can fall to 220 V. The resistance of the heating element does not change. Calculate the power of the shower when the mains voltage is 220 V.
(3)
Show mark scheme
(a)
conversion of 9.2 kW to 9200 W and I = 9200 ÷ 230 (= 40 A);
R = 230 ÷ 40;
evaluation;
e.g. (R =) 5.75 (Ω)
Allow
5.8 / 5.7 / 6 (Ω)
ECF from their current
POT error in kW (e.g. 5750 Ω) scores 2
R = V² ÷ P = 230² ÷ 9200 route for full marks
Do not accept
0.17 (Ω) (9200 ÷ 230², inverted)
0.025 (Ω) (230 ÷ 9200)
(b)
(very) large current (in the cable);
current in (the resistance of) the cable transfers energy (thermally) / heats the cable;
thick cable has lower resistance so less energy is transferred to it / it heats up less / does not overheat (or cause a fire);
Allow
40 A for MP1
thin cable would overheat / melt its insulation / cause a fire, for MP3 if linked to its higher resistance
Do not accept
thick cable stops electricity leaking out
thick cable carries more voltage
(c)
current = 220 ÷ 5.75 (= 38.3 A);
substitution into P = I × V, e.g. P = 38.3 × 220;
evaluation, e.g. (P =) 8400 (W) / 8.4 (kW);
Allow
ECF from (a)
8.42 / 8.41 (kW); 8420 (W)
using R = 5.8 Ω gives 37.9 A and 8.3 (kW): full marks
Do not accept
8.8 (kW) (9.2 × 220 ÷ 230, assumes current unchanged)
25IGCSE-PHY-2-013Evaluating a power bank claim using charge and energyPapers 1 & 2Grade 8–99 marks
A portable power bank is used to recharge mobile phones. The label on the power bank says: 'Capacity 10 400 mAh at 3.7 V — fully charges your phone three times!' A capacity of 1 mAh is the charge that flows when there is a current of 1 mA for 1 hour.
(a)
Show that the charge stored in the power bank is about 37 000 C.
(3)
(b)
Calculate the energy stored in the power bank.
(2)
(c)
To fully charge a phone, the power bank supplies a current of 2.0 A at 5.0 V for 1.5 hours. Evaluate the claim on the label. Your answer should include a calculation.
(4)
Show mark scheme
(a)
conversion of mA to A and hours to seconds;
e.g. 10 400 mA = 10.4 A and 1 h = 3600 s
substitution into Q = I × t;
e.g. Q = 10.4 × 3600
evaluation to at least 3 s.f.;
e.g. (Q =) 37 440 (C)
Allow
37 400 (C)
Q = 10 400 × 3.6 = 37 440 (C)
Do not accept
37 000 (C) as final answer without working (copies target)
(b)
substitution into E = Q × V;
e.g. E = 37 440 × 3.7
evaluation;
e.g. (E =) 139 000 (J)
Allow
138 528 (J)
1.39 × 10⁵ (J)
136 900 (J) using 37 000 C
ECF from (a)
Do not accept
10 100 (J) (37 440 ÷ 3.7)
(c)
substitution into E = I × V × t with time in seconds;
e.g. E = 2.0 × 5.0 × 5400
energy per full charge = 54 000 (J);
number of charges = 139 000 ÷ 54 000 = 2.6;
conclusion: claim not justified / fewer than three full charges (and fewer still because some energy is wasted / transferred to thermal store);
Allow
ECF from (b)
2.5 to 2.6 charges
comparison of 3 × 54 000 = 162 000 J with stored energy for MP3
answer that recognises energy (not charge) must be compared because the output voltage (5.0 V) is higher than the storage voltage (3.7 V) supports MP4
Do not accept
claim is true without calculation
15 (J) using 1.5 h not converted for MP2
comparing charge instead of energy: 37 440 ÷ (2.0 × 5400) = 3.5 charges so claim true (0 marks for MP2–MP4)
using 3.7 V for the phone charge: 2.0 × 3.7 × 5400 = 39 960 J giving 3.5 charges (MP1 not scored; ECF for MP3 only)
27IGCSE-PHY-2-033Energy in the pulses of an electric fence energiserPapers 1 & 2Grade 8–911 marks
A farmer uses an electric fence to keep animals in a field. An energiser powered by a 12 V battery sends a short pulse of charge along the fence wire every 1.5 s. Each pulse transfers a charge of 0.40 mC at a voltage of 7000 V. A small lamp on the energiser flashes each time a pulse is sent.
(a)
Suggest why the energiser has a lamp that flashes with each pulse.
(1)
(b)
Calculate the energy transferred by one pulse.
(3)
(c)
The energiser takes an average current of 0.20 A from the 12 V battery. Calculate the energy supplied by the battery in 12 hours.
(3)
(d)
Calculate the percentage of the energy supplied by the battery in 12 hours that is transferred by the pulses.
(3)
(e)
Compare the energy transferred by each coulomb of charge in a pulse with the energy transferred by each coulomb of charge that leaves the battery.
(1)
Show mark scheme
(a)
to show that (there is) a current / a pulse is being sent / the fence is working;
Allow
to indicate the energiser is switched on
Do not accept
to warn animals (as the only idea)
(b)
conversion of 0.40 mC to 4.0 × 10⁻⁴ C;
substitution;
e.g. E = 4.0 × 10⁻⁴ × 7000
evaluation;
e.g. (E =) 2.8 (J)
Allow
2800 (J) (no conversion) scores 2
Do not accept
17 500 (J) (7000 ÷ 0.40)
5.7 × 10⁻⁸ (J) (charge ÷ voltage)
(c)
conversion of 12 hours to 43 200 s;
substitution into E = I × V × t;
e.g. E = 0.20 × 12 × 43 200
evaluation;
e.g. (E =) 104 000 (J)
Allow
1.0 × 10⁵ / 1.04 × 10⁵ / 103 680 (J)
28.8 (J) (no conversion of hours) scores 2
Do not accept
2.4 (J) (no time)
(d)
number of pulses = 43 200 ÷ 1.5 = 28 800;
energy in pulses = 28 800 × 2.8 = 80 640 (J);
percentage = 78 (%);
Allow
ECF from (b) and (c)
power method: 2.8 ÷ 1.5 = 1.87 W and 0.20 × 12 = 2.4 W for MP1 and MP2
77.8 / 77.78 (%)
Do not accept
0.78 as a final answer without %
128 (%) (inverted ratio)
(e)
7000 J per coulomb in a pulse compared with 12 J per coulomb from the battery;
Allow
each coulomb in a pulse transfers about 580 times more energy
a pulse coulomb transfers more energy because voltage is energy per unit charge
Do not accept
the pulse has more charge / more current (as the comparison)
28IGCSE-PHY-2-034Extended: series and parallel settings of a two-element heaterPapers 1 & 2Grade 8–96 marks
A portable heater has two identical heating elements, X and Y. A switch on the heater connects the elements to the 230 V mains in one of three ways, shown in the diagram as settings 1, 2 and 3. The resistance of each element can be assumed to stay constant.
(a)
Explain which setting gives the highest heating power and which setting gives the lowest heating power. In your answer, refer to the voltage, resistance and current in each setting.
(6)
Show mark scheme
(a)
Indicative content:
MP1. current in an element transfers energy to its thermal store / heats it;
MP2. power = current × voltage and the mains voltage (230 V) is the same for every setting, so larger total current means larger power;
MP3. setting 2: elements in series so the total resistance is (doubled / larger than one element);
MP4. setting 2: the 230 V is shared / each element has (only) 115 V across it;
MP5. setting 2: (total) current is smaller (half of setting 1) so this is the lowest power;
MP6. setting 3: elements in parallel so each element has the full 230 V across it;
MP7. setting 3: current in each element is the same as in setting 1;
MP8. setting 3: total current from the supply is the sum of the two currents / doubled, so this is the highest power;
Allow
award 1 mark per MP to a maximum of 6, then use the levels to check the overall quality of the answer
total resistance of parallel setting is smaller (half of one element) for MP7
correct numerical illustration, e.g. if setting 1 is 1.0 kW then setting 2 is 0.50 kW and setting 3 is 2.0 kW, credited as MP5 and MP8 if supported by reasoning
setting 3 > setting 1 > setting 2 as an overall conclusion
Do not accept
series gives the highest power because the resistance is larger
current is shared equally so each element gets half the current in parallel
each element in series has 230 V across it
Level 0 (0)
No rewardable content.
Level 1 (1–2)
Identifies the correct highest or lowest setting, or makes isolated correct statements about series or parallel resistance, voltage or current, with little linking to power.
Level 2 (3–4)
Correctly explains one setting in terms of voltage, resistance and current leading to power, or identifies both settings correctly with partial reasoning for each.
Level 3 (5–6)
Logically linked explanation of both the series and the parallel settings, using the same mains voltage, the resistance and the current to conclude that setting 3 is highest and setting 2 lowest.
29IGCSE-PHY-2-015Forces between charged rodsPaper 2 onlyGrade 4–51 mark
Three charged rods, P, Q and R, are each hung from a thread so they can move freely. Rod R has a negative charge. When rod Q is brought near rod R, the rods repel each other. When rod P is brought near rod Q, the rods attract each other. Which row gives the charges on rods P and Q?
Show mark scheme
A (P positive, Q negative);
B uses the correct rule for Q and R but thinks attraction happens between like charges for P and Q
C reverses both rules (thinks like charges attract and unlike charges repel)
D uses the reversed rule for Q and R but the correct rule for P and Q
31IGCSE-PHY-2-036Charging on a plastic playground slidePaper 2 onlyGrade 4–55 marks
A child slides down a plastic slide in a playground. The plastic slide becomes negatively charged. At the bottom of the slide, the child's hair stands on end and the hairs spread out.
(a)
State the type of charge on the child's clothes at the bottom of the slide.
(1)
(b)
Explain, in terms of electrons, how the slide becomes negatively charged.
(2)
(c)
Explain why the child's hairs stand on end and spread out.
(2)
Show mark scheme
(a)
positive;
Allow
+
positive for the child
Do not accept
negative
neutral
(b)
friction / rubbing between the clothes and the slide transfers electrons;
electrons move from the child's clothes onto the slide (so the slide has more electrons than protons);
Allow
slide gains electrons for MP2
Do not accept
slide gains protons / positive charge
friction creates charge
(c)
(all) the hairs have the same (type of) charge;
like charges repel (so hairs move apart);
Allow
hairs are positively charged so repel each other
Do not accept
hairs are attracted to the slide (as explanation of spreading)
32IGCSE-PHY-2-016Practical: charging insulating rods by friction with different clothsPaper 2 onlyGrade 6–79 marks
A student investigates how well a polythene rod is charged when it is rubbed with different cloths. She rubs the rod with a cloth, then holds the rod 2 cm above a pile of small pieces of tissue paper and counts how many pieces are lifted by the rod. She repeats this for cotton, wool, silk and nylon cloths. After being rubbed with the wool cloth, the polythene rod has a negative charge.
(a)
Give the independent variable and the dependent variable in this investigation.
(2)
(b)
State two variables the student should control.
(2)
(c)
Explain how rubbing with the wool cloth gives the polythene rod a negative charge.
(2)
(d)
State the charge on the wool cloth after rubbing.
(1)
(e)
The student tries the experiment with a copper rod held in her hand. The copper rod does not lift any pieces of paper. Explain why.
(2)
Show mark scheme
(a)
independent: type / material of cloth;
dependent: number of pieces of paper lifted;
Do not accept
amount of charge as dependent variable without reference to how it is measured
polythene rod as independent variable
(b)
any two from:
number of rubs / time spent rubbing / force of rubbing;
distance of rod above paper;
size / mass of pieces of paper;
same (polythene) rod;
Allow
humidity of air
Do not accept
type of cloth
number of pieces lifted
(c)
(friction) transfers electrons from wool to polythene / polythene gains electrons;
(so polythene has) more electrons than protons / excess of electrons;
Allow
electrons rubbed off the wool onto the rod
Do not accept
positive charges / protons move to the wool
rubbing creates charge
(d)
positive;
Do not accept
negative
neutral
(e)
copper is a (electrical) conductor;
electrons / charge flow through rod and hand / body (to earth) so rod does not stay charged;
33IGCSE-PHY-2-017Electrostatic deflection of droplets in an inkjet printerPaper 2 onlyGrade 6–76 marks
The diagram shows part of an inkjet printer. Tiny ink droplets leave the nozzle and pass a charging electrode, which gives some droplets a negative charge. The droplets then pass between two metal plates, X and Y. Plate X is positively charged and plate Y is negatively charged.
(a)
State and explain the direction in which a negatively charged droplet moves as it passes between the plates.
(2)
(b)
Explain, in terms of electrons, how a droplet becomes negatively charged.
(1)
(c)
Explain how the printer can make different droplets land at different heights on the paper.
(2)
(d)
Some droplets are not given any charge. These droplets go into the gutter. Explain why.
(1)
Show mark scheme
(a)
towards plate X / upwards;
unlike / opposite charges attract (and like charges repel from plate Y);
Allow
attracted to the positive plate
Do not accept
towards plate Y
positive charges move onto the droplet
(b)
(droplet) gains electrons;
Allow
electrons transferred to the droplet
Do not accept
loses protons
gains negative ions with no mention of electrons
(c)
(give droplets) different amounts of charge / change the charge on the plates;
larger charge gives larger force / greater deflection;
Allow
vary voltage of plates / charging electrode
more electrons gives a bigger deflection
Do not accept
make the droplets travel faster without reference to charge
(d)
no (electrostatic) force on them (so they are not deflected / travel straight);
34IGCSE-PHY-2-019Spark from a door handle: charge, current and voltagePaper 2 onlyGrade 6–710 marks
A person walks across a nylon carpet wearing shoes with rubber soles. The person becomes negatively charged. When the person touches a metal door handle, a small spark jumps between their finger and the handle. During the spark, a charge of 0.90 µC flows in a time of 0.30 µs.
(a)
Explain how the person becomes negatively charged.
(2)
(b)
Explain why the spark occurs when the person touches the metal door handle but not when they touch a wooden door.
(2)
(c)
Calculate the average current during the spark.
(3)
(d)
The energy transferred in the spark is 8.1 mJ. Calculate the average voltage between the person and the door handle while the charge flows.
(3)
Show mark scheme
(a)
electrons transferred from carpet to person (by friction);
rubber (soles) is an insulator so electrons / charge cannot flow away (to earth);
Allow
person gains electrons
Do not accept
person gains negative protons
person loses protons / positive charge
(b)
metal is a conductor so electrons can flow (rapidly) from the person (through the handle to earth);
wood is an insulator so electrons / charge cannot flow (so no spark);
Allow
metal handle is earthed
Do not accept
wood absorbs the charge
positive charge flows from the handle to the person
(c)
rearrangement of Q = I × t;
e.g. I = Q ÷ t
substitution with correct powers of ten;
e.g. I = 0.90 × 10⁻⁶ ÷ 0.30 × 10⁻⁶
evaluation;
e.g. (I =) 3.0 (A)
Allow
3 (A)
substitution 0.90 ÷ 0.30 is acceptable as powers of ten cancel
Do not accept
2.7 × 10⁻¹³ (A) (Q × t)
0.33 (A) (t ÷ Q)
(d)
substitution into E = Q × V;
e.g. 8.1 × 10⁻³ = 0.90 × 10⁻⁶ × V
rearrangement;
e.g. V = 8.1 × 10⁻³ ÷ 0.90 × 10⁻⁶
evaluation;
e.g. (V =) 9000 (V)
Allow
9.0 × 10³ (V)
9 kV
POT error giving 9 (V) or 9 × 10⁶ (V) scores 2 marks
35IGCSE-PHY-2-037Practical: number of rubs and repulsion of a charged polythene stripPaper 2 onlyGrade 6–78 marks
A student rubs a polythene strip with a woollen cloth and hangs it from a thread. She rubs a second polythene strip with a woollen cloth and holds it 3 cm from the hanging strip. The hanging strip moves away. She measures the angle through which the thread turns away from the vertical. She repeats this for different numbers of rubs. Number of rubs / –: 5, 10, 15, 20, 25 Angle / °: 8, 15, 19, 21, 22
(a)
State the dependent variable in this investigation.
(1)
(b)
Explain why the hanging strip moves away from the rubbed strip.
(2)
(c)
Describe the pattern shown by the results.
(2)
(d)
Before each new reading, the student removes the charge from the rubbed strip. Explain why she does this.
(2)
(e)
Suggest how the student could measure the angle more accurately.
(1)
Show mark scheme
(a)
angle (of thread / strip) (from the vertical);
Allow
deflection
Do not accept
number of rubs
charge
(b)
both strips have the same charge (negative);
like charges repel;
Allow
same material rubbed with same cloth so same charge for MP1
Do not accept
opposite charges repel
the strips are uncharged
(c)
angle increases as number of rubs increases;
increase gets smaller / angle levels off (at higher numbers of rubs), with data e.g. 7° increase from 5 to 10 rubs but 1° from 20 to 25;
Allow
not proportional for MP2 if supported by data
Do not accept
angle is directly proportional to number of rubs
(d)
(otherwise) charge from the previous reading remains / adds to new charge;
so the charge depends only on the number of rubs for that reading / fair test / results valid;
Allow
each reading starts with no charge
Do not accept
to make the strip safe (only)
(e)
protractor behind the thread read at eye level / photograph against a scale / use of a larger protractor;
Allow
avoid parallax
Do not accept
repeat readings (improves reliability, not accuracy of one measurement)
36IGCSE-PHY-2-038Electrostatic paint spraying of a car panelPaper 2 onlyGrade 6–78 marks
The diagram shows how a car body panel is painted using an electrostatic spray gun. The nozzle of the spray gun is connected to the negative terminal of a high-voltage supply. The car body panel is connected to the positive terminal.
(a)
Explain, in terms of electrons, how the car body panel becomes positively charged.
(1)
(b)
Explain why the paint droplets spread out to form a fine, even spray.
(2)
(c)
Explain why this method wastes less paint than an uncharged spray, and coats parts of the panel that are not facing the spray gun.
(3)
(d)
Suggest why the panel must stay connected to the positive terminal throughout the spraying.
(2)
Show mark scheme
(a)
electrons move from the panel to the (positive terminal of the) supply / panel loses electrons;
Allow
panel has fewer electrons than protons
Do not accept
panel gains protons / positive charge
panel gains electrons
(b)
droplets have the same / like charge;
(like charges) repel (each other);
Do not accept
droplets attract each other
(c)
panel and droplets have opposite / unlike charges;
unlike charges attract / droplets are attracted to the panel;
droplets are pulled onto the panel (even round the back / edges) so fewer miss it / less paint wasted;
Allow
droplets curve / follow the attraction around the panel
Do not accept
droplets are attracted because the panel is metal (only)
opposite charges repel
(d)
(negative) droplets / electrons arriving would cancel / neutralise the panel's positive charge / make it negative;
(then) panel would not attract / would repel the droplets;
Allow
supply keeps removing electrons from the panel for MP1
37IGCSE-PHY-2-018Extended: electrostatic dangers when a fuel tanker unloadsPaper 2 onlyGrade 8–96 marks
A road tanker delivers petrol to a filling station. The petrol flows at high speed through a long rubber hose from the tanker into an underground tank. Before unloading, the driver connects a metal cable from the tanker to a metal post that is connected to earth. The hose has a thin metal wire running along its length.
(a)
Explain why unloading petrol can be dangerous and how the metal cable and the wire in the hose reduce this danger.
(6)
Show mark scheme
(a)
Indicative content:
MP1. petrol rubs against / flows over the inside of the hose (friction);
MP2. electrons are transferred between petrol and hose (so they gain opposite charges);
MP3. rubber is an insulator so charge cannot flow away / charge builds up (on hose / tanker);
MP4. a large voltage / potential difference builds up (between tanker and earth);
MP5. a spark may jump (to earth / to nearby metal);
MP6. spark can ignite petrol vapour causing fire / explosion;
MP7. metal cable / wire is a (good) conductor;
MP8. electrons flow through cable / wire to (or from) earth as soon as charge is produced;
MP9. so charge does not build up / no large voltage / no spark;
Allow
award 1 mark per MP to a maximum of 6, then use the levels to check the overall quality of the answer
charge is discharged / neutralised through the earth connection for MP8
fumes for vapour
Do not accept
friction creates charge
protons / positive charges flow to earth (for MP8)
earth wire absorbs electricity
Level 0 (0)
No rewardable content.
Level 1 (1–2)
Isolated points, e.g. friction causes charge or sparks cause explosions, with little or no linking between how charge builds up and why the cable helps.
Level 2 (3–4)
Explains either the build-up of charge leading to a spark and ignition, or how the conductors remove charge, with some correct use of electron transfer.
Level 3 (5–6)
A logically linked explanation of both the hazard (electron transfer by friction, build-up on an insulator, large voltage, spark, ignition) and the precaution (conductors allow electrons to flow to/from earth so charge cannot build up), using correct terminology throughout.
38IGCSE-PHY-2-020Explaining how a photocopier uses electrostatic chargePaper 2 onlyGrade 8–98 marks
A photocopier contains a rotating drum. Before each copy, the whole surface of the drum is given a positive charge. An image of the page being copied is projected onto the drum. The drum's coating conducts electricity when light falls on it. White parts of the page reflect light onto the drum. Black letters on the page do not reflect light. Negatively charged black toner powder is then brushed over the drum. A sheet of paper is given a positive charge and rolled over the drum to pick up the toner.
(a)
Explain why the toner sticks only to the parts of the drum that match the black letters on the page.
(3)
(b)
Explain why the paper must be given a larger positive charge than the drum.
(2)
(c)
After each copy, bright light is shone over the whole drum. Explain why this is necessary before the next copy is made.
(2)
(d)
Explain why the toner spreads out evenly instead of forming clumps on the drum.
(1)
Show mark scheme
(a)
light (from white areas) makes coating conduct so charge flows away / areas lose their charge;
areas matching (black) letters receive no light so remain positively charged;
negative toner attracted to positive areas (only) / opposite charges attract;
Allow
electrons flow onto the lit parts and neutralise them for MP1
Do not accept
positive charges / protons flow away
toner attracted to the uncharged parts
(b)
(stronger positive charge means) larger (attractive) force on (negative) toner (than from drum);
so toner is pulled off drum onto paper;
Allow
paper attracts toner more strongly than drum does
Do not accept
paper repels the toner
(c)
light makes whole coating conduct so (any remaining) charge flows away / drum discharged;
otherwise old image areas would still attract toner (onto next copy);
Allow
to remove the previous image
Do not accept
to dry the toner
to heat the drum
(d)
(all toner particles have the same / negative charge so) they repel each other;
39IGCSE-PHY-2-039Earthed wrist strap for electronics workersPaper 2 onlyGrade 8–98 marks
Workers who assemble computer circuit boards can become charged by friction as they move. A spark from a charged worker can damage the microchips. Each worker wears a conducting wrist strap. The strap is connected to earth through a wire that contains a 1.0 MΩ resistor.
(a)
A worker becomes positively charged. Explain, in terms of electrons, how the wrist strap removes the charge from the worker.
(2)
(b)
A worker who is not wearing the strap holds a charge of 0.45 µC at a voltage of 3000 V. The worker then puts on the strap. The strap manufacturer claims that a worker is discharged in less than 1 ms. Assume that the current decreases steadily from its initial value to zero, so that the average current is half of the initial current. Evaluate the manufacturer's claim. Your answer should include calculations.
(4)
(c)
Suggest why the strap is connected to earth through a 1.0 MΩ resistor instead of a plain metal wire.
(2)
Show mark scheme
(a)
electrons flow from earth (through strap) to the worker;
until the worker is neutral / uncharged / positive charge is cancelled;
Allow
electrons move onto the worker for MP1
Do not accept
positive charge / protons flow from the worker to earth
40IGCSE-PHY-2-040How an electrostatic precipitator cleans chimney gasesPaper 2 onlyGrade 8–910 marks
A factory chimney contains an electrostatic precipitator to remove smoke particles from waste gases. The smoky gas rises between two vertical metal plates, which are kept positively charged. A wire grid between the plates is kept at a high negative voltage. As the smoke particles pass the wire grid, they become negatively charged. The plates are shaken at regular intervals so that the collected smoke falls into a container. An engineer measures the percentage of the smoke removed for different voltages on the wire grid. Voltage on grid / kV: 10, 20, 30, 40, 50 Smoke removed / %: 60, 80, 90, 95, 97
(a)
Explain why the negatively charged smoke particles move away from the wire grid and towards the metal plates.
(2)
(b)
Explain why the smoke particles fall off the plates when the plates are shaken, instead of staying attracted to them.
(2)
(c)
When the voltage on the grid is increased, each smoke particle gains a larger charge. Explain why a greater percentage of the smoke is removed.
(3)
(d)
The engineer claims that doubling the voltage on the grid from 20 kV to 40 kV halves the amount of smoke that escapes from the chimney. Evaluate the engineer's claim. Use data from the table in your answer.
(3)
Show mark scheme
(a)
particles and grid both negative / have like charges so (particles are) repelled from the grid;
particles and plates have opposite / unlike charges so (particles are) attracted to the plates;
Do not accept
like charges attract
opposite charges repel
particles attracted because the plates are metal (only)
(b)
(extra) electrons flow from the particles to the plate (on contact);
particles become uncharged / neutral so no (electrostatic) attraction;
Allow
particles lose their charge to the plate for MP1
Do not accept
the plate loses its charge (only)
protons flow to the particles
(c)
(larger charge so) larger / stronger force of attraction (to plates);
particles move towards / reach the plates faster / particles accelerate more towards plates;
so more particles reach the plates before leaving the chimney / fewer escape with the gas;
Allow
particles are deflected more for MP2
Do not accept
larger voltage creates more particles
particles are repelled more strongly by the plates
(d)
smoke escaping at 20 kV = 20 % and at 40 kV = 5 %;
5 % is one quarter of 20 % / escaping smoke reduced by a factor of 4;
conclusion: claim is incorrect (or an underestimate) – the escaping smoke is more than halved;
Allow
escaping smoke falls by 15 percentage points / 75 % for MP2
conclusion consistent with their calculated values for MP3
Do not accept
comparison of percentages removed only (80 % → 95 %) without considering smoke escaping, for MP1
claim correct because 40 is double 20 (no use of smoke data)
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