OxfordAQA International GCSE Physics: Charge, current, potential difference and circuit symbols (9203)

This guide follows the OxfordAQA International GCSE Physics (9203) specification, for exams from May/June 2018 onwards. It covers section 3.5.1 Electrical circuits, statements a to g. The qualification is untiered.

Related pages: the revision notes, the practice questions, the OxfordAQA International GCSE Physics hub, the printable checklist, the Energy study guide and the free 10-minute diagnostics.

What this unit covers

Statement	What you must be able to do
3.5.1 a	Explain that charges move easily through some substances, for example the free electrons in a metal
3.5.1 b	Describe static electricity as an imbalance of charge, and explain discharge, for example lightning
3.5.1 c	Use current as how fast charge flows, with Q = It
3.5.1 d	Define the voltage of a source as the energy it gives to charges as it pushes them all the way round a circuit
3.5.1 e	Define the potential difference across a component through the energy that charges transfer in it, in volts
3.5.1 f	Use V = E/Q
3.5.1 g	Interpret and draw circuit diagrams with the standard symbols, including thermistors and LDRs in circuits

Questions use the term potential difference. The specification says correct use of either term potential difference or voltage gains credit.

1. Free charges in metals (3.5.1 a)

Electrical charge moves easily through some substances and hardly at all through others. A metal is a lattice of fixed positive ions with a large number of electrons that are free to move through it. These are the charges that carry current in a wire.

An insulator such as plastic, rubber or dry air has almost no free charges, so nothing flows through it.

A steady current needs a source that pushes the electrons along and a complete conducting loop. An open switch puts a gap in the loop and the flow stops.

Arrows on circuit diagrams show conventional current, which is drawn leaving the positive terminal of the source. The electrons in the metal move the other way, towards the positive terminal.

2. Static electricity and discharge (3.5.1 b)

An object has an imbalance of charge when it holds more of one kind of charge than the other. This is static electricity. The word static means the charge stays where it is, because it has no conducting route along which to travel.

A common way to create an imbalance is to rub two insulators together. Electrons transfer from one surface to the other. In one standard demonstration, a polythene rod rubbed with a dry cloth gains electrons and becomes negative. The cloth has lost electrons, so it is left positive. Only electrons move. The positive ions stay fixed in the material.

If a conducting route appears, the charge flows along it. This is a discharge. Examples are a small spark when you touch a metal door handle, or a crackle as you pull off a jumper.

Lightning

Lightning is a large discharge. You can explain it in four steps:

1. Inside a storm cloud, collisions between particles of ice and water separate charge. One region of the cloud becomes negatively charged and another positively charged.
2. The ground below becomes charged with the opposite sign to the base of the cloud.
3. As the charge builds, the potential difference between the cloud and the ground becomes very large.
4. The air, normally an insulator, can no longer hold the charges apart. A conducting path forms and charge flows along it in a flash, which is the discharge.

Worked example 1: explaining a spark

A student walks across a nylon carpet in rubber-soled shoes, then touches a metal tap and feels a small shock. Rubbing transfers electrons, so she gains an excess of one kind of charge. The rubber insulates her, so the charge stays on her. When she touches the earthed tap there is a conducting route, and the excess charge flows through her hand as a discharge.

3. Current and charge: Q = It (3.5.1 c)

Electric current is how quickly electric charge flows past a point. Charge Q is measured in coulombs (C), current I in amperes (A) and time t in seconds (s).

I = Q / t        Q = I × t        t = Q / I

One ampere is one coulomb passing a point every second. Convert the time to seconds and milliamperes to amperes (divide by 1000) before you substitute.

Worked example 2: charge from current

A current of 0.45 A flows in a lamp for 3.5 minutes. Calculate the charge that flows.

t = 3.5 × 60 = 210 s
Q = I t = 0.45 × 210 = 94.5 C

The charge is 94.5 C.

Worked example 3: a very large current

A lightning flash transfers 18 C in 0.50 ms. Calculate the current.

t = 0.50 ms = 0.50 × 10⁻³ s = 0.00050 s
I = Q / t = 18 / 0.00050 = 36 000 A

The current is 3.6 × 10⁴ A.

Worked example 4: finding time

A charging unit supplies a steady 0.80 A. How long does it take to deliver 9000 C? Give the answer in hours.

t = Q / I = 9000 / 0.80 = 11 250 s
t = 11 250 / 3600 = 3.125 h

The time is 3.1 hours to 2 significant figures.

4. Voltage of a source and potential difference (3.5.1 d and e)

These two statements use similar words, so keep them apart.

• The voltage of a source means the energy the source gives to charges as it drives them right round the circuit. A 12 V battery gives 12 J to each coulomb.
• The potential difference across a component measures how much energy the charges transfer while passing through it. If the potential difference across a lamp is 4 V, each coulomb that passes through it transfers 4 joules.

Both are measured in volts (V). The energy a source supplies equals the energy the components transfer, which is conservation of energy.

5. The relationship V = E/Q (3.5.1 f)

Potential difference V (volts), energy transferred E (joules) and charge Q (coulombs) are linked by:

V = E / Q        E = Q V        Q = E / V

One volt is one joule per coulomb. Rearrange the equation first, then substitute.

Worked example 5: energy supplied

A 12 V battery drives 35 C of charge round a circuit. How much energy does it supply?

E = Q V = 35 × 12 = 420 J

The battery supplies 420 J.

Worked example 6: charge from energy

A lamp transfers 56 J of energy while 4.0 V is across it. How much charge passes through it?

Q = E / V = 56 / 4.0 = 14 C

The charge is 14 C.

Worked example 7: combining both equations

A current of 0.30 A flows for 2.0 minutes through a heater with a potential difference of 6.0 V across it. Calculate the energy transferred.

t = 2.0 × 60 = 120 s
Q = I t = 0.30 × 120 = 36 C
E = Q V = 36 × 6.0 = 216 J

The energy transferred is 220 J to 2 significant figures (216 J before rounding).

Worked example 8: energy shared in a circuit

A 12 V battery drives 20 C round a circuit containing a lamp and a motor. The potential difference across the lamp is 7.5 V. Find the energy transferred by the lamp, and the potential difference across the motor.

Energy supplied by battery = Q V = 20 × 12 = 240 J
Energy transferred by lamp = 20 × 7.5 = 150 J
Energy left for motor      = 240 - 150 = 90 J
Potential difference across motor = E / Q = 90 / 20 = 4.5 V

The motor has a potential difference of 4.5 V.

6. Circuit diagrams and symbols (3.5.1 g)

You must be able to interpret and draw circuit diagrams that use the standard symbols. Draw with a ruler and keep corners square.

Component	What the symbol looks like
Cell	Two parallel lines of different lengths. The longer, thinner line marks the positive terminal.
Battery	Two or more cell symbols in a row
Switch (open)	A gap in the line with a lever lifted away from the second contact
Switch (closed)	The lever rests on the second contact, so the line is complete
Lamp	A circle with a cross inside
Fuse	A rectangle with a line running through its length
Resistor	A plain rectangle
Variable resistor	A rectangle with a sloping arrow through it
Diode	A triangle pointing at a bar, drawn inside a circle, with a wire at each end
LED	The diode symbol with two arrows pointing away from it
Thermistor	A rectangle with a sloping line through it, ending in a bar
LDR	A rectangle inside a circle, with arrows pointing towards it
Ammeter	The letter A inside a circle
Voltmeter	The letter V inside a circle

Reading a circuit

An ammeter measures current, so it goes in the loop where the current is to be measured. A voltmeter measures the potential difference across a component, so its terminals connect to either side of that component. Trace the loop from the positive terminal of the cell and check that no open switch breaks it.

7. Thermistors and LDRs in circuits (3.5.1 g)

The resistance of a thermistor decreases as its temperature increases. The resistance of an LDR (light-dependent resistor) decreases as light intensity increases. Because of this, the current in a circuit containing one of them depends on the temperature or the light level. This makes them useful as sensors.

Worked example 9: thermistor readings

A thermistor is connected in series with a 3.0 V cell and an ammeter. The readings are:

Temperature (°C)	Current (mA)
20	5.0
40	12.0
60	31.0

(a) Describe the trend. The current increases as the temperature increases.

(b) Explain it. The potential difference stays at 3.0 V, so a larger current means a smaller resistance. This matches a thermistor, whose resistance decreases as the temperature rises.

(c) Find the charge that passes in 10 minutes at 60 °C.

I = 31.0 mA = 0.0310 A
t = 10 × 60 = 600 s
Q = I t = 0.0310 × 600 = 18.6 C

A thermostat

A thermostat uses a thermistor to sense temperature. As the room warms, the thermistor’s resistance falls and the current in the sensing circuit rises. When it reaches a set value, the circuit switches the heater off. When the room cools, the resistance rises, the current falls and the heater switches back on.

Switching a light on in the dark

An LDR can switch a lamp on when it gets dark. In bright light the LDR has a low resistance. As the light falls its resistance increases, so the current in the sensing part of the circuit changes. That change is used to operate a switch in the lamp circuit.

Worked example 10: LDR readings

An LDR in series with a cell and an ammeter gives 9.0 mA in daylight, 3.5 mA at dusk and 0.80 mA in the dark. Describe what happens to its resistance as it gets darker, and find the charge in 10 minutes in the dark.

The current falls as the light falls with a constant potential difference, so the resistance increases.

I = 0.80 mA = 0.00080 A
Q = I t = 0.00080 × 600 = 0.48 C

Common errors

• Writing that current is “used up” by a lamp. The charge is not used up. The energy it carries is transferred.
• Leaving the time in minutes in Q = It. The time must be in seconds.
• Mixing up the voltage of a source (energy supplied per coulomb) with the potential difference across a component (energy transferred per coulomb).
• Saying charge is created in a cloud. Charge is separated, not created.
• Drawing a cell the wrong way round. The long line is positive.
• Saying an LDR’s resistance goes up in bright light.

Official syllabus

OxfordAQA International GCSE Physics (9203), specification version 5.2, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.5.1 Electrical circuits, statements a to g.