Revision Notes
OxfordAQA International GCSE Physics: Charge, current, potential difference and circuit symbols (9203) – Revision Notes
Revision notes on charge, static electricity, current, potential difference, circuit symbols, thermistors and LDRs for OxfordAQA IGCSE Physics 9203.
- Subject
- Physics
- Level
- IGCSE
- Topic
- Charge, current, potential difference and circuit symbols
- Author
- Marlbridge Academic Team
- Updated
Aligned to OxfordAQA IGCSE Physics (9203), For exams May/June 2018 onwards. Official specification .
Syllabus page (what it covers and how it is assessed): OxfordAQA IGCSE Physics.
Syllabus points this page covers
9203
- 3.5.1 Electrical circuits
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These notes condense section 3.5.1 Electrical circuits, statements a to g, of the OxfordAQA International GCSE Physics (9203) specification, for exams from May/June 2018 onwards. For full explanations, read the study guide. Then try the practice questions. See also the OxfordAQA International GCSE Physics hub, the printable checklist and the free 10-minute diagnostics.
Key definitions
| Term | Meaning (3.5.1) |
|---|---|
| Free charges | In a metal, many electrons are free to move through the lattice (a) |
| Static electricity | Unbalanced charge that has nowhere to flow because no conducting route exists (b) |
| Discharge | Charge flowing when a conducting route is provided (b) |
| Electric current | The rate of flow of electric charge, in amperes (c) |
| Voltage of a source | The energy the source gives to charges as it drives them round the whole circuit, in volts (d) |
| Potential difference | The energy transferred by charges in a component, in volts (e) |
Questions use the term potential difference. The specification says correct use of either potential difference or voltage gains credit.
Equations
| Equation | Quantities and units | Rearrangements |
|---|---|---|
| I = Q / t | current I in A, charge Q in C, time t in s | Q = I t, t = Q / I |
| V = E / Q | V potential difference (V), E energy (J), Q charge (C) | E = Q V, Q = E / V |
One ampere is one coulomb per second. One volt is one joule per coulomb.
Two-step link: charge from the current first, then energy from the charge.
E = Q V = (I t) V
Unit prefixes to recall: 1 mA = 0.001 A, 1 ms = 0.001 s, 1 kJ = 1000 J, 1 minute = 60 s and 1 hour = 3600 s.
Method in steps
- Write the quantity you want and list the data.
- Convert time to seconds and milliamperes to amperes.
- Rearrange the equation, then substitute.
- Finish with the unit, and round sensibly.
- If there is a second part, check whether you can use your first answer.
Free charges and static electricity
- Metals conduct because they contain many free electrons. Insulators have almost none.
- A current needs a source and a complete conducting loop.
- Conventional current leaves the positive terminal. The electrons in the metal move the opposite way.
- Rubbing two insulators transfers electrons from one to the other. The surface that gains electrons is negative and the surface that loses them is positive.
- The charge stays put because it has no conducting route. A conductor held in the hand loses its charge because charge flows away through you.
Lightning in four steps
- Collisions inside a storm cloud separate charge.
- The ground becomes charged with the opposite sign to the base of the cloud.
- The potential difference between cloud and ground becomes very large.
- The air stops insulating, a conducting path forms and charge flows in a flash. This is the discharge.
Worked reminders
Charge from current. A current of 1.8 A flows for 2.5 minutes.
t = 2.5 × 60 = 150 s
Q = I t = 1.8 × 150 = 270 C
Current from charge. 45 C passes in 0.30 ms during a spark.
I = Q / t = 45 / 0.00030 = 150 000 A = 1.5 × 10⁵ A
Energy from charge. 22 C is driven round a circuit by a 9.0 V battery.
E = Q V = 22 × 9.0 = 198 J
Potential difference from energy. A motor transfers 330 J from 55 C.
V = E / Q = 330 / 55 = 6.0 V
Energy shared round a circuit. A 6.0 V battery drives 40 C round a circuit with a lamp and a buzzer. The lamp has 2.5 V across it.
Energy supplied = 40 × 6.0 = 240 J
Lamp transfers = 40 × 2.5 = 100 J
Buzzer transfers = 240 - 100 = 140 J, so V = 140 / 40 = 3.5 V
Reading thermistor data. A thermistor on a fixed supply carries 8.0 mA at 25 °C and 26 mA at 55 °C. The current rises, so the resistance falls. In 5.0 minutes at 55 °C, Q = 0.026 × 300 = 7.8 C.
Circuit symbols
| Component | Symbol in words |
|---|---|
| Cell | The longer, thinner line is positive and the shorter, thicker line is negative |
| Battery | Two or more cells in a row |
| Switch, open or closed | Gap with a lever away from the contact, or lever resting on it |
| Lamp | Circle with a cross |
| Fuse | Rectangle with a line through its length |
| Resistor | Plain rectangle |
| Variable resistor | Rectangle with a sloping arrow through it |
| Diode | Triangle pointing at a bar, inside a circle |
| LED | Diode symbol with arrows leaving |
| Thermistor | Rectangle with a sloping line ending in a bar |
| LDR | Rectangle in a circle, arrows arriving |
| Ammeter | A in a circle, in the loop being measured |
| Voltmeter | V in a circle, across the component being measured |
Thermistors and LDRs
| Device | Rule | Example use |
|---|---|---|
| Thermistor | Resistance decreases as temperature increases | Thermostat |
| LDR | Resistance decreases as light intensity increases | Switching a light on when it gets dark |
- With a fixed potential difference, lower resistance means a larger current. So a thermistor circuit carries more current when hot, and an LDR circuit carries more current when bright.
- Thermostat: the sensing circuit responds as the thermistor warms or cools, and the heater is switched off or on.
- Dark switch: as the light falls, the LDR’s resistance rises and the current in the sensing circuit changes. That change operates the switch for the lamp.
Model sentences for explanation questions
- Current: “Current is the rate of flow of charge, so more charge passes each second when the current is larger.”
- Potential difference: “The potential difference across the lamp is 3.0 V, so each coulomb transfers 3.0 J in the lamp.”
- Static electricity: “Electrons were transferred by rubbing, so the object has an excess of charge. It cannot flow away because the object is an insulator.”
- Discharge: “When a conducting route was provided, the charge flowed along it.”
- LDR: “In the dark the resistance of the LDR is high, so the current is small.”
- Thermistor: “As the temperature rises the resistance of the thermistor falls, so the current increases.”
Must-know distinctions
| Pair | Difference |
|---|---|
| Voltage of a source and potential difference | Energy supplied per coulomb, and energy transferred per coulomb |
| Charge and current | Coulombs, and coulombs per second |
| Static and current electricity | Charge held in place, and charge flowing |
| Conducting and insulating | Many free charges, and almost none |
| Thermistor and LDR | Responds to temperature, and responds to light |
| Ammeter and voltmeter | Measures the current through a component, and the potential difference across it |
Quick self-test
- A current of 2.5 A flows for 4.0 minutes. Calculate the charge.
- 72 C flows in 90 s. Calculate the current.
- A current of 40 mA delivers 120 C. How many minutes does it take?
- A 5.0 V supply drives 18 C round a circuit. Calculate the energy supplied.
- A lamp transfers 120 J from 15 C of charge. Calculate the potential difference.
- A current of 0.20 A flows for 5.0 minutes through a component with 4.5 V across it. Calculate the energy transferred.
- Why do metals conduct but plastics do not?
- Describe what is meant by static electricity.
- Which has the lower resistance, a thermistor at 80 °C or at 20 °C?
- Describe the symbol for an LDR.
- In the symbol for a cell, which line is the positive terminal?
Answers
- t = 240 s, so Q = 2.5 × 240 = 600 C.
- I = 72 / 90 = 0.80 A.
- 40 mA = 0.040 A, so t = 120 / 0.040 = 3000 s = 50 minutes.
- E = 18 × 5.0 = 90 J.
- V = 120 / 15 = 8.0 V.
- t = 300 s, Q = 0.20 × 300 = 60 C, E = 60 × 4.5 = 270 J.
- Metals contain many electrons that are free to move. Plastics have almost no free charges.
- Unbalanced charge on an object or in a region, where the charge has no conducting path to move along.
- The thermistor at 80 °C, because its resistance decreases as the temperature increases.
- A rectangle inside a circle, with arrows pointing towards it.
- The long thin line.
Where marks are usually lost
- Substituting a time in minutes or hours into Q = It.
- Forgetting to convert milliamperes to amperes.
- Writing “current is used up” instead of saying that energy is transferred.
- Writing “voltage flows” or “current across a component”. Current flows through, and potential difference is across.
- Describing lightning without saying that charge builds up and the air then conducts.
- Answering a discharge question without mentioning a conducting route.
- Stating the thermistor and LDR rules the wrong way round.
- Drawing the voltmeter in the loop or the ammeter across a component.
- Missing the unit, or writing C for current and A for charge.
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.
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