What are the basic electrical quantities studied in the Cambridge IGCSE Physics curriculum?

In the Cambridge IGCSE Physics curriculum, the basic electrical quantities include current, voltage (potential difference), resistance, and charge. These quantities are fundamental to understanding how electric circuits function and are crucial for solving problems related to electricity and magnetism.

How is electric current defined and measured in IGCSE Physics?

Electric current is defined as the rate of flow of electric charge through a conductor. It is measured in amperes (A) using an ammeter. In the IGCSE Physics syllabus, students learn that current is calculated using the formula I = Q/t, where I is the current, Q is the charge in coulombs, and t is the time in seconds.

What is the relationship between voltage, current, and resistance in an electric circuit?

The relationship between voltage, current, and resistance in an electric circuit is described by Ohm's Law, which states that V = IR. Here, V represents voltage in volts, I is the current in amperes, and R is the resistance in ohms. This fundamental principle is essential for analyzing and solving circuit problems in the Cambridge IGCSE Physics course.

How does resistance affect the flow of current in a circuit?

Resistance is a measure of how much a material opposes the flow of electric current. In a circuit, higher resistance results in a lower current flow, assuming the voltage remains constant. This concept is crucial in the IGCSE Physics curriculum, where students learn to calculate resistance using the formula R = V/I and understand its impact on circuit behavior.

What is the significance of electrical charge in the study of electricity and magnetism?

Electrical charge is a fundamental property of matter that causes it to experience a force when placed in an electric or magnetic field. In the context of IGCSE Physics, understanding charge is essential for explaining how electric fields are created and how they interact with charged particles. Charge is measured in coulombs (C), and its study is integral to grasping the principles of electricity and magnetism.

Why is "Confusing electron flow direction with conventional current" a common mistake in Electrical Quantities?

Why it happens: Two conventions used in different sources. How to avoid it: Conventional current: + → −. Electron flow: − → +. Use whichever the question asks; default in 0625 is conventional. Source: 0625/42 — recurring

Why is "Quoting voltage in amps (or current in volts)" a common mistake in Electrical Quantities?

Why it happens: Slip. How to avoid it: V (volts) for pd; A (amps) for current; Ω (ohms) for resistance.

Why is "Using minutes/hours directly in $Q = It$ or $E = Pt$" a common mistake in Electrical Quantities?

Why it happens: Forgetting to convert. How to avoid it: Always convert to SECONDS first.

Why is "Saying thicker wire has more resistance" a common mistake in Electrical Quantities?

Why it happens: Mixing 'big' with 'resistive'. How to avoid it: Resistance is INVERSELY proportional to area. Thicker wire = lower resistance.

Why is "Saying positive charge moves during charging by friction" a common mistake in Electrical Quantities?

Why it happens: Thinking both kinds of charge can move. How to avoid it: Only electrons (negative charge) move. The object that gains electrons becomes negative; the one that loses them becomes positive. Source: 0625 Examiner Reports — recurring

Why is "Concluding a body is charged because it is attracted to a charged rod" a common mistake in Electrical Quantities?

Why it happens: Forgetting that an uncharged body is also attracted (by induction). How to avoid it: Only REPULSION proves a body carries charge of a definite sign.

Cambridge IGCSE Physics (0625)

Electrical Quantities

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Short Notes - Electrical Quantities

Detailed Study Notes

Detailed notes on Electricity and Magnetism for Cambridge IGCSE Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Electrical Quantities — Cambridge IGCSE 0625 Physics Extended (2026)

Charge, current, voltage, resistance, plus power and energy. The four big formulae: $Q = It$ , $V = IR$ , $P = IV$ , $E = IVt$ .

At a glance

Charge $Q = It$ , units coulombs ( $\text{C}$ ). There are positive and negative charges.
Like charges repel; unlike charges attract. Charging by friction transfers electrons only.
Conductors have free electrons (charge flows); insulators do not.
Electric field = a region where a charge feels a force; direction = force on a positive charge.
Current $I$ : rate of flow of charge ( $\text{A}$ ). Conventional current = positive charges flowing from + to −.
Voltage / p.d. $V$ : energy per unit charge ( $\text{V} = \text{J/C}$ ).
Resistance $R = V/I$ , units ohms ( $\Omega$ ).
Power $P = IV$ ; energy $E = IVt$ or $E = Pt$ .

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

4.2.1 — State that there are positive and negative charges; like charges repel, unlike attract.
4.2.1 — Explain charging by friction as a transfer of electrons (negative charge) only.
4.2.1 — Distinguish conductors and insulators using a simple electron model.
4.2.1 — Describe an electric field as a region where a charge feels a force, and state field direction.
4.2.1 — Describe the electric field patterns around a point charge, a charged sphere and between parallel plates.
4.2 — Recall and use $Q = It$ , $V = IR$ , $P = IV$ and $E = IVt$ .
4.2 — Describe how resistance varies with length and cross-section.

Electrostatics — charge, friction and conductors

Two kinds of charge. Like repel, unlike attract. Friction transfers electrons only.

There are two kinds of electric charge: positive and negative. The basic rule:

Like charges repel (two positive, or two negative, push apart).
Unlike charges attract (a positive and a negative pull together).

Charging by friction. When two insulators are rubbed together (e.g. a polythene rod and a dry cloth), electrons are transferred from one to the other. The positive nuclei do not move — only the negatively charged electrons.

The material that gains electrons becomes negatively charged.
The material that loses electrons becomes positively charged.

Because the SAME electrons move from one body to the other, the two end up with equal and opposite charges. Charge is transferred, never created or destroyed.

Only electrons transfer: the object that gains them is negative, the one that loses them is positive — equal and opposite.

Detecting charge. A charged rod brought near small pieces of paper, or a suspended charged ball, shows a force. Remember: only repulsion proves the sign of a charge — a charged rod also attracts an uncharged object, so attraction alone is not proof.

Conductors and insulators.

A conductor (e.g. a metal) contains free electrons that can move through the material, so charge flows easily.
An insulator (e.g. plastic, rubber, dry air) has no free electrons — its electrons are bound to atoms — so charge cannot flow.

Charging and discharging. An insulated charged object keeps its charge. Earthing (connecting it by a conductor to the ground) lets electrons flow on or off until the object is neutral — this is discharging. This is why a charged metal sphere is neutralised when you touch it: your body provides a conducting path to earth.

Worked. A glass rod is rubbed with silk and becomes positive. Which way did electrons move? — The rod is positive, so it LOST electrons; electrons moved from the rod to the silk, leaving the silk negative.

Two charges: positive and negative.
Like charges repel; unlike attract.
Friction transfers electrons (negative charge) only.
Conductor: free electrons. Insulator: none.
Earthing discharges a charged object.

Electric fields and field patterns

An electric field is a region where a charge feels a force. Field lines run + to −.

An electric field is a region in which an electric charge experiences a force.

Direction of the field. The direction of the electric field at a point is the direction of the force on a positive charge placed at that point. So field lines:

point away from positive charge,
point towards negative charge.

Field patterns you must be able to draw:

Around a positive point charge — straight field lines pointing radially outwards in all directions, evenly spaced. (For a negative point charge they point radially inwards.)
Around a charged conducting sphere — outside the sphere the pattern is the same as a point charge: radial lines from the centre. The field lines start at right angles to the sphere's surface.
Between two oppositely charged parallel plates — parallel, equally spaced straight lines running from the positive plate to the negative plate. This is a uniform field (the field has the same strength and direction everywhere between the plates). End effects — the curved lines near the plate edges — are not examined.

The spacing of the field lines shows the field strength: lines close together mean a strong field; lines far apart mean a weak field.

A point charge (and a charged sphere, outside it) gives a radial field; two charged plates give a uniform field from + to −.

Worked. A small positive charge is released from rest between two parallel plates. Which way does it move? — It feels a force in the direction of the field, i.e. from the positive plate towards the negative plate, so it accelerates towards the negative plate.

Electric field = region where a charge feels a force.
Field direction = force on a POSITIVE charge.
Point charge / sphere: radial lines (out for +, in for −).
Parallel plates: uniform field, + plate to − plate.
Closer lines = stronger field.

Charge and current

$Q = It$ . Current = rate of flow of charge.

Electric charge $Q$ is measured in coulombs ( $\text{C}$ ). Carried by electrons (each $-1.6 \times 10^{-19}\,\text{C}$ ).

Electric current $I$ is the rate of flow of charge: $I = \dfrac{Q}{t}, \quad Q = It.$

Units: amperes ( $\text{A}$ ). $1\,\text{A} = 1\,\text{C/s}$ .

Conventional current flows from + to − (the opposite direction to actual electron flow). When we draw circuit diagrams, the arrows mark conventional current.

Worked. A current of $0.5\,\text{A}$ flows for $2\,\text{minutes}$ . Charge?

$t = 120\,\text{s}$ .
$Q = 0.5 \times 120 = 60\,\text{C}$ .

Measure current with an ammeter connected in SERIES (current flows THROUGH it).

$Q = It$ .
$1\,\text{A} = 1\,\text{C/s}$ .
Conventional current: + to −.
Ammeter: series.

Voltage and Ohm's law

Voltage = energy per charge. $V = IR$ for ohmic conductors.

Potential difference (voltage) $V$ : the energy transferred per unit charge between two points. $V = \dfrac{E}{Q}, \quad E = QV.$

Units: volts ( $\text{V} = \text{J/C}$ ).

Ohm's law. $V = IR.$

For an ohmic conductor (constant temperature), current is proportional to voltage. The graph of $V$ vs $I$ is a straight line through the origin; gradient = resistance.

Ohmic: constant resistance, straight line. Lamp: resistance climbs with temperature, so the curve bends over.

Non-ohmic. Filament lamps and diodes don't obey Ohm's law:

Filament lamp: as current rises, the wire heats up, resistance increases. $V$ vs $I$ curve bends — gradient steeper at higher currents.
Diode: only allows current to flow in one direction. Curve nearly zero in reverse, then rises sharply once forward voltage is reached.

Measure voltage with a voltmeter connected in PARALLEL across the component.

Ammeter in series (current flows through it); voltmeter in parallel across the component.

Worked. $V = 12\,\text{V}$ , $I = 0.5\,\text{A}$ . Find resistance.

$R = V/I = 12/0.5 = 24\,\Omega$ .

$V = IR$ .
Ohmic: linear $V$ - $I$ .
Filament lamp: curves up.
Voltmeter: parallel.

Resistance — what affects it

Longer wire → more resistance. Wider wire → less resistance. Hotter wire → more resistance.

Resistance $R$ measures how much a component opposes current flow.

For a wire, resistance depends on:

Length ( $L$ ): doubling the length doubles the resistance ( $R \propto L$ ).
Cross-section area ( $A$ ): doubling the area HALVES the resistance ( $R \propto 1/A$ ).
Material: copper has low resistance; nichrome has high resistance.
Temperature: hotter metals have HIGHER resistance (more atomic vibrations impede electron flow).

For 0625, you only need the two proportionalities $R \propto L$ and $R \propto 1/A$ (§4.2.4 is qualitative on this) — not a combined formula.

Resistance rises with length and falls with cross-sectional area.

Worked. Two wires of the same material. Wire A: length $1\,\text{m}$ , area $0.5\,\text{mm}^{2}$ , resistance $4\,\Omega$ . Wire B: length $2\,\text{m}$ , area $0.25\,\text{mm}^{2}$ . Find $R_{B}$ .

B is twice as long ( $\times 2$ ) and half the area ( $\times 2$ ). So $R_{B} = 4 \times 2 \times 2 = 16\,\Omega$ .

Cambridge tip. Long thin wires are useful when you NEED resistance (heating elements, rheostats). Short fat wires are used for low-loss connections.

$R \propto L$ (length).
$R \propto 1/A$ (area).
Material matters (resistivity).
Hotter conductor → higher resistance (metals).

Electrical power and energy

$P = IV$ . $E = IVt$ (or $Pt$ ).

Electrical power. $P = IV.$

Units: watts ( $\text{W} = \text{J/s}$ ).

Combined with $V = IR$ : $P = I^{2}R = \dfrac{V^{2}}{R}.$

Use whichever form has the variables you know.

Energy. $E = Pt = IVt.$

Units: joules. For domestic billing, energy is in kilowatt-hours ( $\text{kWh}$ ): $1\,\text{kWh} = 1000\,\text{W} \times 3600\,\text{s} = 3.6 \times 10^{6}\,\text{J}$ .

Worked. A $60\,\text{W}$ lamp left on for $5$ hours.

$E = 60 \times 5 \times 3600 = 1{,}080{,}000\,\text{J}$ .
Or: $E = 0.06\,\text{kW} \times 5\,\text{h} = 0.3\,\text{kWh}$ .

Worked. A heater rated $230\,\text{V}$ , $1500\,\text{W}$ . Find current.

$I = P/V = 1500/230 \approx 6.5\,\text{A}$ .

$P = IV$ .
$E = IVt = Pt$ .
Cross-form: $P = I^{2}R = V^{2}/R$ .
Domestic energy: $\text{kWh}$ .

Quick recap

Two charges: like repel, unlike attract; friction transfers electrons only.
Conductor = free electrons; insulator = none. Earthing discharges.
Electric field = region where a charge feels a force; direction = force on a + charge.
$Q = It$ , $V = IR$ , $P = IV = I^{2}R = V^{2}/R$ , $E = IVt$ .
Resistance: ↑ with $L$ and $T$ ; ↓ with $A$ .
Ammeter: series. Voltmeter: parallel.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Like charges repel; unlike charges attract.
Charging by friction — transfers electrons (negative charge) only.
Conductor — has free electrons that allow charge to flow; insulator — has none.
Electric field — a region in which a charge experiences a force.
Charge — quantity of electricity, measured in coulombs.
Current — rate of flow of charge, $I = Q/t$ .
Potential difference (voltage) — energy transferred per unit charge.
Resistance — opposition to current flow, $R = V/I$ .
Ohm's law — $V = IR$ for an ohmic conductor at constant temperature.
Electrical power — rate of energy transfer, $P = IV$ .

How it’s examined

Electrical quantities appear on every Paper 2 (3-5 marks: $V = IR$ , $P = IV$ calculations) and every Paper 4 (8-10 marks: combined circuit + power + energy chain). Examiner reports flag putting an ammeter in parallel and forgetting to convert hours to seconds in $E = Pt$ (or vice versa).

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (4.2); 0625/42 Oct/Nov 2024 — Q16 ($V = IR$ + $P = IV$); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

Take this whole topic with you

Step-by-step worked examples — Electrical Quantities

Step-by-step solutions to past-paper-style questions on electrical quantities, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Electrical Quantities

Almost every electrical quantities exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find, calculate — with the quantities handed to you for one formula: $Q=It$ , $V=IR$ , $P=VI$ or $E=Pt$ .

How to approach it

Write the formula, rearrange before substituting, convert every quantity to SI units, then substitute. For power, pick the form ( $P=VI$ , $I^2R$ or $V^2/R$ ) that matches the variables you are given.

Common trap

Using minutes or hours directly in $Q=It$ or $E=Pt$ . Examiner reports flag missing time conversions — always convert to seconds first.

Show that / prove

Recognise it by

Compare, explain why or show that — a proportionality argument or qualitative reasoning, e.g. comparing the resistance of two wires of the same material.

How to approach it

Start from the proportionality ( $R \propto l/A$ ), take the ratio of the two cases so common factors cancel, then state the conclusion clearly.

Common trap

Thinking a thicker wire has more resistance. Resistance is inversely proportional to area — examiner reports note candidates routinely invert the area relationship.

Multi-step problem

Recognise it by

Several linked parts in one question — e.g. state the e.m.f., calculate the terminal pd, then explain the difference — where each part feeds the next.

How to approach it

Handle the parts in order: identify the source quantity, apply $V = \varepsilon - Ir$ for the terminal pd, then explain the 'lost volts' as energy dissipated in the internal resistance.

Common trap

Treating e.m.f. and terminal pd as the same thing. Examiner reports flag candidates using 'e.m.f.' and 'voltage' interchangeably — the difference is the pd across the internal resistance.

Graph or diagram

Recognise it by

A current-time graph or a $V$ - $I$ graph to read or interpret — finding charge transferred, or judging whether a component is ohmic.

How to approach it

For a current-time graph, charge is the area underneath (split into triangles and rectangles). For a $V$ - $I$ graph, resistance is $V/I$ — the gradient of the line from the origin to the point.

Common trap

Using the tangent gradient $\mathrm{d}V/\mathrm{d}I$ as the resistance of a non-ohmic component. Examiner reports stress that resistance is always $R=V/I$ at the chosen point, not the slope of the curve.

1Charge from current and time
CoreDirect calculation• Q=It
Question
A current of $0.5\,\text{A}$ flows for $4$ minutes. Find the charge transferred.
Step-by-step solution
1. Step 1
  Convert minutes to seconds.
  $t = 4 \times 60 = 240\,\text{s}$
2. Step 2
  $Q = I t$ .
  $Q = 0.5 \times 240 = 120\,\text{C}$
Answer
$120\,\text{C}$
2Ohm's law
CoreDirect calculation• Adapted from 0625/22 May/Jun 2024 Q16• V=IR
Question
A $12\,\text{V}$ supply drives a current of $0.4\,\text{A}$ through a resistor. Find the resistance.
Step-by-step solution
1. Step 1
  $V = IR \Rightarrow R = V/I$ .
  $R = \dfrac{12}{0.4} = 30\,\Omega$
Answer
$30\,\Omega$
3Power in a circuit
ExtendedDirect calculation• P=VI
Question
Find the power dissipated by a $230\,\text{V}$ kettle that draws $9.0\,\text{A}$ .
Step-by-step solution
1. Step 1
  $P = VI$ .
  $P = 230 \times 9.0 = 2070\,\text{W}$
Answer
$2070\,\text{W} \approx 2.1\,\text{kW}$
4Electrical energy
ExtendedDirect calculation• E=Pt
Question
An $1100\,\text{W}$ heater is left on for $15$ minutes. Find the energy used.
Step-by-step solution
1. Step 1
  Convert time.
  $t = 15 \times 60 = 900\,\text{s}$
2. Step 2
  $E = P t$ .
  $E = 1100 \times 900 = 9.9 \times 10^{5}\,\text{J}$
Answer
$9.9 \times 10^{5}\,\text{J} = 990\,\text{kJ}$
5Resistance of a wire
ExtendedShow that / prove• resistivity
Question
Two wires of the same material. Wire A has length $L$ and area $A$ ; wire B has length $2L$ and area $A/2$ . Compare resistances.
Step-by-step solution
1. Step 1
  $R \propto \dfrac{l}{A}$ .
  $\dfrac{R_{B}}{R_{A}} = \dfrac{2L}{A/2} \div \dfrac{L}{A} = 4$
Answer
$R_{B} = 4 R_{A}$
6Beyond 0625 — EMF versus terminal pd (internal resistance)
ExtendedMulti-step problem• Adapted from 0625/42 Oct/Nov 2023 Q8• emf, pd
Question
A battery of e.m.f. $6.0\,\text{V}$ has internal resistance $0.5\,\Omega$ . It drives a current of $2.0\,\text{A}$ through an external resistor. State the e.m.f., calculate the terminal pd, and explain the difference.
Step-by-step solution
1. Step 1
  The e.m.f. is the energy per coulomb supplied BY the battery: $\varepsilon = 6.0\,\text{V}$ .
2. Step 2
  The terminal pd is the energy per coulomb delivered to the external circuit: $V = \varepsilon - Ir$ .
  $V = 6.0 - (2.0 \times 0.5) = 5.0\,\text{V}$
3. Step 3
  The 1 V 'missing' is energy per coulomb dissipated inside the battery against its internal resistance.
Answer
$\varepsilon = 6.0\,\text{V}$ ; terminal pd $= 5.0\,\text{V}$ . The difference (1 V) is the pd across the internal resistance.
Examiner tip
Enrichment beyond Cambridge IGCSE 0625 — internal resistance and the equation V = ε − Ir are NOT in the 0625 syllabus (§4.2.3 covers e.m.f. and p.d. only via E = W/Q). This material belongs to AS/A-level physics. It is included as enrichment to deepen understanding of why e.m.f. and terminal p.d. differ; it will not be examined on 0625. The 0625-relevant takeaway is the qualitative distinction: e.m.f. is the source's energy per coulomb; terminal p.d. is what the external circuit actually receives.
7Charge from a current-time graph
ExtendedGraph or diagram• Q=It, graph
Question
On a current-time graph, the current rises linearly from $0$ to $2.0\,\text{A}$ in $10\,\text{s}$ , then stays at $2.0\,\text{A}$ for a further $10\,\text{s}$ . Find the total charge transferred.
Step-by-step solution
1. Step 1
  Total charge = area UNDER the $I$ - $t$ graph.
2. Step 2
  Triangle (rising section): $Q_{1} = \tfrac{1}{2} \times 10 \times 2.0$ .
  $Q_{1} = 10\,\text{C}$
3. Step 3
  Rectangle (constant section): $Q_{2} = 2.0 \times 10$ .
  $Q_{2} = 20\,\text{C}$
4. Step 4
  Total $Q = Q_{1} + Q_{2}$ .
  $Q = 30\,\text{C}$
Answer
$30\,\text{C}$
8Power dissipated in a resistor — choose the right form
ExtendedDirect calculation• P=I^2R
Question
A heater coil of resistance $25\,\Omega$ carries a current of $4.0\,\text{A}$ . Find the power dissipated.
Step-by-step solution
1. Step 1
  Use the form of $P$ that matches the given variables ( $I$ and $R$ ): $P = I^{2}R$ .
2. Step 2
  Substitute.
  $P = (4.0)^{2} \times 25 = 16 \times 25 = 400\,\text{W}$
Answer
$400\,\text{W}$
9Resistors in series vs parallel — quick comparison
CoreDirect calculation• series, parallel
Question
Two identical $6\,\Omega$ resistors are connected (a) in series and (b) in parallel. Find the total resistance in each case.
Step-by-step solution
1. Step 1
  (a) Series: add directly.
  $R_{T} = 6 + 6 = 12\,\Omega$
2. Step 2
  (b) Parallel of two equal resistors: $R_{T} = R/2$ .
  $R_{T} = 6/2 = 3\,\Omega$
Answer
(a) $12\,\Omega$ in series. (b) $3\,\Omega$ in parallel.
10Ohmic vs non-ohmic from a V-I graph
ChallengeGraph or diagram• V-I graph, ohmic
Question
Two components are tested. Component A gives a straight line through the origin on a $V$ - $I$ graph. Component B (a filament lamp) gives a curve that bends, with the gradient $V/I$ rising as $I$ rises. For each: (a) state whether it obeys Ohm's law and (b) explain what happens to its resistance.
Step-by-step solution
1. Step 1
  Resistance at any point = $V/I$ = gradient from the ORIGIN to the point on a $V$ vs $I$ plot.
2. Step 2
  Component A: straight line through origin → constant gradient → constant resistance → OBEYS Ohm's law.
3. Step 3
  Component B (filament): gradient rises with $I$ → resistance INCREASES. Cause: as current rises, the filament heats up; metal ions vibrate more, electrons scatter more, so resistance rises. Does NOT obey Ohm's law.
Answer
A: ohmic (constant $R$ ). B: non-ohmic — resistance rises with current because the filament heats up.
Examiner tip
The examiner report flags candidates often using 'gradient of the curve' (i.e. $\mathrm{d}V/\mathrm{d}I$ ) for resistance — for a non-ohmic component the correct definition is still $R = V/I$ at a particular point.
11Charging by friction — which way do electrons move?
Core• electrostatics, charge, friction
Question
A polythene rod is rubbed with a dry cloth. After rubbing, the rod is found to be negatively charged. (a) State which way electrons have moved. (b) State the charge on the cloth. (c) Explain why the rod and cloth carry equal and opposite amounts of charge.
Step-by-step solution
1. Step 1
  Only negative charge (electrons) can move; the positive nuclei stay fixed. The rod gained negative charge, so electrons were transferred FROM the cloth TO the rod.
2. Step 2
  The cloth has lost electrons, so it is left with a net positive charge.
3. Step 3
  Every electron gained by the rod is an electron lost by the cloth — charge is transferred, not created. So the rod's negative charge equals the cloth's positive charge in size.
Answer
(a) Electrons move from the cloth to the rod. (b) The cloth is positively charged. (c) The rod's negative charge equals the cloth's positive charge because the same electrons moved off one and onto the other.
Examiner tip
Mark schemes insist that friction transfers negative charge (electrons) only — never write that 'positive charge moved to the cloth'.
12Predicting attraction and repulsion
Core• electrostatics, charge
Question
A positively charged rod is brought near a small suspended ball. The ball swings TOWARDS the rod. State two possible charge states of the ball, and explain how a further test could decide between them.
Step-by-step solution
1. Step 1
  Like charges repel; unlike charges attract. Attraction towards a positive rod means the ball is either negatively charged (unlike charges attract) OR uncharged (the rod induces opposite charge on the near side).
2. Step 2
  Bring a known NEGATIVE rod near the ball. If the ball is repelled, it must be negatively charged. If it is attracted again, the ball is uncharged (an uncharged object is attracted to any charged rod).
Answer
The ball is either negatively charged or uncharged. A negatively charged rod will repel a negative ball but attract an uncharged one — only repulsion proves the ball carries charge.
Examiner tip
Repulsion is the only definite test for the sign of a charge — attraction can be caused by induction on a neutral body.
13Conductors, insulators and discharging
Core• electrostatics, conductor, insulator
Question
A metal sphere on an insulating stand is charged positively. (a) Explain, using the electron model, why a metal conducts charge but the stand does not. (b) Describe what happens to the charge if you touch the sphere with your hand.
Step-by-step solution
1. Step 1
  A metal has free (delocalised) electrons that can move easily through the material, so charge flows — it is a conductor. An insulator (e.g. plastic) has no free electrons; its electrons are bound to atoms, so charge cannot flow.
2. Step 2
  Touching the sphere connects it through your body to earth — a conducting path. Electrons flow from earth onto the positive sphere to neutralise it.
3. Step 3
  The sphere is discharged (earthed): it becomes neutral.
Answer
(a) Metals conduct because they have free electrons that can move; insulators have no free electrons. (b) Touching it earths the sphere — electrons flow from earth and discharge it to neutral.
14Electric field patterns and field direction
Extended• electric field, field lines
Question
(a) Define an electric field. (b) State the rule for the direction of an electric field at a point. (c) Describe the field pattern (i) around an isolated positive point charge and (ii) between two oppositely charged parallel plates.
Step-by-step solution
1. Step 1
  An electric field is a region in which an electric charge experiences a force.
2. Step 2
  The direction of the field at a point is the direction of the force on a POSITIVE charge placed at that point. Field lines therefore point away from positive charge and towards negative charge.
3. Step 3
  (i) Around a positive point charge the field lines are straight, radial lines pointing directly OUTWARDS, evenly spread in all directions (the pattern around a charged conducting sphere is the same shape outside the sphere).
4. Step 4
  (ii) Between two oppositely charged parallel plates the field lines are parallel, equally spaced straight lines running from the positive plate to the negative plate — a uniform field (ignoring the edges).
Answer
(a) A region where a charge feels a force. (b) The direction of the force on a positive charge. (c)(i) Radial lines pointing outwards from a positive point charge / charged sphere; (ii) parallel, evenly spaced lines from + plate to − plate — a uniform field.
Examiner tip
For the parallel-plate field, the syllabus states end effects (the curved lines at the plate edges) will NOT be examined — draw the central region as straight, evenly spaced lines.

Model Answers — Electrical Quantities

High-scoring sample answers for electrical quantities on the Cambridge IGCSE 0625 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2/4 short-answer style1 mark
Define electric current and state its SI unit.
Model answer
Electric current is the rate of flow of electric charge. Its SI unit is the ampere (A), where $1\,\text{A} = 1\,\text{C/s}$ .
Why this scores
One mark for 'rate of flow of charge' with the unit ampere. Conventional current flows from + to −, opposite to the electron flow.
Question 2
Paper 2/4 style2 marks
A current of $2.0\,\text{A}$ flows in a circuit for $5.0$ minutes. Calculate the charge that passes.
Model answer
Convert the time to seconds: $t = 5.0 \times 60 = 300\,\text{s}$ . Then

$Q = It = 2.0 \times 300 = \mathbf{600\,\text{C}}.$
Why this scores
One mark for converting minutes to seconds and using $Q = It$ , one for $600\,\text{C}$ . Forgetting the time conversion is the standard error.
Question 3
Paper 4 structured style3 marks
A $12\,\text{V}$ supply drives a current of $2.0\,\text{A}$ through a heating element. (a) Calculate the resistance of the element. (b) Calculate the power it dissipates.
Model answer
(a) $R = \dfrac{V}{I} = \dfrac{12}{2.0} = \mathbf{6.0\,\Omega}$ .

(b) $P = VI = 12 \times 2.0 = \mathbf{24\,\text{W}}$ (equivalently $P = I^{2}R = 2.0^{2} \times 6.0 = 24\,\text{W}$ ).
Why this scores
Three marks: $R = V/I = 6.0\,\Omega$ (1); use of a correct power formula (1); $24\,\text{W}$ (1). Choosing the power equation that matches the given quantities saves work.
Question 4
Paper 4 structured style4 marks
A polythene rod is rubbed with a dry cloth and becomes negatively charged. (a) Explain, in terms of electrons, how the rod becomes negatively charged. (b) State the charge on the cloth. (c) Explain why the charges on the rod and cloth are equal in size.
Model answer
(a) Rubbing transfers electrons from the cloth onto the rod. Electrons are negatively charged, so the rod gains negative charge (only electrons move — the positive nuclei stay fixed).

(b) The cloth has lost electrons, so it is left positively charged.

(c) Every electron gained by the rod is one lost by the cloth — charge is transferred, not created — so the rod's negative charge is equal in size to the cloth's positive charge.
Why this scores
Four marks: electrons transferred from cloth to rod (1); rod negative because it gains electrons (1); cloth positive (1); equal charges because the same electrons moved across / charge is conserved (1). 'Positive charge moved to the cloth' is wrong — only electrons move.
Question 5
Paper 4 explanation style5 marks
A fixed resistor and a filament lamp are each tested. The resistor gives a straight line through the origin on a current–voltage graph; the lamp gives a curve that bends over (the current rises less steeply at higher voltages). (a) State which component obeys Ohm's law. (b) Explain what happens to the resistance of the filament lamp as the voltage increases, and why.
Model answer
(a) The fixed resistor obeys Ohm's law — the straight line through the origin shows that current is proportional to voltage (constant resistance).

(b) For the filament lamp the resistance increases as the voltage increases. As more current flows, the filament gets hotter, so the metal ions vibrate more vigorously; the electrons collide with them more often and find it harder to move through the metal, so the resistance rises. This is why the current rises less steeply than the voltage (it does not obey Ohm's law).
Why this scores
Five marks: resistor obeys Ohm's law because $I \propto V$ (1); lamp resistance increases (1); filament heats up (1); ions vibrate more (1); electrons collide more / harder to pass so resistance rises (1). Resistance at any point is still $R = V/I$ , not the slope of the curve.
Question 6
Paper 4 multi-part structured style6 marks
(a) Define potential difference. (b) A $2.0\,\text{kW}$ heater is connected to the $230\,\text{V}$ mains. Calculate the current it draws and the resistance of its element. (c) Calculate the electrical energy it transfers in $5.0$ minutes.
Model answer
(a) The potential difference between two points is the energy transferred (work done) per unit charge moving between them; unit volt ( $1\,\text{V} = 1\,\text{J/C}$ ).

(b) Current: $I = \dfrac{P}{V} = \dfrac{2000}{230} \approx \mathbf{8.7\,\text{A}}$ . Resistance: $R = \dfrac{V^{2}}{P} = \dfrac{230^{2}}{2000} \approx \mathbf{26\,\Omega}$ (or $R = V/I = 230/8.7 \approx 26\,\Omega$ ).

(c) $t = 5.0 \times 60 = 300\,\text{s}$ , so $E = Pt = 2000 \times 300 = \mathbf{6.0 \times 10^{5}\,\text{J}}$ ( $= 600\,\text{kJ}$ ).
Why this scores
Six marks: definition of pd as energy per unit charge (1); $I = P/V \approx 8.7\,\text{A}$ (1); $R \approx 26\,\Omega$ (1); time converted to seconds (1); $E = Pt$ (1); $6.0 \times 10^{5}\,\text{J}$ (1).

Key Formulae — Electrical Quantities

The formulae you need to memorise for electrical quantities on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Charge
$Q = It$
$Q$
charge in coulombs (C)
$I$
current in amps (A)
$t$
time in seconds (s)
When to use
Steady current.
Ohm's law
$V = IR$
$V$
potential difference (V)
$R$
resistance (Ω)
When to use
Component obeying Ohm's law (e.g. fixed resistor at constant T).
Electrical power
$P = VI = I^{2}R = \dfrac{V^{2}}{R}$
When to use
Choose the form that matches what is given.
Electrical energy
$E = Pt = VIt$
When to use
Energy transferred in time $t$ at constant power.
Resistance of a wire
$R \propto \dfrac{l}{A}$
$l$
length of wire
$A$
cross-sectional area
When to use
Comparing wires of the same material.

Key Definitions and Keywords — Electrical Quantities

Definitions to memorise and the exact keywords mark schemes credit for electrical quantities answers — sharpened from recent examiner reports for the 2026 0625 sitting.

Current
Examiner keyword
Rate of flow of charge. SI unit ampere (A) = C/s. Conventional current is from + to −.
Potential difference / voltage
Examiner keyword
Energy transferred per unit charge between two points. Measured in volts (V) = J/C.
Resistance
Examiner keyword
Ratio of pd to current. Unit ohm ( $\Omega$ ).
Electromotive force (e.m.f.)
Examiner keyword
Energy transferred per unit charge by a source (battery, generator). Unit volt (V).
Coulomb (C)
Examiner keyword
$1\,\text{C}$ = charge transferred by a current of $1\,\text{A}$ in $1\,\text{s}$ .
Electric field
Examiner keyword
A region in which an electric charge experiences a force. Its direction at a point is the direction of the force on a positive charge there.
Conductor and insulator
Examiner keyword
A conductor (e.g. metal) has free electrons that allow charge to flow; an insulator (e.g. plastic) has no free electrons, so charge cannot flow.

Common Mistakes and Misconceptions — Electrical Quantities

The traps other students keep falling into on electrical quantities questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Confusing electron flow direction with conventional current
0625/42 — recurring
Why it happens
Two conventions used in different sources.
How to avoid it
Conventional current: + → −. Electron flow: − → +. Use whichever the question asks; default in 0625 is conventional.
✕Quoting voltage in amps (or current in volts)
Why it happens
Slip.
How to avoid it
V (volts) for pd; A (amps) for current; Ω (ohms) for resistance.
✕Using minutes/hours directly in $Q = It$ or $E = Pt$
Why it happens
Forgetting to convert.
How to avoid it
Always convert to SECONDS first.
✕Saying thicker wire has more resistance
Why it happens
Mixing 'big' with 'resistive'.
How to avoid it
Resistance is INVERSELY proportional to area. Thicker wire = lower resistance.
✕Saying positive charge moves during charging by friction
0625 Examiner Reports — recurring
Why it happens
Thinking both kinds of charge can move.
How to avoid it
Only electrons (negative charge) move. The object that gains electrons becomes negative; the one that loses them becomes positive.
✕Concluding a body is charged because it is attracted to a charged rod
Why it happens
Forgetting that an uncharged body is also attracted (by induction).
How to avoid it
Only REPULSION proves a body carries charge of a definite sign.

Electrical Quantities — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

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Short Notes - Electrical Quantities

Detailed Study Notes

Detailed notes on Electricity and Magnetism for Cambridge IGCSE Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Electrical Quantities — Cambridge IGCSE 0625 Physics Extended (2026)

Charge, current, voltage, resistance, plus power and energy. The four big formulae: $Q = It$ , $V = IR$ , $P = IV$ , $E = IVt$ .

At a glance

Charge $Q = It$ , units coulombs ( $\text{C}$ ). There are positive and negative charges.
Like charges repel; unlike charges attract. Charging by friction transfers electrons only.
Conductors have free electrons (charge flows); insulators do not.
Electric field = a region where a charge feels a force; direction = force on a positive charge.
Current $I$ : rate of flow of charge ( $\text{A}$ ). Conventional current = positive charges flowing from + to −.
Voltage / p.d. $V$ : energy per unit charge ( $\text{V} = \text{J/C}$ ).
Resistance $R = V/I$ , units ohms ( $\Omega$ ).
Power $P = IV$ ; energy $E = IVt$ or $E = Pt$ .

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

4.2.1 — State that there are positive and negative charges; like charges repel, unlike attract.
4.2.1 — Explain charging by friction as a transfer of electrons (negative charge) only.
4.2.1 — Distinguish conductors and insulators using a simple electron model.
4.2.1 — Describe an electric field as a region where a charge feels a force, and state field direction.
4.2.1 — Describe the electric field patterns around a point charge, a charged sphere and between parallel plates.
4.2 — Recall and use $Q = It$ , $V = IR$ , $P = IV$ and $E = IVt$ .
4.2 — Describe how resistance varies with length and cross-section.

Electrostatics — charge, friction and conductors

Two kinds of charge. Like repel, unlike attract. Friction transfers electrons only.

There are two kinds of electric charge: positive and negative. The basic rule:

Like charges repel (two positive, or two negative, push apart).
Unlike charges attract (a positive and a negative pull together).

The material that gains electrons becomes negatively charged.
The material that loses electrons becomes positively charged.

Because the SAME electrons move from one body to the other, the two end up with equal and opposite charges. Charge is transferred, never created or destroyed.

Only electrons transfer: the object that gains them is negative, the one that loses them is positive — equal and opposite.

Conductors and insulators.

A conductor (e.g. a metal) contains free electrons that can move through the material, so charge flows easily.
An insulator (e.g. plastic, rubber, dry air) has no free electrons — its electrons are bound to atoms — so charge cannot flow.

Two charges: positive and negative.
Like charges repel; unlike attract.
Friction transfers electrons (negative charge) only.
Conductor: free electrons. Insulator: none.
Earthing discharges a charged object.

Electric fields and field patterns

An electric field is a region where a charge feels a force. Field lines run + to −.

An electric field is a region in which an electric charge experiences a force.

Direction of the field. The direction of the electric field at a point is the direction of the force on a positive charge placed at that point. So field lines:

point away from positive charge,
point towards negative charge.

Field patterns you must be able to draw:

Around a positive point charge — straight field lines pointing radially outwards in all directions, evenly spaced. (For a negative point charge they point radially inwards.)
Around a charged conducting sphere — outside the sphere the pattern is the same as a point charge: radial lines from the centre. The field lines start at right angles to the sphere's surface.
Between two oppositely charged parallel plates — parallel, equally spaced straight lines running from the positive plate to the negative plate. This is a uniform field (the field has the same strength and direction everywhere between the plates). End effects — the curved lines near the plate edges — are not examined.

The spacing of the field lines shows the field strength: lines close together mean a strong field; lines far apart mean a weak field.

A point charge (and a charged sphere, outside it) gives a radial field; two charged plates give a uniform field from + to −.

Electric field = region where a charge feels a force.
Field direction = force on a POSITIVE charge.
Point charge / sphere: radial lines (out for +, in for −).
Parallel plates: uniform field, + plate to − plate.
Closer lines = stronger field.

Charge and current

$Q = It$ . Current = rate of flow of charge.

Electric charge $Q$ is measured in coulombs ( $\text{C}$ ). Carried by electrons (each $-1.6 \times 10^{-19}\,\text{C}$ ).

Electric current $I$ is the rate of flow of charge: $I = \dfrac{Q}{t}, \quad Q = It.$

Units: amperes ( $\text{A}$ ). $1\,\text{A} = 1\,\text{C/s}$ .

Conventional current flows from + to − (the opposite direction to actual electron flow). When we draw circuit diagrams, the arrows mark conventional current.

Worked. A current of $0.5\,\text{A}$ flows for $2\,\text{minutes}$ . Charge?

$t = 120\,\text{s}$ .
$Q = 0.5 \times 120 = 60\,\text{C}$ .

Measure current with an ammeter connected in SERIES (current flows THROUGH it).

$Q = It$ .
$1\,\text{A} = 1\,\text{C/s}$ .
Conventional current: + to −.
Ammeter: series.

Voltage and Ohm's law

Voltage = energy per charge. $V = IR$ for ohmic conductors.

Potential difference (voltage) $V$ : the energy transferred per unit charge between two points. $V = \dfrac{E}{Q}, \quad E = QV.$

Units: volts ( $\text{V} = \text{J/C}$ ).

Ohm's law. $V = IR.$

For an ohmic conductor (constant temperature), current is proportional to voltage. The graph of $V$ vs $I$ is a straight line through the origin; gradient = resistance.

Ohmic: constant resistance, straight line. Lamp: resistance climbs with temperature, so the curve bends over.

Non-ohmic. Filament lamps and diodes don't obey Ohm's law:

Filament lamp: as current rises, the wire heats up, resistance increases. $V$ vs $I$ curve bends — gradient steeper at higher currents.
Diode: only allows current to flow in one direction. Curve nearly zero in reverse, then rises sharply once forward voltage is reached.

Measure voltage with a voltmeter connected in PARALLEL across the component.

Ammeter in series (current flows through it); voltmeter in parallel across the component.

Worked. $V = 12\,\text{V}$ , $I = 0.5\,\text{A}$ . Find resistance.

$R = V/I = 12/0.5 = 24\,\Omega$ .

$V = IR$ .
Ohmic: linear $V$ - $I$ .
Filament lamp: curves up.
Voltmeter: parallel.

Resistance — what affects it

Longer wire → more resistance. Wider wire → less resistance. Hotter wire → more resistance.

Resistance $R$ measures how much a component opposes current flow.

For a wire, resistance depends on:

Length ( $L$ ): doubling the length doubles the resistance ( $R \propto L$ ).
Cross-section area ( $A$ ): doubling the area HALVES the resistance ( $R \propto 1/A$ ).
Material: copper has low resistance; nichrome has high resistance.
Temperature: hotter metals have HIGHER resistance (more atomic vibrations impede electron flow).

For 0625, you only need the two proportionalities $R \propto L$ and $R \propto 1/A$ (§4.2.4 is qualitative on this) — not a combined formula.

Resistance rises with length and falls with cross-sectional area.

B is twice as long ( $\times 2$ ) and half the area ( $\times 2$ ). So $R_{B} = 4 \times 2 \times 2 = 16\,\Omega$ .

Cambridge tip. Long thin wires are useful when you NEED resistance (heating elements, rheostats). Short fat wires are used for low-loss connections.

$R \propto L$ (length).
$R \propto 1/A$ (area).
Material matters (resistivity).
Hotter conductor → higher resistance (metals).

Electrical power and energy

$P = IV$ . $E = IVt$ (or $Pt$ ).

Electrical power. $P = IV.$

Units: watts ( $\text{W} = \text{J/s}$ ).

Combined with $V = IR$ : $P = I^{2}R = \dfrac{V^{2}}{R}.$

Use whichever form has the variables you know.

Energy. $E = Pt = IVt.$

Units: joules. For domestic billing, energy is in kilowatt-hours ( $\text{kWh}$ ): $1\,\text{kWh} = 1000\,\text{W} \times 3600\,\text{s} = 3.6 \times 10^{6}\,\text{J}$ .

Worked. A $60\,\text{W}$ lamp left on for $5$ hours.

$E = 60 \times 5 \times 3600 = 1{,}080{,}000\,\text{J}$ .
Or: $E = 0.06\,\text{kW} \times 5\,\text{h} = 0.3\,\text{kWh}$ .

Worked. A heater rated $230\,\text{V}$ , $1500\,\text{W}$ . Find current.

$I = P/V = 1500/230 \approx 6.5\,\text{A}$ .

$P = IV$ .
$E = IVt = Pt$ .
Cross-form: $P = I^{2}R = V^{2}/R$ .
Domestic energy: $\text{kWh}$ .

Quick recap

Two charges: like repel, unlike attract; friction transfers electrons only.
Conductor = free electrons; insulator = none. Earthing discharges.
Electric field = region where a charge feels a force; direction = force on a + charge.
$Q = It$ , $V = IR$ , $P = IV = I^{2}R = V^{2}/R$ , $E = IVt$ .
Resistance: ↑ with $L$ and $T$ ; ↓ with $A$ .
Ammeter: series. Voltmeter: parallel.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Like charges repel; unlike charges attract.
Charging by friction — transfers electrons (negative charge) only.
Conductor — has free electrons that allow charge to flow; insulator — has none.
Electric field — a region in which a charge experiences a force.
Charge — quantity of electricity, measured in coulombs.
Current — rate of flow of charge, $I = Q/t$ .
Potential difference (voltage) — energy transferred per unit charge.
Resistance — opposition to current flow, $R = V/I$ .
Ohm's law — $V = IR$ for an ohmic conductor at constant temperature.
Electrical power — rate of energy transfer, $P = IV$ .

How it’s examined

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (4.2); 0625/42 Oct/Nov 2024 — Q16 ($V = IR$ + $P = IV$); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

Take this whole topic with you

Step-by-step worked examples — Electrical Quantities

Step-by-step solutions to past-paper-style questions on electrical quantities, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Electrical Quantities

Almost every electrical quantities exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find, calculate — with the quantities handed to you for one formula: $Q=It$ , $V=IR$ , $P=VI$ or $E=Pt$ .

How to approach it

Common trap

Using minutes or hours directly in $Q=It$ or $E=Pt$ . Examiner reports flag missing time conversions — always convert to seconds first.

Show that / prove

Recognise it by

Compare, explain why or show that — a proportionality argument or qualitative reasoning, e.g. comparing the resistance of two wires of the same material.

How to approach it

Start from the proportionality ( $R \propto l/A$ ), take the ratio of the two cases so common factors cancel, then state the conclusion clearly.

Common trap

Thinking a thicker wire has more resistance. Resistance is inversely proportional to area — examiner reports note candidates routinely invert the area relationship.

Multi-step problem

Recognise it by

Several linked parts in one question — e.g. state the e.m.f., calculate the terminal pd, then explain the difference — where each part feeds the next.

How to approach it

Handle the parts in order: identify the source quantity, apply $V = \varepsilon - Ir$ for the terminal pd, then explain the 'lost volts' as energy dissipated in the internal resistance.

Common trap

Treating e.m.f. and terminal pd as the same thing. Examiner reports flag candidates using 'e.m.f.' and 'voltage' interchangeably — the difference is the pd across the internal resistance.

Graph or diagram

Recognise it by

A current-time graph or a $V$ - $I$ graph to read or interpret — finding charge transferred, or judging whether a component is ohmic.

How to approach it

For a current-time graph, charge is the area underneath (split into triangles and rectangles). For a $V$ - $I$ graph, resistance is $V/I$ — the gradient of the line from the origin to the point.

Common trap

1Charge from current and time
CoreDirect calculation• Q=It
Question
A current of $0.5\,\text{A}$ flows for $4$ minutes. Find the charge transferred.
Step-by-step solution
1. Step 1
  Convert minutes to seconds.
  $t = 4 \times 60 = 240\,\text{s}$
2. Step 2
  $Q = I t$ .
  $Q = 0.5 \times 240 = 120\,\text{C}$
Answer
$120\,\text{C}$
2Ohm's law
CoreDirect calculation• Adapted from 0625/22 May/Jun 2024 Q16• V=IR
Question
A $12\,\text{V}$ supply drives a current of $0.4\,\text{A}$ through a resistor. Find the resistance.
Step-by-step solution
1. Step 1
  $V = IR \Rightarrow R = V/I$ .
  $R = \dfrac{12}{0.4} = 30\,\Omega$
Answer
$30\,\Omega$
3Power in a circuit
ExtendedDirect calculation• P=VI
Question
Find the power dissipated by a $230\,\text{V}$ kettle that draws $9.0\,\text{A}$ .
Step-by-step solution
1. Step 1
  $P = VI$ .
  $P = 230 \times 9.0 = 2070\,\text{W}$
Answer
$2070\,\text{W} \approx 2.1\,\text{kW}$
4Electrical energy
ExtendedDirect calculation• E=Pt
Question
An $1100\,\text{W}$ heater is left on for $15$ minutes. Find the energy used.
Step-by-step solution
1. Step 1
  Convert time.
  $t = 15 \times 60 = 900\,\text{s}$
2. Step 2
  $E = P t$ .
  $E = 1100 \times 900 = 9.9 \times 10^{5}\,\text{J}$
Answer
$9.9 \times 10^{5}\,\text{J} = 990\,\text{kJ}$
5Resistance of a wire
ExtendedShow that / prove• resistivity
Question
Two wires of the same material. Wire A has length $L$ and area $A$ ; wire B has length $2L$ and area $A/2$ . Compare resistances.
Step-by-step solution
1. Step 1
  $R \propto \dfrac{l}{A}$ .
  $\dfrac{R_{B}}{R_{A}} = \dfrac{2L}{A/2} \div \dfrac{L}{A} = 4$
Answer
$R_{B} = 4 R_{A}$
6Beyond 0625 — EMF versus terminal pd (internal resistance)
ExtendedMulti-step problem• Adapted from 0625/42 Oct/Nov 2023 Q8• emf, pd
Question
A battery of e.m.f. $6.0\,\text{V}$ has internal resistance $0.5\,\Omega$ . It drives a current of $2.0\,\text{A}$ through an external resistor. State the e.m.f., calculate the terminal pd, and explain the difference.
Step-by-step solution
1. Step 1
  The e.m.f. is the energy per coulomb supplied BY the battery: $\varepsilon = 6.0\,\text{V}$ .
2. Step 2
  The terminal pd is the energy per coulomb delivered to the external circuit: $V = \varepsilon - Ir$ .
  $V = 6.0 - (2.0 \times 0.5) = 5.0\,\text{V}$
3. Step 3
  The 1 V 'missing' is energy per coulomb dissipated inside the battery against its internal resistance.
Answer
$\varepsilon = 6.0\,\text{V}$ ; terminal pd $= 5.0\,\text{V}$ . The difference (1 V) is the pd across the internal resistance.
Examiner tip
Enrichment beyond Cambridge IGCSE 0625 — internal resistance and the equation V = ε − Ir are NOT in the 0625 syllabus (§4.2.3 covers e.m.f. and p.d. only via E = W/Q). This material belongs to AS/A-level physics. It is included as enrichment to deepen understanding of why e.m.f. and terminal p.d. differ; it will not be examined on 0625. The 0625-relevant takeaway is the qualitative distinction: e.m.f. is the source's energy per coulomb; terminal p.d. is what the external circuit actually receives.
7Charge from a current-time graph
ExtendedGraph or diagram• Q=It, graph
Question
On a current-time graph, the current rises linearly from $0$ to $2.0\,\text{A}$ in $10\,\text{s}$ , then stays at $2.0\,\text{A}$ for a further $10\,\text{s}$ . Find the total charge transferred.
Step-by-step solution
1. Step 1
  Total charge = area UNDER the $I$ - $t$ graph.
2. Step 2
  Triangle (rising section): $Q_{1} = \tfrac{1}{2} \times 10 \times 2.0$ .
  $Q_{1} = 10\,\text{C}$
3. Step 3
  Rectangle (constant section): $Q_{2} = 2.0 \times 10$ .
  $Q_{2} = 20\,\text{C}$
4. Step 4
  Total $Q = Q_{1} + Q_{2}$ .
  $Q = 30\,\text{C}$
Answer
$30\,\text{C}$
8Power dissipated in a resistor — choose the right form
ExtendedDirect calculation• P=I^2R
Question
A heater coil of resistance $25\,\Omega$ carries a current of $4.0\,\text{A}$ . Find the power dissipated.
Step-by-step solution
1. Step 1
  Use the form of $P$ that matches the given variables ( $I$ and $R$ ): $P = I^{2}R$ .
2. Step 2
  Substitute.
  $P = (4.0)^{2} \times 25 = 16 \times 25 = 400\,\text{W}$
Answer
$400\,\text{W}$
9Resistors in series vs parallel — quick comparison
CoreDirect calculation• series, parallel
Question
Two identical $6\,\Omega$ resistors are connected (a) in series and (b) in parallel. Find the total resistance in each case.
Step-by-step solution
1. Step 1
  (a) Series: add directly.
  $R_{T} = 6 + 6 = 12\,\Omega$
2. Step 2
  (b) Parallel of two equal resistors: $R_{T} = R/2$ .
  $R_{T} = 6/2 = 3\,\Omega$
Answer
(a) $12\,\Omega$ in series. (b) $3\,\Omega$ in parallel.
10Ohmic vs non-ohmic from a V-I graph
ChallengeGraph or diagram• V-I graph, ohmic
Question
Two components are tested. Component A gives a straight line through the origin on a $V$ - $I$ graph. Component B (a filament lamp) gives a curve that bends, with the gradient $V/I$ rising as $I$ rises. For each: (a) state whether it obeys Ohm's law and (b) explain what happens to its resistance.
Step-by-step solution
1. Step 1
  Resistance at any point = $V/I$ = gradient from the ORIGIN to the point on a $V$ vs $I$ plot.
2. Step 2
  Component A: straight line through origin → constant gradient → constant resistance → OBEYS Ohm's law.
3. Step 3
  Component B (filament): gradient rises with $I$ → resistance INCREASES. Cause: as current rises, the filament heats up; metal ions vibrate more, electrons scatter more, so resistance rises. Does NOT obey Ohm's law.
Answer
A: ohmic (constant $R$ ). B: non-ohmic — resistance rises with current because the filament heats up.
Examiner tip
The examiner report flags candidates often using 'gradient of the curve' (i.e. $\mathrm{d}V/\mathrm{d}I$ ) for resistance — for a non-ohmic component the correct definition is still $R = V/I$ at a particular point.
11Charging by friction — which way do electrons move?
Core• electrostatics, charge, friction
Question
A polythene rod is rubbed with a dry cloth. After rubbing, the rod is found to be negatively charged. (a) State which way electrons have moved. (b) State the charge on the cloth. (c) Explain why the rod and cloth carry equal and opposite amounts of charge.
Step-by-step solution
1. Step 1
  Only negative charge (electrons) can move; the positive nuclei stay fixed. The rod gained negative charge, so electrons were transferred FROM the cloth TO the rod.
2. Step 2
  The cloth has lost electrons, so it is left with a net positive charge.
3. Step 3
  Every electron gained by the rod is an electron lost by the cloth — charge is transferred, not created. So the rod's negative charge equals the cloth's positive charge in size.
Answer
(a) Electrons move from the cloth to the rod. (b) The cloth is positively charged. (c) The rod's negative charge equals the cloth's positive charge because the same electrons moved off one and onto the other.
Examiner tip
Mark schemes insist that friction transfers negative charge (electrons) only — never write that 'positive charge moved to the cloth'.
12Predicting attraction and repulsion
Core• electrostatics, charge
Question
A positively charged rod is brought near a small suspended ball. The ball swings TOWARDS the rod. State two possible charge states of the ball, and explain how a further test could decide between them.
Step-by-step solution
1. Step 1
  Like charges repel; unlike charges attract. Attraction towards a positive rod means the ball is either negatively charged (unlike charges attract) OR uncharged (the rod induces opposite charge on the near side).
2. Step 2
  Bring a known NEGATIVE rod near the ball. If the ball is repelled, it must be negatively charged. If it is attracted again, the ball is uncharged (an uncharged object is attracted to any charged rod).
Answer
The ball is either negatively charged or uncharged. A negatively charged rod will repel a negative ball but attract an uncharged one — only repulsion proves the ball carries charge.
Examiner tip
Repulsion is the only definite test for the sign of a charge — attraction can be caused by induction on a neutral body.
13Conductors, insulators and discharging
Core• electrostatics, conductor, insulator
Question
A metal sphere on an insulating stand is charged positively. (a) Explain, using the electron model, why a metal conducts charge but the stand does not. (b) Describe what happens to the charge if you touch the sphere with your hand.
Step-by-step solution
1. Step 1
  A metal has free (delocalised) electrons that can move easily through the material, so charge flows — it is a conductor. An insulator (e.g. plastic) has no free electrons; its electrons are bound to atoms, so charge cannot flow.
2. Step 2
  Touching the sphere connects it through your body to earth — a conducting path. Electrons flow from earth onto the positive sphere to neutralise it.
3. Step 3
  The sphere is discharged (earthed): it becomes neutral.
Answer
(a) Metals conduct because they have free electrons that can move; insulators have no free electrons. (b) Touching it earths the sphere — electrons flow from earth and discharge it to neutral.
14Electric field patterns and field direction
Extended• electric field, field lines
Question
(a) Define an electric field. (b) State the rule for the direction of an electric field at a point. (c) Describe the field pattern (i) around an isolated positive point charge and (ii) between two oppositely charged parallel plates.
Step-by-step solution
1. Step 1
  An electric field is a region in which an electric charge experiences a force.
2. Step 2
  The direction of the field at a point is the direction of the force on a POSITIVE charge placed at that point. Field lines therefore point away from positive charge and towards negative charge.
3. Step 3
  (i) Around a positive point charge the field lines are straight, radial lines pointing directly OUTWARDS, evenly spread in all directions (the pattern around a charged conducting sphere is the same shape outside the sphere).
4. Step 4
  (ii) Between two oppositely charged parallel plates the field lines are parallel, equally spaced straight lines running from the positive plate to the negative plate — a uniform field (ignoring the edges).
Answer
(a) A region where a charge feels a force. (b) The direction of the force on a positive charge. (c)(i) Radial lines pointing outwards from a positive point charge / charged sphere; (ii) parallel, evenly spaced lines from + plate to − plate — a uniform field.
Examiner tip
For the parallel-plate field, the syllabus states end effects (the curved lines at the plate edges) will NOT be examined — draw the central region as straight, evenly spaced lines.

Model Answers — Electrical Quantities

High-scoring sample answers for electrical quantities on the Cambridge IGCSE 0625 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2/4 short-answer style1 mark
Define electric current and state its SI unit.
Model answer
Electric current is the rate of flow of electric charge. Its SI unit is the ampere (A), where $1\,\text{A} = 1\,\text{C/s}$ .
Why this scores
One mark for 'rate of flow of charge' with the unit ampere. Conventional current flows from + to −, opposite to the electron flow.
Question 2
Paper 2/4 style2 marks
A current of $2.0\,\text{A}$ flows in a circuit for $5.0$ minutes. Calculate the charge that passes.
Model answer
Convert the time to seconds: $t = 5.0 \times 60 = 300\,\text{s}$ . Then

$Q = It = 2.0 \times 300 = \mathbf{600\,\text{C}}.$
Why this scores
One mark for converting minutes to seconds and using $Q = It$ , one for $600\,\text{C}$ . Forgetting the time conversion is the standard error.
Question 3
Paper 4 structured style3 marks
A $12\,\text{V}$ supply drives a current of $2.0\,\text{A}$ through a heating element. (a) Calculate the resistance of the element. (b) Calculate the power it dissipates.
Model answer
(a) $R = \dfrac{V}{I} = \dfrac{12}{2.0} = \mathbf{6.0\,\Omega}$ .

(b) $P = VI = 12 \times 2.0 = \mathbf{24\,\text{W}}$ (equivalently $P = I^{2}R = 2.0^{2} \times 6.0 = 24\,\text{W}$ ).
Why this scores
Three marks: $R = V/I = 6.0\,\Omega$ (1); use of a correct power formula (1); $24\,\text{W}$ (1). Choosing the power equation that matches the given quantities saves work.
Question 4
Paper 4 structured style4 marks
A polythene rod is rubbed with a dry cloth and becomes negatively charged. (a) Explain, in terms of electrons, how the rod becomes negatively charged. (b) State the charge on the cloth. (c) Explain why the charges on the rod and cloth are equal in size.
Model answer
(a) Rubbing transfers electrons from the cloth onto the rod. Electrons are negatively charged, so the rod gains negative charge (only electrons move — the positive nuclei stay fixed).

(b) The cloth has lost electrons, so it is left positively charged.

(c) Every electron gained by the rod is one lost by the cloth — charge is transferred, not created — so the rod's negative charge is equal in size to the cloth's positive charge.
Why this scores
Four marks: electrons transferred from cloth to rod (1); rod negative because it gains electrons (1); cloth positive (1); equal charges because the same electrons moved across / charge is conserved (1). 'Positive charge moved to the cloth' is wrong — only electrons move.
Question 5
Paper 4 explanation style5 marks
A fixed resistor and a filament lamp are each tested. The resistor gives a straight line through the origin on a current–voltage graph; the lamp gives a curve that bends over (the current rises less steeply at higher voltages). (a) State which component obeys Ohm's law. (b) Explain what happens to the resistance of the filament lamp as the voltage increases, and why.
Model answer
(a) The fixed resistor obeys Ohm's law — the straight line through the origin shows that current is proportional to voltage (constant resistance).

(b) For the filament lamp the resistance increases as the voltage increases. As more current flows, the filament gets hotter, so the metal ions vibrate more vigorously; the electrons collide with them more often and find it harder to move through the metal, so the resistance rises. This is why the current rises less steeply than the voltage (it does not obey Ohm's law).
Why this scores
Five marks: resistor obeys Ohm's law because $I \propto V$ (1); lamp resistance increases (1); filament heats up (1); ions vibrate more (1); electrons collide more / harder to pass so resistance rises (1). Resistance at any point is still $R = V/I$ , not the slope of the curve.
Question 6
Paper 4 multi-part structured style6 marks
(a) Define potential difference. (b) A $2.0\,\text{kW}$ heater is connected to the $230\,\text{V}$ mains. Calculate the current it draws and the resistance of its element. (c) Calculate the electrical energy it transfers in $5.0$ minutes.
Model answer
(a) The potential difference between two points is the energy transferred (work done) per unit charge moving between them; unit volt ( $1\,\text{V} = 1\,\text{J/C}$ ).

(b) Current: $I = \dfrac{P}{V} = \dfrac{2000}{230} \approx \mathbf{8.7\,\text{A}}$ . Resistance: $R = \dfrac{V^{2}}{P} = \dfrac{230^{2}}{2000} \approx \mathbf{26\,\Omega}$ (or $R = V/I = 230/8.7 \approx 26\,\Omega$ ).

(c) $t = 5.0 \times 60 = 300\,\text{s}$ , so $E = Pt = 2000 \times 300 = \mathbf{6.0 \times 10^{5}\,\text{J}}$ ( $= 600\,\text{kJ}$ ).
Why this scores
Six marks: definition of pd as energy per unit charge (1); $I = P/V \approx 8.7\,\text{A}$ (1); $R \approx 26\,\Omega$ (1); time converted to seconds (1); $E = Pt$ (1); $6.0 \times 10^{5}\,\text{J}$ (1).

Key Formulae — Electrical Quantities

The formulae you need to memorise for electrical quantities on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Charge
$Q = It$
$Q$
charge in coulombs (C)
$I$
current in amps (A)
$t$
time in seconds (s)
When to use
Steady current.
Ohm's law
$V = IR$
$V$
potential difference (V)
$R$
resistance (Ω)
When to use
Component obeying Ohm's law (e.g. fixed resistor at constant T).
Electrical power
$P = VI = I^{2}R = \dfrac{V^{2}}{R}$
When to use
Choose the form that matches what is given.
Electrical energy
$E = Pt = VIt$
When to use
Energy transferred in time $t$ at constant power.
Resistance of a wire
$R \propto \dfrac{l}{A}$
$l$
length of wire
$A$
cross-sectional area
When to use
Comparing wires of the same material.

Key Definitions and Keywords — Electrical Quantities

Definitions to memorise and the exact keywords mark schemes credit for electrical quantities answers — sharpened from recent examiner reports for the 2026 0625 sitting.

Current
Examiner keyword
Rate of flow of charge. SI unit ampere (A) = C/s. Conventional current is from + to −.
Potential difference / voltage
Examiner keyword
Energy transferred per unit charge between two points. Measured in volts (V) = J/C.
Resistance
Examiner keyword
Ratio of pd to current. Unit ohm ( $\Omega$ ).
Electromotive force (e.m.f.)
Examiner keyword
Energy transferred per unit charge by a source (battery, generator). Unit volt (V).
Coulomb (C)
Examiner keyword
$1\,\text{C}$ = charge transferred by a current of $1\,\text{A}$ in $1\,\text{s}$ .
Electric field
Examiner keyword
A region in which an electric charge experiences a force. Its direction at a point is the direction of the force on a positive charge there.
Conductor and insulator
Examiner keyword
A conductor (e.g. metal) has free electrons that allow charge to flow; an insulator (e.g. plastic) has no free electrons, so charge cannot flow.

Common Mistakes and Misconceptions — Electrical Quantities

The traps other students keep falling into on electrical quantities questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Confusing electron flow direction with conventional current
0625/42 — recurring
Why it happens
Two conventions used in different sources.
How to avoid it
Conventional current: + → −. Electron flow: − → +. Use whichever the question asks; default in 0625 is conventional.
✕Quoting voltage in amps (or current in volts)
Why it happens
Slip.
How to avoid it
V (volts) for pd; A (amps) for current; Ω (ohms) for resistance.
✕Using minutes/hours directly in $Q = It$ or $E = Pt$
Why it happens
Forgetting to convert.
How to avoid it
Always convert to SECONDS first.
✕Saying thicker wire has more resistance
Why it happens
Mixing 'big' with 'resistive'.
How to avoid it
Resistance is INVERSELY proportional to area. Thicker wire = lower resistance.
✕Saying positive charge moves during charging by friction
0625 Examiner Reports — recurring
Why it happens
Thinking both kinds of charge can move.
How to avoid it
Only electrons (negative charge) move. The object that gains electrons becomes negative; the one that loses them becomes positive.
✕Concluding a body is charged because it is attracted to a charged rod
Why it happens
Forgetting that an uncharged body is also attracted (by induction).
How to avoid it
Only REPULSION proves a body carries charge of a definite sign.

Electrical Quantities — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Electrical Quantities

Short Notes - Electrical Quantities

Detailed Study Notes

At a glance

What you’ll learn

Electrostatics — charge, friction and conductors

Electric fields and field patterns

Charge and current

Voltage and Ohm's law

Resistance — what affects it

Electrical power and energy

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Electrical Quantities

1Charge from current and time

2Ohm's law

3Power in a circuit

4Electrical energy

5Resistance of a wire

6Beyond 0625 — EMF versus terminal pd (internal resistance)

7Charge from a current-time graph

8Power dissipated in a resistor — choose the right form

9Resistors in series vs parallel — quick comparison

10Ohmic vs non-ohmic from a V-I graph

11Charging by friction — which way do electrons move?

12Predicting attraction and repulsion

13Conductors, insulators and discharging

14Electric field patterns and field direction

Charge

Ohm's law

Electrical power

Electrical energy

Resistance of a wire

Current

Potential difference / voltage

Resistance

Electromotive force (e.m.f.)

Coulomb (C)

Electric field

Conductor and insulator

✕Confusing electron flow direction with conventional current

✕Quoting voltage in amps (or current in volts)

✕Using minutes/hours directly in Q=ItQ = ItQ=It or E=PtE = PtE=Pt

✕Saying thicker wire has more resistance

✕Saying positive charge moves during charging by friction

✕Concluding a body is charged because it is attracted to a charged rod

Electrical Quantities — frequently asked questions

Electrical Quantities

Short Notes - Electrical Quantities

Detailed Study Notes

At a glance

What you’ll learn

Electrostatics — charge, friction and conductors

Electric fields and field patterns

Charge and current

Voltage and Ohm's law

Resistance — what affects it

Electrical power and energy

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Electrical Quantities

1Charge from current and time

2Ohm's law

3Power in a circuit

4Electrical energy

5Resistance of a wire

6Beyond 0625 — EMF versus terminal pd (internal resistance)

7Charge from a current-time graph

8Power dissipated in a resistor — choose the right form

9Resistors in series vs parallel — quick comparison

10Ohmic vs non-ohmic from a V-I graph

11Charging by friction — which way do electrons move?

12Predicting attraction and repulsion

13Conductors, insulators and discharging

14Electric field patterns and field direction

✕Using minutes/hours directly in $Q = It$ or $E = Pt$