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.

Electricity and MagnetismCambridge IGCSE Physics (0625)

Electrical Quantities

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Electrical Quantities — Cambridge IGCSE 0625 Physics Extended (2026)

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

What you’ll learn

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

4.2 — Recall and use $Q = It$ .
4.2 — Recall and use $V = IR$ .
4.2 — Recall and use $P = IV$ .
4.2 — Calculate electrical energy using $E = IVt$ .
4.2 — Describe how resistance varies with length and cross-section.

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.

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.

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).

Combined for a wire: $R = \dfrac{\rho L}{A},$ where $\rho$ is the resistivity (material property; not always required at IGCSE Extended but the relationship is).

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: $\text{kWh}$ .

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.

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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.

1Charge from current and time
Core• 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
Core• 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
Extended• 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
Extended• 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
Extended• 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
$R_{B} = 4 R_{A}$

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 $1\,\text{s}$ .

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.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Electrical Quantities — frequently asked questions

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

Electrical Quantities

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Charge from current and time

2Ohm's law

3Power in a circuit

4Electrical energy

5Resistance of a wire

Charge

Ohm's law

Electrical power

Electrical energy

Resistance of a wire

Current

Potential difference / voltage

Resistance

Electromotive force (e.m.f.)

Coulomb (C)

✕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

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Electrical Quantities — frequently asked questions

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