What you'll learn
Electric current and circuit components form a core practical and theoretical topic in CIE IGCSE Physics, accounting for significant marks in both Paper 2 (Extended) and Paper 4 (Practical). This guide covers charge flow, current definitions, potential difference, resistance, component behaviour, circuit analysis, and the mathematical relationships tested consistently across exam series.
Key terms and definitions
Electric current — the rate of flow of charge, measured in amperes (A), where 1 ampere equals 1 coulomb of charge passing a point per second.
Charge — a property of matter measured in coulombs (C); electrons carry negative charge of approximately 1.6 × 10⁻¹⁹ C each.
Potential difference (voltage) — the energy transferred per unit charge between two points in a circuit, measured in volts (V), where 1 volt equals 1 joule per coulomb.
Resistance — the opposition to current flow in a component, measured in ohms (Ω), calculated as voltage divided by current.
Conventional current — the model showing current flowing from positive to negative terminals, opposite to electron flow direction.
Electromotive force (e.m.f.) — the energy supplied by a cell or power supply per unit charge, measured in volts.
Series circuit — a circuit arrangement where components connect end-to-end, forming a single path for current.
Parallel circuit — a circuit arrangement where components connect across common points, providing multiple paths for current.
Core concepts
Electric charge and current
Electric current flows when charge moves through a conductor. In metals, free electrons act as charge carriers, moving from the negative terminal toward the positive terminal. However, conventional current direction is defined as positive to negative, established before electron discovery.
The fundamental equation relating charge, current, and time:
Q = I × t
Where:
- Q = charge in coulombs (C)
- I = current in amperes (A)
- t = time in seconds (s)
CIE examiners frequently test this relationship through calculations requiring rearrangement. For example, a current of 2.5 A flowing for 4 minutes transfers: Q = 2.5 × (4 × 60) = 600 C
Current measurement requires an ammeter connected in series with the component. Ammeters have very low resistance to minimise circuit disruption. Digital multimeters set to ammeter mode serve the same function in practical assessments.
Potential difference and energy transfer
Potential difference (p.d.) quantifies energy transfer in circuits. When charge moves through a component, electrical energy converts to other forms. A 6 V potential difference across a lamp means each coulomb of charge transfers 6 joules of energy, converting to light and heat.
The energy transfer equation:
E = Q × V
Where:
- E = energy in joules (J)
- Q = charge in coulombs (C)
- V = potential difference in volts (V)
Combining with Q = I × t gives the power equation:
P = I × V
Where P = power in watts (W).
Voltmeters measure potential difference and must connect in parallel across components. High resistance prevents significant current through the voltmeter itself. CIE practical papers regularly require correct ammeter and voltmeter placement in circuit diagrams.
Resistance and Ohm's law
Ohm's law states that for a conductor at constant temperature, current is directly proportional to potential difference:
V = I × R
Where R = resistance in ohms (Ω).
This relationship holds for ohmic conductors — components showing a straight-line graph through the origin when voltage is plotted against current. Fixed resistors and metal wires at constant temperature exhibit ohmic behaviour.
Factors affecting wire resistance:
- Length: resistance increases proportionally with length (R ∝ l)
- Cross-sectional area: resistance decreases as area increases (R ∝ 1/A)
- Material: different materials have different resistivities
- Temperature: higher temperatures increase resistance in metals
Non-ohmic components
Several components show non-linear voltage-current relationships:
Filament lamps: As current increases, temperature rises significantly. Higher temperature means greater resistance, producing a curved I-V characteristic. The graph shows current increasing less steeply at higher voltages.
Diodes: Allow current flow in one direction only (forward bias) while blocking reverse current. The I-V graph shows near-zero current for negative voltages, then sharp increase above approximately 0.6 V (for silicon diodes). CIE questions often require sketching or interpreting diode characteristics.
Thermistors (NTC type): Resistance decreases as temperature increases. These find applications in temperature sensors and automatic circuits. Exam questions link thermistor behaviour to practical contexts like fire alarms or electronic thermometers.
Light-dependent resistors (LDRs): Resistance decreases with increasing light intensity. Used in automatic lighting circuits, LDRs typically show resistance falling from megohms in darkness to hundreds of ohms in bright light.
Series and parallel circuits
Series circuits exhibit three key properties:
- Current is identical at all points: I₁ = I₂ = I₃
- Total voltage equals sum of component voltages: V_total = V₁ + V₂ + V₃
- Total resistance equals sum of resistances: R_total = R₁ + R₂ + R₃
The single current path means breaking the circuit anywhere stops all current flow. If one bulb fails, all bulbs go out.
Parallel circuits follow different rules:
- Voltage across each branch equals supply voltage: V₁ = V₂ = V₃ = V_supply
- Total current equals sum of branch currents: I_total = I₁ + I₂ + I₃
- Combined resistance: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃
Multiple current paths mean one component failing doesn't affect others. Household lighting uses parallel circuits so individual lights can operate independently.
For two resistors in parallel, a useful formula: R_total = (R₁ × R₂)/(R₁ + R₂)
CIE examiners expect clear circuit analysis distinguishing series from parallel arrangements, particularly in Paper 2 calculations requiring multi-step solutions.
Circuit symbols and diagrams
Standard symbols appear across all CIE papers:
- Cell (single): short and long parallel lines
- Battery (multiple cells): several cell symbols in series
- Switch (open/closed): gap in line or connected line
- Fixed resistor: rectangle
- Variable resistor: rectangle with arrow
- Lamp: circle with cross
- Voltmeter: circle with V
- Ammeter: circle with A
- Diode: triangle with line (arrow shows conventional current direction)
- Thermistor: rectangle with temperature symbol
- LDR: rectangle with light arrows
- Fuse: rectangle with line through centre
Circuit diagrams must show neat, straight lines meeting at junctions, with components evenly spaced. Wires are assumed to have negligible resistance unless stated.
Worked examples
Example 1: Charge and current calculation
Question: A current of 3.5 A flows through a motor for 5 minutes. Calculate: (a) the charge transferred [2] (b) the energy transferred if the motor operates at 12 V. [2]
Solution: (a) Q = I × t Convert time: 5 minutes = 5 × 60 = 300 s Q = 3.5 × 300 = 1050 C ✓✓
(b) E = Q × V E = 1050 × 12 = 12 600 J (or 12.6 kJ) ✓✓
Mark scheme notes: First mark in (a) for correct conversion to seconds, second for answer with unit. In (b), first mark for correct formula or substitution, second for answer.
Example 2: Series and parallel resistance
Question: Calculate the total resistance when: (a) a 6 Ω and 12 Ω resistor are connected in series [2] (b) the same resistors are connected in parallel. [3]
Solution: (a) R_total = R₁ + R₂ R_total = 6 + 12 = 18 Ω ✓✓
(b) 1/R_total = 1/R₁ + 1/R₂ 1/R_total = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 ✓ R_total = 12/3 = 4 Ω ✓✓
Mark scheme notes: Part (a) awards marks for formula and answer. Part (b) gives one mark for correct formula/method, one for correct fractions manipulation, one for final answer.
Example 3: Current-voltage characteristics
Question: A student investigates a filament lamp, measuring current at different voltages. At 2 V, current is 0.5 A. At 6 V, current is 1.2 A. (a) Calculate the resistance at each voltage. [4] (b) Explain why the resistance changes. [2]
Solution: (a) At 2 V: R = V/I = 2/0.5 = 4 Ω ✓✓ At 6 V: R = V/I = 6/1.2 = 5 Ω ✓✓
(b) As current increases, the filament temperature rises ✓. Higher temperature causes greater resistance (in metals) ✓.
Mark scheme notes: Each resistance calculation earns two marks (method and answer). Explanation requires linking current/voltage to temperature AND temperature to resistance.
Common mistakes and how to avoid them
• Confusing current and voltage: Students often describe voltage "flowing" or current "across" components. Correction: Current flows through components; voltage (potential difference) exists across them. Use precise language matching the definitions.
• Incorrect ammeter/voltmeter placement: Placing voltmeters in series or ammeters in parallel. Correction: Ammeters measure current and go in series (current must pass through them); voltmeters measure potential difference and connect in parallel (across the component).
• Forgetting time conversion: Using minutes or hours directly in Q = I × t without converting to seconds. Correction: Always convert time to seconds before calculation. Show this conversion step explicitly for method marks.
• Misapplying series/parallel rules: Adding voltages in parallel or currents in series. Correction: In series, current stays constant but voltages add. In parallel, voltage stays constant but currents add. Learn the three rules for each circuit type separately.
• Stating resistance is "used up": Describing resistance as consumed in circuits. Correction: Resistance is a property of components that opposes current. Energy is transferred, charge flows, but resistance remains constant (unless temperature changes).
• Reversing thermistor behaviour: Claiming thermistor resistance increases with temperature. Correction: NTC (negative temperature coefficient) thermistors, the type specified in CIE syllabuses, decrease resistance with increasing temperature. Always state "NTC thermistor" if required.
Exam technique for Electric current and circuit components
• Command word "calculate" requires formula, substitution with units, and final answer with unit. Show working even if using a calculator. Marks are awarded for method, so incorrect arithmetic may still earn partial credit if the approach is sound.
• Circuit analysis questions often require multiple steps: identify series/parallel sections, calculate combined resistances, find total resistance, then use Ohm's law for currents or voltages. Work systematically through the circuit, labelling values clearly.
• "Explain" questions need cause-and-effect statements. For example, explaining lamp brightness requires linking resistance to current (using V = I × R), then current to power (P = I × V or P = I²R), then power to brightness. Each link earns marks.
• Practical questions (Paper 4 or Paper 6) test circuit construction skills. Draw circuits with ruler-straight lines, ensure ammeters are in series with components being measured, and place voltmeters directly across the component terminals. Incorrect meter placement loses method marks even if readings are recorded correctly.
Quick revision summary
Electric current (I = Q/t) measures charge flow in coulombs per second. Potential difference (V) quantifies energy transfer per coulomb. Resistance (R = V/I) opposes current; it increases with length and temperature, decreases with cross-sectional area. Series circuits: same current throughout, voltages add, resistances add. Parallel circuits: same voltage across branches, currents add, reciprocal resistances add. Ohmic conductors obey V ∝ I; filament lamps, diodes, thermistors and LDRs show non-linear behaviour. Always use ammeters in series and voltmeters in parallel.