What you'll learn
Series and parallel circuits form a fundamental component of the CIE IGCSE Physics electricity syllabus. Understanding how current, voltage and resistance behave differently in these two circuit configurations is essential for answering calculation questions, circuit analysis problems and practical investigation tasks. This topic typically accounts for 8-12 marks across Paper 2 and Paper 4.
Key terms and definitions
Series circuit — a circuit where components are connected end-to-end in a single loop, so the same current flows through each component.
Parallel circuit — a circuit where components are connected across common points, creating separate branches for current to flow through.
Current (I) — the rate of flow of charge, measured in amperes (A), which remains constant in a series circuit but divides in parallel branches.
Potential difference (V) — the energy transferred per unit charge between two points, measured in volts (V), which divides in series circuits but remains constant across parallel branches.
Resistance (R) — the opposition to current flow, measured in ohms (Ω), which adds directly in series but combines differently in parallel configurations.
Ammeter — a device connected in series to measure current, with very low resistance to avoid affecting the circuit.
Voltmeter — a device connected in parallel across a component to measure potential difference, with very high resistance to prevent current diversion.
Total resistance — the combined opposition to current flow in a circuit, calculated differently for series (sum) and parallel (reciprocal sum) arrangements.
Core concepts
Current behaviour in series and parallel circuits
In a series circuit, current has only one path to follow. The same current flows through every component because charge cannot accumulate at any point — what flows into a component must flow out. This principle means:
- Current is identical at all points: I₁ = I₂ = I₃ = I_total
- An ammeter placed anywhere in the loop reads the same value
- If one component fails (bulb breaks), the entire circuit stops functioning
- Adding more components increases total resistance, decreasing current throughout
In a parallel circuit, current splits at junctions where the circuit divides into branches. The total current leaving the power supply equals the sum of currents in all branches:
- Current divides: I_total = I₁ + I₂ + I₃
- Each branch operates independently — if one component fails, others continue working
- Current takes all available paths, with more current flowing through lower-resistance branches
- Adding parallel branches provides more pathways, decreasing total resistance and increasing total current
CIE IGCSE exam papers frequently test this concept by asking students to compare ammeter readings at different positions or explain why adding bulbs in parallel increases battery current while adding them in series decreases it.
Voltage distribution in series and parallel circuits
Potential difference in series circuits divides between components. The sum of voltage drops across individual components equals the supply voltage:
- V_total = V₁ + V₂ + V₃
- Components with higher resistance receive a larger share of the voltage
- Energy is transferred progressively as charge moves through each component
- Voltmeters across different components show different readings
This voltage division explains why bulbs in series appear dimmer than when connected individually — each receives only a fraction of the battery voltage.
Potential difference in parallel circuits remains constant across all branches:
- V₁ = V₂ = V₃ = V_total
- Every component receives the full supply voltage
- Voltmeters connected across any branch read the same value
- Each component operates at full brightness/power
Exam questions often present circuit diagrams with voltmeters at various positions, requiring students to state readings or calculate missing values using these rules.
Resistance calculations and combinations
Total resistance in series is found by simple addition:
R_total = R₁ + R₂ + R₃
This makes intuitive sense — charge must pass through each resistor in turn, encountering cumulative opposition. A 2Ω resistor followed by a 3Ω resistor presents the same opposition as a single 5Ω resistor.
Total resistance in parallel follows the reciprocal rule:
1/R_total = 1/R₁ + 1/R₂ + 1/R₃
For two resistors only, this simplifies to:
R_total = (R₁ × R₂)/(R₁ + R₂)
The total resistance of parallel branches is always lower than the smallest individual resistor. This occurs because parallel paths provide multiple routes for current, reducing overall opposition.
Special cases that appear in CIE papers:
- Two identical resistors in parallel: R_total = R/2
- Three identical resistors in parallel: R_total = R/3
- One resistor much smaller than others: R_total ≈ smallest resistance
Practical applications and circuit analysis
Christmas tree lights traditionally used series connections — when one bulb failed, the entire string went dark. Modern sets use parallel wiring so individual bulbs can fail without affecting others. This real-world example frequently appears in exam questions asking students to explain circuit behaviour or suggest improvements.
Household electrical circuits use parallel wiring because:
- Each appliance receives full mains voltage (230V in most countries)
- Appliances operate independently
- Switching one device off doesn't affect others
- Total current capacity can be controlled by proper fuse/circuit breaker selection
Battery life considerations differ between configurations. In series, identical bulbs draw less current (due to higher total resistance), potentially extending battery life but reducing brightness. In parallel, higher current drains batteries faster but maintains full brightness.
Circuit analysis for CIE IGCSE requires systematic application of rules:
- Identify whether components are in series or parallel
- Apply appropriate current rules (same or dividing)
- Apply appropriate voltage rules (dividing or same)
- Calculate total resistance using the correct formula
- Use V = IR to find unknown quantities
Component behaviour in different configurations
Identical lamps demonstrate configuration effects clearly:
Series connection:
- Each lamp receives reduced voltage
- Each carries the same reduced current
- Lamps appear dim
- Total power consumption is lower
Parallel connection:
- Each lamp receives full battery voltage
- Each carries full current
- Lamps appear bright
- Total power consumption is higher
- Battery drains more quickly
Ammeters and voltmeters must be connected correctly:
- Ammeters in series with components (to measure current through)
- Voltmeters in parallel across components (to measure voltage across)
- Incorrect connections can damage meters or give meaningless readings
- Ideal ammeters have zero resistance; ideal voltmeters have infinite resistance
Combined series-parallel circuits
More complex circuits contain both series and parallel sections. Analysis requires identifying each section separately:
- Locate parallel branches and calculate their combined resistance
- Treat this combined resistance as a single component
- Add all series resistances (including the combined parallel value)
- Calculate total current from total voltage and total resistance
- Work backwards to find current and voltage in each branch
CIE IGCSE questions typically limit complexity to one parallel section within an otherwise series circuit, or vice versa.
Worked examples
Example 1: Series circuit calculation
A circuit contains a 6V battery and three resistors of 2Ω, 3Ω and 5Ω connected in series.
(a) Calculate the total resistance. [1 mark]
R_total = R₁ + R₂ + R₃ = 2 + 3 + 5 = 10Ω
(b) Calculate the current flowing through the circuit. [2 marks]
Using V = IR, rearranged to I = V/R I = 6/10 = 0.6A
(c) Calculate the potential difference across the 5Ω resistor. [2 marks]
V = IR = 0.6 × 5 = 3V
(Alternative: Calculate voltages across 2Ω and 3Ω, then subtract from 6V)
Example 2: Parallel circuit calculation
Two resistors, 4Ω and 6Ω, are connected in parallel across a 12V battery.
(a) Calculate the total resistance of the circuit. [2 marks]
Using 1/R_total = 1/R₁ + 1/R₂ 1/R_total = 1/4 + 1/6 = 3/12 + 2/12 = 5/12 R_total = 12/5 = 2.4Ω
Alternative method: R_total = (R₁ × R₂)/(R₁ + R₂) = (4 × 6)/(4 + 6) = 24/10 = 2.4Ω
(b) Calculate the total current from the battery. [2 marks]
I_total = V/R_total = 12/2.4 = 5A
(c) Calculate the current through the 4Ω resistor. [2 marks]
V across the 4Ω resistor = 12V (same as battery in parallel) I = V/R = 12/4 = 3A
Example 3: Comparing configurations
Three identical 3V, 0.5A lamps are available with a 9V battery.
(a) The lamps are connected in series. Calculate:
(i) The total resistance when each lamp is operating at normal brightness. [2 marks]
R of one lamp = V/I = 3/0.5 = 6Ω R_total = 6 + 6 + 6 = 18Ω
(ii) The current in the circuit. [2 marks]
I = V/R = 9/18 = 0.5A
(b) Explain whether the lamps will achieve normal brightness in this series circuit. [2 marks]
Yes, they will achieve normal brightness [1]. Each lamp receives 3V (9V ÷ 3 lamps) and 0.5A flows through each, which are the lamp's rated values [1].
(c) State what would happen to the other lamps if one lamp failed in this series circuit. [1 mark]
The other lamps would go out / the circuit would be broken [1].
Common mistakes and how to avoid them
Mistake: Assuming current is "used up" in series circuits. Correction: Current is the same at all points in a series circuit. Charge cannot accumulate or disappear — the same number of electrons per second pass through each component. Energy is transferred, but current remains constant.
Mistake: Adding resistances directly in parallel circuits using R_total = R₁ + R₂. Correction: Parallel resistances combine using the reciprocal formula: 1/R_total = 1/R₁ + 1/R₂. Total resistance in parallel is always less than the smallest individual resistor because multiple paths reduce overall opposition.
Mistake: Stating that voltage is the same everywhere in series, or that it divides in parallel. Correction: The opposite is true. Voltage divides between components in series (V_total = V₁ + V₂), but remains constant across all branches in parallel (V₁ = V₂ = V_total).
Mistake: Drawing ammeters in parallel or voltmeters in series when sketching circuits. Correction: Ammeters must be in series with the component (current flows through them), while voltmeters must be in parallel across the component (measuring voltage difference between two points).
Mistake: Forgetting that identical resistors in parallel give R_total = R/n. Correction: When n identical resistors R are connected in parallel, total resistance is simply R divided by the number of resistors. For two 6Ω resistors in parallel: R_total = 6/2 = 3Ω. This shortcut saves time in exams.
Mistake: Not recognising combined series-parallel sections in complex circuits. Correction: Systematically identify which components share the same current (series) and which share the same voltage (parallel). Calculate parallel sections first, then treat the result as a single series component.
Exam technique for Series and parallel circuits
Command word awareness: "Calculate" requires numerical answers with working and units (typically 2-3 marks). "Explain" needs a statement plus reasoning (2 marks). "State" or "Give" need only the answer (1 mark). "Compare" requires discussion of both similarities and differences.
Circuit diagram questions: When asked to add ammeters or voltmeters, use the correct symbol (A in a circle, V in a circle) and position carefully. Ammeters break the circuit and are inserted in series; voltmeters connect across components without breaking the circuit. Mark schemes penalise incorrect placement even if the symbol is correct.
Showing calculation steps: CIE mark schemes award method marks even when the final answer is wrong. Always show: formula used, substitution of values with units, and final answer with units. For resistance calculations, explicitly state whether using series (sum) or parallel (reciprocal) method.
Comparing brightness of bulbs: Questions often show identical bulbs in different positions. Brightness depends on power (P = VI or P = I²R). The bulb with greater current or greater voltage has higher power and appears brighter. Structure answers by: calculating or comparing currents, then stating which is brighter and why.
Quick revision summary
Series circuits: current identical throughout (I₁ = I₂), voltage divides (V_total = V₁ + V₂), resistance adds (R_total = R₁ + R₂). One component failure breaks entire circuit. Parallel circuits: current divides (I_total = I₁ + I₂), voltage same across branches (V₁ = V₂ = V_total), resistance uses reciprocal formula (1/R_total = 1/R₁ + 1/R₂). Branches operate independently. Ammeters connect in series; voltmeters connect in parallel. Apply V = IR systematically to solve problems.