You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
This lesson covers the rules for series and parallel circuits — as required by the Edexcel GCSE Physics specification (1PH0). You need to understand how current, voltage and resistance behave in each type of circuit, and be able to carry out calculations using these rules.
In a series circuit, all components are connected one after another in a single loop. There is only one path for the current to take.
graph LR
A["Battery"] --> B["Lamp 1"]
B --> C["Lamp 2"]
C --> D["Lamp 3"]
D --> A
style A fill:#2c3e50,color:#fff
style B fill:#e67e22,color:#fff
style C fill:#e67e22,color:#fff
style D fill:#e67e22,color:#fff
The current is the same at every point in a series circuit. There is only one path, so all the charge must flow through every component.
Itotal=I1=I2=I3
If the ammeter reads 0.5 A anywhere in the circuit, it reads 0.5 A everywhere in that circuit.
The total voltage from the battery is shared (divided) between the components. The voltages across each component add up to the battery voltage.
Vtotal=V1+V2+V3
A component with a higher resistance takes a larger share of the voltage.
Rtotal=R1+R2+R3
Adding more resistors in series increases the total resistance, which decreases the current.
Exam Tip: In a series circuit, if one component breaks (e.g., a lamp filament snaps), the entire circuit is broken and all components stop working. This is why series circuits are rarely used for household lighting.
In a parallel circuit, components are connected on separate branches. There are multiple paths for the current to take.
graph TD
A["Battery +"] --> B["Junction"]
B --> C["Branch 1: Lamp 1"]
B --> D["Branch 2: Lamp 2"]
B --> E["Branch 3: Lamp 3"]
C --> F["Junction"]
D --> F
E --> F
F --> G["Battery −"]
style A fill:#2c3e50,color:#fff
style B fill:#7f8c8d,color:#fff
style C fill:#e67e22,color:#fff
style D fill:#e67e22,color:#fff
style E fill:#e67e22,color:#fff
style F fill:#7f8c8d,color:#fff
style G fill:#2c3e50,color:#fff
The potential difference is the same across each branch in a parallel circuit. Every branch is connected directly across the battery.
Vtotal=V1=V2=V3
The total current from the battery splits at junctions. The current through each branch depends on the resistance of that branch.
Itotal=I1+I2+I3
A branch with lower resistance carries a larger current.
Adding components in parallel provides more paths for current, so the total resistance decreases. The total resistance is always less than the smallest individual resistance.
Higher Tier formula:
Rtotal1=R11+R21
For two equal resistors in parallel:
Rtotal=2R
Exam Tip: A common mistake is thinking total resistance in parallel is the sum of the individual resistances. It is NOT. In parallel, total resistance is always less than the smallest resistor. If you are on Higher tier, you need to use the 1/R formula.
| Feature | Series Circuit | Parallel Circuit |
|---|---|---|
| Current | Same through all components | Splits between branches; total = sum of branch currents |
| Voltage | Shared between components; total = sum of individual voltages | Same across each branch |
| Resistance | R_total = R₁ + R₂ + R₃ | 1/R_total = 1/R₁ + 1/R₂ (Higher) |
| If one component breaks | Whole circuit stops | Other branches continue to work |
| Adding components | Increases total resistance, decreases current | Decreases total resistance, increases total current |
Three resistors of 4 Ω, 6 Ω and 10 Ω are connected in series to a 12 V battery.
a) Total resistance: R_total = 4 + 6 + 10 = 20 Ω
b) Current through the circuit: I = V / R = 12 / 20 = 0.6 A
c) Voltage across the 6 Ω resistor: V = I × R = 0.6 × 6 = 3.6 V
d) Check: Do the voltages add up to 12 V? V₁ = 0.6 × 4 = 2.4 V V₂ = 0.6 × 6 = 3.6 V V₃ = 0.6 × 10 = 6.0 V Total = 2.4 + 3.6 + 6.0 = 12.0 V ✔
Two resistors, 6 Ω and 12 Ω, are connected in parallel across a 12 V battery.
a) Voltage across each resistor: V = 12 V (same across both branches in parallel)
b) Current through the 6 Ω resistor: I₁ = V / R₁ = 12 / 6 = 2 A
c) Current through the 12 Ω resistor: I₂ = V / R₂ = 12 / 12 = 1 A
d) Total current from the battery: I_total = I₁ + I₂ = 2 + 1 = 3 A
e) Total resistance (Higher): 1/R_total = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 R_total = 12/3 = 4 Ω
Note: 4 Ω is less than the smallest individual resistance (6 Ω), confirming our answer is sensible.
A circuit has a battery connected to two lamps. One lamp has 3 V across it and the other has 3 V across it. The battery provides 6 V.
Since the voltages add up to the battery voltage, the lamps are in series.
If both lamps had 6 V across them, they would be in parallel.
Exam Tip: To decide whether components are in series or parallel: if the voltages across the components add up to the supply voltage, they are in series. If each component has the same voltage as the supply, they are in parallel.
Household appliances are connected in parallel because:
Subscribe to continue reading
Get full access to this lesson and all 10 lessons in this course.