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This lesson covers the rules for series and parallel circuits, including how current, potential difference and resistance behave in each type of circuit. This is a core part of the AQA GCSE Physics specification (4.2.1.4) and is heavily examined.
In a series circuit, all components are connected one after another in a single loop. There is only one path for the current to follow.
graph LR
A[Battery] --> B[Lamp 1]
B --> C[Lamp 2]
C --> D[Lamp 3]
D --> A
The current is the same at every point in a series circuit. This is because there is only one path for the charge to flow through — the same charge passes through every component.
I(total) = I(1) = I(2) = I(3)
If you place ammeters at different points in a series circuit, they will all read the same value.
The total potential difference from the power supply is shared between the components in a series circuit. The p.d. across each component depends on its resistance — components with higher resistance have a greater share of the total p.d.
V(total) = V(1) + V(2) + V(3)
This makes sense because the total energy supplied by the battery per coulomb must equal the total energy transferred by the components per coulomb.
The total resistance of a series circuit is the sum of the individual resistances:
R(total) = R(1) + R(2) + R(3)
Adding more resistors in series increases the total resistance, which decreases the current (for a given p.d.).
| Property | Rule in Series |
|---|---|
| Current | Same everywhere: I(total) = I(1) = I(2) = I(3) |
| Potential Difference | Shared: V(total) = V(1) + V(2) + V(3) |
| Resistance | Adds up: R(total) = R(1) + R(2) + R(3) |
Exam Tip: A key consequence of the series rules is that if one component breaks or is removed, the circuit is broken and ALL components stop working. This is why old-fashioned Christmas tree lights used to go out entirely when one bulb failed — they were wired in series.
In a parallel circuit, components are connected on separate branches. Each branch provides a separate path for the current.
graph LR
A[Battery] --> B{Junction}
B --> C[Lamp 1]
B --> D[Lamp 2]
B --> E[Lamp 3]
C --> F{Junction}
D --> F
E --> F
F --> A
The total current from the power supply is divided between the branches. The current through each branch depends on the resistance of that branch — branches with lower resistance carry more current.
I(total) = I(1) + I(2) + I(3)
At a junction, the current splits. The total current entering a junction equals the total current leaving it (this is known as Kirchhoff's first law or the conservation of charge).
The potential difference across each branch of a parallel circuit is the same and equals the p.d. of the power supply.
V(total) = V(1) = V(2) = V(3)
This is because each branch is connected directly to the power supply.
Adding more resistors in parallel decreases the total resistance of the circuit. This is because there are more paths for the current to flow through.
For two resistors in parallel:
1/R(total) = 1/R(1) + 1/R(2)
For three resistors in parallel:
1/R(total) = 1/R(1) + 1/R(2) + 1/R(3)
The total resistance of a parallel combination is always less than the smallest individual resistance.
| Property | Rule in Parallel |
|---|---|
| Current | Shared: I(total) = I(1) + I(2) + I(3) |
| Potential Difference | Same everywhere: V(total) = V(1) = V(2) = V(3) |
| Resistance | Decreases: 1/R(total) = 1/R(1) + 1/R(2) + 1/R(3) |
Exam Tip: A very common exam question gives you a parallel circuit with two resistors and asks for the total resistance. Use the formula 1/R(total) = 1/R(1) + 1/R(2), then take the reciprocal. For example, two 6-ohm resistors in parallel: 1/R = 1/6 + 1/6 = 2/6, so R = 6/2 = 3 ohms. The total is always less than the smallest individual value.
| Feature | Series | Parallel |
|---|---|---|
| Number of paths | One | Multiple |
| Current | Same through all components | Splits between branches |
| Potential difference | Shared between components | Same across all branches |
| Total resistance | Sum of all resistances | Less than smallest individual resistance |
| Effect of adding components | Total resistance increases; current decreases | Total resistance decreases; total current increases |
| If one component fails | Whole circuit stops | Other branches still work |
| Brightness of identical lamps | Dimmer (less p.d. per lamp) | Brighter (full p.d. across each lamp) |
Three resistors of 4 ohms, 6 ohms and 10 ohms are connected in series to a 12 V battery.
a) Calculate the total resistance.
R(total) = 4 + 6 + 10 = 20 ohms
b) Calculate the current in the circuit.
I = V / R = 12 / 20 = 0.6 A
c) Calculate the p.d. across the 6-ohm resistor.
V = I x R = 0.6 x 6 = 3.6 V
d) Verify that the p.d. values add up to the total.
V(4 ohm) = 0.6 x 4 = 2.4 V V(6 ohm) = 0.6 x 6 = 3.6 V V(10 ohm) = 0.6 x 10 = 6.0 V Total = 2.4 + 3.6 + 6.0 = 12.0 V (matches the battery p.d.)
Two resistors of 12 ohms and 6 ohms are connected in parallel to a 6 V battery.
a) Calculate the total resistance.
1/R(total) = 1/12 + 1/6 = 1/12 + 2/12 = 3/12
R(total) = 12/3 = 4 ohms
b) Calculate the total current from the battery.
I(total) = V / R(total) = 6 / 4 = 1.5 A
c) Calculate the current through each resistor.
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