Parallel Circuits (Oxford AQA IGCSE Physics)
Revision Note
Parallel Circuits
For components connected in parallel:
The current from the supply splits in the branches
The potential difference across each branch is the same
The combined resistance is less than that of either component
Current in parallel
A junction in a parallel circuit is where two or more wires meet
The current is always split at a circuit junction
The current is conserved
This means the amount of current flowing into the junction is equal to the amount of current flowing out of it
This is because the charge is conserved
Current does not always split equally – often there will be more current in some branches than in others
The current in each branch will only be identical if the resistance of the components along each branch is identical
Current behaves in this way because it is the flow of electrons:
Electrons are physical matter – they cannot be created or destroyed
This means the total number of electrons (and hence current) going around a circuit must remain the same
When the electrons reach a junction, however, some of them will go one way and the rest will go the other
Current is split
Potential difference in parallel
The potential difference across each branch is the same
Voltage is the same
Resistance in parallel
The combined resistance is less than the resistance of any of the individual components
If two resistors of equal resistance are connected in parallel, then the combined resistance will halve
Charge has more paths to travel along, so its flow is resisted less than if only one path existed
Resistance of two resistors
Worked Example
In the circuit below, ammeter A0 shows a reading of 10 A, and ammeter A1 shows a reading of 6 A.
What is the reading on ammeter A2?
Answer:
Step 1: Recall that at a junction, the current is conserved
This means that the total amount of current flowing into a junction is equal to the total amount flowing out
Step 2: Consider the first junction in the circuit where the current splits
The diagram below shows the first junction in the circuit
Step 3: Calculate the missing amount of current
Since 10 A flows into the junction (the total current from the battery), 10 A must flow out of the junction
The question says that 6 A flows through ammeter A1 so the remaining current flowing through ammeter A2 must be:
10 A − 6 A = 4 A
Therefore, 4 A flows through ammeter A2
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