Series circuits for AQA GCSE Physics
This page covers the following topics:
1. Current in series circuits
2. Potential difference in series circuits
3. Resistance in series circuits
4. Series circuits
In a series circuit, the current at any point will be equal. As more lamps are added to a circuit, their brightness will decrease as less current will flow through them, although the current in the circuit will still be the same everywhere. If a component in a series circuit breaks, the circuit will become an open one and current will not flow.
In a series circuit, the total potential difference is equal to the sum of the potential differences across each component of the circuit. This is usually equal to the voltage of the cell or battery of the circuit.
The total resistance in the circuit will be the sum of all the individual resistances of the components.
All equations for the current, potential difference and resistance in a series circuit can be used together for calculations.
A lamp and a resistor are connected in series with a cell. Given that the current through the lamp is 5 A, what is the current flowing through the resistor?
In a series circuit, the current at any point will be equal, therefore the current through the resistor will be equal to the current through the lamp. So, the current through the resistor is 5 A.
Two resistors, both of resistance 10 Ω, are connected in series with a cell and an ammeter, which shows a value of 3 A. Calculate the potential difference across the cell.
Since the resistors are connected in series, the total resistance in the circuit will be equal to the sum of the individual resistances. Therefore, Total Resistance = 10 Ω + 10 Ω = 20 Ω. Using Ohm's Law, V = 3 Α × 20 Ω = 60 V.
Two resistors are connected in series with a cell of 14 V. Given that the total resistance of the resistors is 7 Ω, calculate the current through each resistor.
In a series circuit, the current at any point will be equal, therefore the current through the two resistors will be equal. Using V = IR, 14 V = I × 7 Ω, therefore I = 14 V/7 Ω = 2 A.
A variable resistor is connected in series with a cell and a fixed resistor of resistance 12 Ω. Given that the total resistance of the circuit is 20 Ω, calculate the resistance of the variable resistor.
The total resistance in a series circuit is the sum of all the individual resistances. So, Resistance = 20 Ω − 12 Ω = 8 Ω.
Use the diagram provided to find the current passing through resistor R if the potential difference provided across the first resistor is 4 V.
Ohm's law states that V = IR.
4 V = I × 12 Ω
I = 4 V ÷ 12 Ω = 0.333 A (3 s. f.)
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