⚡ Current: add
Current splits at a junction. Add the branch currents to get the supply current.
Energy and Voltage in Circuits » Current and Voltage Rules in Circuits
What you'll learn this session
Study time: 30 minutes
Edexcel spec: 2.17, 2.18
- Why current is conserved at a junction in a circuit
- How to use ammeter readings to find a missing current
- Why two components connected in parallel have the same voltage
- How to use both rules together in exam questions
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Unlock This CourseWhat happens to current at a junction?
A junction is a point in a circuit where three or more wires meet. Parallel circuits have junctions: one where the current splits into the branches, and one where the branches join up again.
The rule for a junction is short:
The total current into a junction equals the total current out of the junction.
We say that current is conserved at a junction. None is lost and none is gained.
The spec asks you to understand why. The reason comes from what a current is. Remember, current is the rate of flow of charge. So think about the charge:
- Charge cannot be created or destroyed. A junction is just a join between wires, so it cannot make charge and it cannot use charge up.
- Charge does not pile up at the junction or leak out of it.
- So the charge that arrives at the junction each second must equal the charge that leaves it each second.
- Charge per second is current. So the current in equals the current out.
Picture a river that splits to flow around an island. Some water goes left and some goes right. No water appears or vanishes at the split, so the two flows add up to the flow of the river before the island. When the two flows meet again, the full river carries on.
Key terms:
- Conserved: the total stays the same. Nothing is lost and nothing is created.
- Supply current: the current that flows through the battery or power supply.
- Branch current: the current in one branch of a parallel circuit.
Checking the rule with ammeters
Here is a circuit to draw. Start with a battery. From its positive terminal, draw a wire to ammeter A1, then on to a junction. At the junction the wire splits into two branches. The top branch has ammeter A2 and lamp X. The bottom branch has ammeter A3 and lamp Y. The two branches meet at a second junction, and one wire runs back to the negative terminal of the battery.
A1 measures the supply current. A2 and A3 measure the branch currents. Whatever lamps you use, the readings always fit this pattern:
supply current = sum of the branch currents
A1 = A2 + A3
At the second junction the branch currents join up again. An ammeter placed between this junction and the battery gives the same reading as A1. No current has been used up by the lamps.
Worked example 1: finding a branch current
In the circuit above, A1 reads 0.50 A and A2 reads 0.30 A. Calculate the reading on A3.
Rule: current into the junction = current out of the junction
Substitution: 0.50 = 0.30 + A3
A3 = 0.50 − 0.30
Answer: A3 = 0.20 A
Worked example 2: finding the supply current
Three resistors are connected in parallel with a power supply. The branch currents are 0.2 A, 0.4 A and 0.6 A. Calculate the supply current.
Rule: supply current = sum of the branch currents
Substitution: supply current = 0.2 + 0.4 + 0.6
Answer: supply current = 1.2 A
Notice that the branch currents in these examples are not equal. The current only splits equally when the branches are identical. If the branches are different, the branch with the lower resistance takes the larger current. The rule still works every time: the branch currents add up to the supply current.
Voltage across components in parallel
The second rule is one to know by heart:
The voltage across two components connected in parallel is the same.
Go back to lamps X and Y. Connect a voltmeter across lamp X and another across lamp Y. If the battery is 6 V, both voltmeters read 6 V. A voltmeter across the battery reads 6 V too.
You can see why from the circuit diagram. One end of lamp X and one end of lamp Y are joined by wire to the same junction, which leads to the positive terminal. Their other ends are joined to the second junction, which leads to the negative terminal. So both lamps are connected between the same two points. A voltmeter across either lamp is really measuring the voltage between those same two points, so the readings must match.
This means each branch gets the full supply voltage. The voltage is not shared out between the branches.
| In a parallel circuit | Rule | Example with a 6 V battery |
|---|---|---|
| Current | The branch currents add up to the supply current | 0.30 A + 0.20 A = 0.50 A |
| Voltage | The voltage across each branch is the same | 6 V across lamp X and 6 V across lamp Y |
🔋 Voltage: same
Voltage does not split between branches. Each branch has the same voltage across it.
Using both rules together
Exam questions often mix these two rules with voltage = current × resistance, which you learnt in Voltage, Current and Resistance.
Worked example 3: two resistors in parallel
A 20 Ω resistor and a 30 Ω resistor are connected in parallel with a 12 V battery. Calculate the current in each resistor and the supply current.
The resistors are in parallel, so the voltage across each one is 12 V.
20 Ω resistor: I = V ÷ R = 12 ÷ 20 = 0.6 A
30 Ω resistor: I = V ÷ R = 12 ÷ 30 = 0.4 A
Supply current = sum of the branch currents = 0.6 + 0.4
Answer: supply current = 1.0 A
The resistor with the lower resistance has the larger current, as expected.
Common mistakes
- Saying current is used up. Components transfer energy, but the charge keeps flowing. The current that returns to the battery equals the current that left it.
- Always halving the current. The current splits equally only if the branches are identical. Use the readings you are given.
- Sharing the voltage between branches. With a 6 V battery and two lamps in parallel, each lamp has 6 V across it, not 3 V.
- Mixing up the rules. In parallel, you add currents. You do not add the branch voltages.
- Answering "why" with only the rule. For an explain question, talk about charge: it cannot be created or destroyed, so the charge flowing in each second equals the charge flowing out.
Exam-style question
A student connects a lamp and a motor in parallel with a 6 V battery. An ammeter next to the battery reads 0.8 A. An ammeter in the lamp branch reads 0.3 A.
(a) State the voltage across the motor. [1 mark]
(b) Calculate the current in the motor. [2 marks]
(c) Explain why current is conserved at a junction in a circuit. [2 marks]
(d) Calculate the resistance of the motor. [3 marks]
Model answer
(a) 6 V (1)
(b) current in motor = supply current − current in lamp = 0.8 − 0.3 (1) = 0.5 A (1)
(c) Charge cannot be created or destroyed at a junction (1), so the charge flowing into the junction each second equals the charge flowing out each second, and current is the rate of flow of charge (1).
(d) R = V ÷ I (1) = 6 ÷ 0.5 (1) = 12 Ω (1)
Exam tip
In part (d), use the motor's own values: the 6 V across the motor and the 0.5 A through it. A common slip is to divide by the 0.8 A supply current, which is not the current in the motor.