Follow one series path
Series current is the same; component potential differences add to the supply and need not be equal.
Continue where you stopped
The core idea
Electricity · Lesson 3 of 6 · about 25–35 min
What you need to understand
Series current is the same; component potential differences add to the supply and need not be equal.
Definitions
- series
- connected on one unbranched pathFor example: The same current passes through two series lamps.
Key idea
- LookA 6.0 V supply is connected to two series components. One has 2.0 V across it and the current is 0.30 A. Find the other potential difference and its current.
- ThinkSeries current is the same; component potential differences add to the supply and need not be equal.
- DoUse current, potential-difference and resistance evidence to explain or complete a series circuit.
Explanation
In one unbranched series path, the same current enters and leaves every component. Components transfer energy, but current is not consumed, so successive current readings in that path are equal.
The potential differences across the series components sum to the supply potential difference. The individual values are not necessarily equal: for a 6.0 V supply, 2.0 V and 4.0 V are a valid pair. Adding another series resistor increases total resistance.
Pause and say it: Series current is the same; component potential differences add to the supply and need not be equal.
Common mistake
Tempting wrong idea: Current is divided between successive components in a series circuit.
Why it fails: A series circuit has only one path, so all the charge that flows through one component must also flow through the next. None is split off along the way, so the current is the same at every point.
Use this instead: Current is the same at every point in a series circuit because there is one unbranched path.
Practical work
What to show: Use current, potential-difference and resistance evidence to explain or complete a series circuit.
Before you finish: Components transfer energy but do not consume current.
Practical link: The topic investigation is taught in “Choose the meter job before connecting it”.
6. Worked Examples
Modelled example 1
Use an unequal potential-difference pair
Problem
A 6.0 V supply is connected to two series components. One has 2.0 V across it and the current is 0.30 A. Find the other component’s potential difference and current.
Study the worked solution
Apply both series rules
Method
Subtract the known component potential difference from the supply and keep the current unchanged.
Reason
Series component potential differences sum to the supply, while one unbranched path carries the same current throughout.
Working
V₂ = 6.0 V − 2.0 V = 4.0 V; I₂ = 0.30 A
Guided practice 2
Complete a series record
Problem
Three components form one series path across an 8.0 V supply. The first two potential differences are 1.5 V and 2.0 V. An ammeter before the first component reads 0.40 A. Find the third potential difference and the current after the third component.
Apply the two series rules separately
Hints
Hint 1: potential difference
Subtract both known component values from the supply value.
Hint 2: current
There is no junction in the stated path.
View solution step by step
Use the additive potential-difference rule
Method
Subtract both known component potential differences from the supply value.
Reason
All component potential differences in the one path must sum to the supply potential difference.
Working
V₃ = 8.0 V − 1.5 V − 2.0 V = 4.5 VKeep the one-path current unchanged
Method
Carry the measured current through every successive component.
Reason
Current is not consumed by successive series components.
Working
The current after the third component is 0.40 A.
Challenge 3
Add a resistor to one path
Problem
A resistor is added in series with a lamp. Predict the change in total resistance and explain whether the steady current can have different values immediately before and after the new resistor.
Link path, current and resistance
Hints
Hint 1: inspect the path
Adding the resistor does not create a junction.
Hint 2: separate the rules
Apply the series-resistance rule and the same-current rule independently.
View solution step by step
Follow the unchanged topology
Method
The added series resistor increases total resistance.
Reason
It adds opposition on the only complete path without creating a junction.
Working
The same steady current enters and leaves the new resistor.
Guided practice
Try it with support
A 6.0 V supply is connected to two series lamps. One lamp has 2.0 V across it and the current is 0.30 A. Find the other voltage and state its current.
- Series voltages add to the supply.
- Series current is the same everywhere.
Check the guided answer
Answer: The other voltage is 6.0 − 2.0 = 4.0 V, and its current is 0.30 A.
Check: The component potential differences add to the supply; they are not necessarily equal.
Practise and continue
Practise this
Two successive ammeters in one series path read 0.42 A and 0.21 A. What should the learner conclude?
Need a hint?
- Current has no junction at which to split.
Check your answer
Answer: At least one connection or reading is wrong: steady current should be 0.42 A at both positions or 0.21 A at both, not different values.
Check: Components transfer energy but do not consume current.
Think like a scientist
Predict and explain what happens to total resistance when another resistor is added to the only series path.
Check the reasoning
Total resistance increases because the added resistor extends the same path through which current must pass.
Remember: Do not use the parallel-path rule in a circuit with no branch.
One-minute check
- Hide the page and explain follow one series path in your own words.
- Give a new example that is different from the worked example.
- Correct this common mistake: “Current is divided between successive components in a series circuit.”