Cells In Series (Total E.M.F.)
Key idea: Learn how to find total e.m.f. for cells in series (aiding and opposing), and solve typical O Level questions using algebraic addition.
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The core idea
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Learning objectives
- State current as rate of charge flow measured in amperes
- Distinguish conventional current from electron flow
- Apply charge equals current multiplied by time
- Define source e.m.f. as work done per unit charge around a circuit
- Calculate total e.m.f. for sources in series
- Define component potential difference as work done per unit charge
- State resistance as potential difference divided by current
- Apply resistance equals potential difference divided by current
- Apply wire-resistance proportionalities for length and cross-sectional area
- Describe the effect of temperature on metallic resistance
- Sketch and interpret required current–voltage characteristics
1. Definition
A. Cells in series
Cells are connected in series when the positive terminal of one cell is connected to the negative terminal of the next cell.
B. Total e.m.f.
In a series arrangement, the total e.m.f. is the algebraic sum of the individual e.m.f.s (add aiding cells, subtract opposing cells).
You should be able to calculate the total e.m.f. when several sources are arranged in series.
2. Key Ideas
- If several cells are connected in series so that they aid one another, add their e.m.f.s:
εₜₒₜₐₗ = ε₁ + ε₂ + …
- If one (or more) cell is reversed so it opposes, subtract its e.m.f.:
εₜₒₜₐₗ = (sum of aiding e.m.f.s) - (sum of opposing e.m.f.s)
- Total e.m.f. is measured in volts (V).
3. Detailed Explanations
A. Cells that aid vs oppose
- Aiding: the cells are connected in the same orientation (their e.m.f.s add).
- Opposing: one cell is reversed (its e.m.f. is subtracted).
In exams, you can think of it as adding the e.m.f.s algebraically (with signs).
B. Why e.m.f. adds in series
In a complete circuit, each coulomb of charge gains energy from the source(s). If the charge passes through multiple cells in series, it gains energy from each cell, so the energy gained per coulomb (total e.m.f.) increases.
4. Common Mistakes
- Adding e.m.f.s even when one cell is reversed (remember to subtract opposing cells).
- Mixing up e.m.f. (energy supplied per coulomb by sources) with p.d. (energy transferred per coulomb in components).
- Forgetting to include the unit V in the final answer.
5. Exam Tips
- If all the cells face the same way: just add the numbers.
- If a cell is reversed: treat it as a minus and subtract.
- If the question gives a diagram, mark each cell as +ε (aiding) or −ε (opposing) before adding.
- State the final answer with unit V.
6. Worked Examples
Modelled example 1
Total e.m.f. in series
Problem
Study the worked solution
Assign source signs
Method
Treat every cell as positive because all orientations aid.Reason
Charge gains energy in the same sense through each source.Working
+ 1.5 + 1.5 + 1.5 VAdd and interpret
Method
Sum the three e.m.f.s to obtain 4.5 V.Reason
Each coulomb gains 1.5 J from each cell.Working
εₜₒₜₐₗ = 4.5 V = 4.5 J C⁻¹
Guided practice 2
One cell opposing
Problem
Write a signed e.m.f. sum
Hints
Hint 1: mark orientation first
Hint 2: combine the aiding pair
View solution step by step
Add algebraically
Method
Add the aiding e.m.f.s and subtract the opposing e.m.f.Reason
The reversed cell removes energy per coulomb in the chosen circuit direction.Working
εₜₒₜₐₗ = 1.5 + 1.5-1.2 = 1.8 V
Common misconception 3
How many cells?
Learner response
Keep count distinct from voltage
View solution step by step
Set up the repeated-source equation
Method
Multiply the number of identical aiding cells by 1.5 V.Reason
Every cell adds the same e.m.f.Working
1.5n = 9.0Solve and label the count
Method
Divide by 1.5 to obtain six cells.Reason
Volt units cancel in the ratio, leaving a dimensionless count.Working
n = 9.0/1.5 = 6 cells
Examiner practice 4
Mixed sources
Examination question
Show signed method, magnitude and dominant direction
View solution step by step
Calculate the net e.m.f.
2 marksMethod
Subtract the opposing 1.5 V from 6.0 V.Reason
The sources drive charge in opposite senses.Working
εₜₒₜₐₗ = 6.0-1.5 = 4.5 VState the net direction
1 markMethod
State that the resultant orientation is that of the 6.0 V battery.Reason
The larger e.m.f. dominates the algebraic sum.Working
Net source direction: 6.0 V battery direction.
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark signed operation, magnitude and resultant direction.
Challenge 5
Two aiding, one opposing
Multi-source transfer
Assign every source a sign before calculating
Hints
Hint 1: group the aiding cells
Hint 2: apply the opposing sign
View solution step by step
Calculate the algebraic sum
Method
Add the two aiding sources and subtract the opposing source.Reason
Energy supplied per coulomb is added with orientation signs.Working
εₜₒₜₐₗ = 2.0 + 1.5-1.2 = 2.3 V
7. Mind Stretchers
Mind stretcher 1: Zero total e.m.f.Extension
Three cells have e.m.f.s of 1.5 V, 1.5 V and 3.0 V. They are connected in series, but the 3.0 V cell opposes the other two. Find the total e.m.f.
Show Answer
εₜₒₜₐₗ = 1.5 + 1.5 - 3.0 = 0.0 V
Mind stretcher 2: Reasoning checkExtension
If the total e.m.f. is 0 V, does that mean there is definitely no current in a circuit? Explain.
Show Answer
In a simple d.c. circuit, a total e.m.f. of 0 V means there is no net energy supplied per coulomb, so there is no driving voltage for a steady current. So you would expect no current.
8. Practice and next step
Practise identifying source orientation before adding signed e.m.f.s in Structured Current Electricity. Then continue to resistance, where the resulting p.d. is related to current through a component.
Continue with the next resource in this course.
Course and syllabus information
- Course
- SEC G3 Physics
- Edition
- SEC G3 Physics 2027