The shape of the lesson
By the end of the lesson, learners can
- Load the cell: Draw more current from the cell and watch the terminal voltage fall below the e.m.f. by the lost volts Ir.
- E.M.F. versus terminal p.d: Energy is conserved around the whole loop, inside the cell included.
- Why batteries get warm: Internal resistance is where a cell loses energy as heat.
- E.M.F.: energy per charge supplied.
- Terminal: V = E − Ir.
- Lost: Ir inside the cell.
One picture every idea returns to
The electromotive force (e.m.f.) is the energy supplied per unit charge by a source, the total energy each coulomb gains. A real source has internal resistance r, so when a current I flows, some energy is lost inside it: the terminal potential difference is V = E − Ir. The e.m.f. equals the terminal p.d. only when no current flows.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, The cost of being a battery.. A real cell is not a perfect source. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Load the cell | Draw more current from the cell and watch the terminal voltage fall below the e.m.f. by the lost volts Ir. Project tab 5 of the lab (E.m.f. and internal resistance) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: E.M.F. versus terminal p.d | Energy is conserved around the whole loop, inside the cell included. | Pairs |
| 21 to 30 min | Teach and check: Why batteries get warm | Internal resistance is where a cell loses energy as heat. | Jigsaw |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 16 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Why batteries get warm, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The e.m.f. is not simply the voltage you read across a battery in use; that is the terminal p.d., which is lower by Ir. | Use a minus sign: V = E − Ir. The internal resistance is in series with the external circuit, so the same current flows through it, and the cell is warmest when delivering a large current into a small external resistance. |
E.M.F. and internal resistance: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Define the electromotive force of a source and give its unit.
A cell of e.m.f. 6.0 V and internal resistance 0.40 Ω supplies a current of 3.0 A. Find the terminal potential difference.
A battery of e.m.f. 9.0 V and internal resistance 1.0 Ω is connected to a 5.0 Ω resistor. Find the current and the terminal potential difference.
State the condition under which the terminal potential difference of a cell equals its e.m.f.
A cell of e.m.f. 1.5 V and internal resistance 0.30 Ω is short-circuited (external resistance about zero). Find the current.
A graph of terminal p.d. against current for a cell is a straight line with intercept 6.0 V and gradient −0.50 V A⁻¹. State the e.m.f. and the internal resistance.
Answer key and marking notes
- The e.m.f. is the energy supplied per unit charge by the source ✓
- The unit is the volt (V) ✓
- V = E − Ir = 6.0 − (3.0 × 0.40) ✓
- V = 6.0 − 1.2 = 4.8 V ✓
- I = E/(R + r) = 9.0/(5.0 + 1.0) = 1.5 A ✓
- V = IR = 1.5 × 5.0 = 7.5 V ✓
- When no current is drawn from the cell ✓
- because then the lost volts Ir = 0, so V = E ✓
- With R ≈ 0, I = E/r ✓
- I = 1.5/0.30 = 5.0 A ✓
- The intercept gives the e.m.f.: E = 6.0 V ✓
- The magnitude of the gradient gives the internal resistance: r = 0.50 Ω ✓
14 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The electromotive force of a source is the:
- Aenergy supplied per unit charge
- Benergy lost per unit charge
- Ccurrent supplied
- Dcharge stored
The terminal potential difference of a cell is:
- AV = E + Ir
- BV = Ir
- CV = E − Ir
- DV = E/r
A cell of e.m.f. 12 V and internal resistance 0.50 Ω supplies 2.0 A. The terminal p.d. is:
- A13 V
- B11 V
- C1.0 V
- D6.0 V
The terminal p.d. equals the e.m.f. when:
- Athe current is maximum
- Bthe cell is short-circuited
- Cr is large
- Dno current is drawn
A current of 3.0 A flows through an internal resistance of 0.50 Ω. The lost volts are:
- A1.5 V
- B6.0 V
- C0.17 V
- D3.5 V
A cell of e.m.f. 6.0 V and internal resistance 1.0 Ω is connected to a 2.0 Ω resistor. The current is:
- A3.0 A
- B6.0 A
- C2.0 A
- D1.0 A
Energy is dissipated inside a cell at a rate of:
- AI²R
- BI²r
- CE/r
- DIr²
A cell of e.m.f. 1.5 V and internal resistance 0.50 Ω is short-circuited. The current is:
- A0.75 A
- B1.5 A
- C0.33 A
- D3.0 A
On a graph of terminal p.d. against current, the magnitude of the gradient gives the:
- Ainternal resistance
- Be.m.f.
- Cexternal resistance
- Dpower
On a graph of terminal p.d. against current, the intercept on the voltage axis gives the:
- Ainternal resistance
- Bcurrent
- Ce.m.f.
- Dexternal resistance
A cell with high internal resistance, compared with one of low internal resistance, has a terminal p.d. that:
- Arises with current
- Bfalls more sharply as current increases
- Cis unaffected by current
- Dis always zero
The terminal p.d. is less than the e.m.f. because:
- Athe e.m.f. decreases
- Bcharge is lost
- Cthe voltmeter is faulty
- Dsome energy is dissipated inside the cell
A cell has terminal p.d. 5.4 V when supplying 2.0 A through an internal resistance of 0.30 Ω. Its e.m.f. is:
- A6.0 V
- B4.8 V
- C5.7 V
- D10.8 V
The unit of electromotive force is the:
- Aampere
- Bohm
- Cvolt
- Dwatt
Quiz key
E.m.f. is energy supplied per coulomb.
V = E − Ir.
V = E − Ir = 12 − (2.0 × 0.50) = 11 V.
With I = 0, V = E.
Lost volts = Ir = 3.0 × 0.50 = 1.5 V.
I = E/(R+r) = 6.0/(2.0+1.0) = 2.0 A.
Power lost internally = I²r.
I = E/r = 1.5/0.50 = 3.0 A.
V = E − Ir, gradient = −r.
At I = 0, V = E, the intercept.
Larger r means larger Ir, so the terminal p.d. sags more under load.
Energy lost across the internal resistance lowers the terminal p.d.
E = V + Ir = 5.4 + (2.0 × 0.30) = 6.0 V.
E.m.f. is energy per charge, measured in volts.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.