The shape of the lesson
By the end of the lesson, learners can
- Turn up the power: Set the voltage across a resistor and the current follows from V = IR; watch the power dissipated track P = VI and the three equivalent forms agree at every setting.
- Three forms of the same power: Pick the form that matches the quantities you are given.
- Choosing the right form: All three power formulas are equivalent through V = IR.
- P.D.: V = W/Q.
- Power: P = VI = I²R = V²/R.
- Unit: volt = J C⁻¹.
One picture every idea returns to
The potential difference across a component is the energy transferred per unit charge, V = W / Q, measured in volts (one joule per coulomb). The electrical power, the rate of energy transfer, is P = VI, which combines with V = IR to give the equivalent forms P = I²R and P = V² / R.
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, Energy per charge.. Potential difference measures how much energy each coulomb of charge gives up as it passes through a component. 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: Turn up the power | Set the voltage across a resistor and the current follows from V = IR; watch the power dissipated track P = VI and the three equivalent forms agree at every setting. Project tab 2 of the lab (P.d. and power) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Three forms of the same power | Pick the form that matches the quantities you are given. | Pairs |
| 21 to 30 min | Teach and check: Choosing the right form | All three power formulas are equivalent through V = IR. | Anticipation / Reaction Guide |
| 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, 14 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 Choosing the right form, 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.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
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 three power formulas are equivalent, not different physics: pick the one matching your known quantities. | Watch the squares: in P = I²R doubling the current quadruples the power, and in P = V²/R doubling the voltage quadruples it. Potential difference is energy per charge (J C⁻¹), not energy per second. |
Potential difference and power: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
Define potential difference and state its unit.
A charge of 5.0 C transfers 60 J of energy through a component. Find the potential difference.
A device operates at 240 V and draws a current of 0.50 A. Find its power.
A current of 3.0 A flows through an 8.0 Ω resistor. Find the power dissipated and the energy transferred in 2.0 minutes.
A 12 V supply is connected across a 6.0 Ω resistor. Find the power dissipated using P = V²/R.
An electric heater rated 1150 W runs from a 230 V supply. Find the current it draws and its resistance.
Answer key and marking notes
- Potential difference is the energy transferred per unit charge, V = W/Q ✓
- The unit is the volt (V), equal to one joule per coulomb ✓
- V = W/Q = 60/5.0 ✓
- V = 12 V ✓
- P = VI = 240 × 0.50 ✓
- P = 120 W ✓
- P = I²R = 3.0² × 8.0 = 72 W ✓
- Energy = Pt = 72 × 120 = 8640 J ✓
- P = V²/R = 12²/6.0 ✓
- P = 144/6.0 = 24 W ✓
- I = P/V = 1150/230 = 5.0 A ✓
- R = V/I = 230/5.0 = 46 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.
Potential difference is the:
- Acharge per unit time
- Benergy transferred per unit charge
- Cforce per unit charge
- Dpower per unit current
The volt is equivalent to:
- Aone coulomb per second
- Bone watt per volt
- Cone joule per second
- Done joule per coulomb
A component has 6.0 V across it and carries 3.0 A. The power is:
- A18 W
- B2.0 W
- C9.0 W
- D0.5 W
A current of 2.0 A flows through a 5.0 Ω resistor. The power dissipated is:
- A10 W
- B2.5 W
- C20 W
- D40 W
A 10 V supply is across a 5.0 Ω resistor. The power is:
- A2.0 W
- B20 W
- C50 W
- D0.5 W
A charge of 4.0 C transfers 24 J through a lamp. The p.d. is:
- A96 V
- B0.17 V
- C20 V
- D6.0 V
Doubling the current through a fixed resistor changes the power to:
- Afour times
- Bdouble
- Chalf
- Dunchanged
To measure the p.d. across a component, a voltmeter is connected:
- Ain series with it
- Bto earth
- Cin parallel with it
- Dthrough an ammeter
A 100 W lamp runs for 60 s. The energy transferred is:
- A1.7 J
- B6000 J
- C160 J
- D100 J
A 12 V battery moves 5.0 C of charge through a circuit. The energy transferred is:
- A2.4 J
- B17 J
- C0.42 J
- D60 J
The unit of electrical power is the:
- Awatt
- Bjoule
- Cvolt
- Dohm
A 60 W lamp operates at 12 V. The current it draws is:
- A720 A
- B0.20 A
- C5.0 A
- D48 A
Doubling the voltage across a fixed resistor changes the power to:
- Adouble
- Bfour times
- Chalf
- Dunchanged
Which power formula uses only the current and the resistance?
- AP = VI
- BP = V²/R
- CP = V/I
- DP = I²R
Quiz key
V = W/Q.
1 V = 1 J C⁻¹.
P = VI = 6.0 × 3.0 = 18 W.
P = I²R = 2.0² × 5.0 = 20 W.
P = V²/R = 100/5.0 = 20 W.
V = W/Q = 24/4.0 = 6.0 V.
P = I²R, so doubling I gives 4 times the power.
A voltmeter goes in parallel so it reads the p.d. across the component.
E = Pt = 100 × 60 = 6000 J.
W = VQ = 12 × 5.0 = 60 J.
Power is measured in watts.
I = P/V = 60/12 = 5.0 A.
P = V²/R, so doubling V gives 4 times the power.
P = I²R uses current and resistance.
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.