The shape of the lesson
By the end of the lesson, learners can
- Series or parallel: Switch resistors between series and parallel and watch the currents split and recombine while the totals obey both Kirchhoff laws and the combination rules.
- Two laws, two combinations: Each law leads to one resistor rule.
- Splitting the current: Kirchhoff laws tell you how current and voltage divide.
- First law: currents balance at a junction.
- Series: R = R₁ + R₂.
- Parallel: 1/R = 1/R₁ + 1/R₂.
One picture every idea returns to
Kirchhoff first law states that the total current into a junction equals the total current out, a statement of conservation of charge. Kirchhoff second law states that around any closed loop the sum of the e.m.f.s equals the sum of the p.d.s, a statement of conservation of energy. From these, resistors in series add (R = R₁ + R₂ + ...), while in parallel the reciprocals add (1/R = 1/R₁ + 1/R₂ + ...).
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, Charge in, charge out.. Two simple bookkeeping rules govern every circuit. 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: Series or parallel | Switch resistors between series and parallel and watch the currents split and recombine while the totals obey both Kirchhoff laws and the combination rules. Project tab 6 of the lab (Kirchhoff’s laws) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Two laws, two combinations | Each law leads to one resistor rule. | Pairs |
| 21 to 30 min | Teach and check: Splitting the current | Kirchhoff laws tell you how current and voltage divide. | Corners |
| 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 Splitting the current, 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.
Corners
Put one answer or viewpoint in each corner of the room. Students move to the corner they agree with, justify the choice to a partner there, then hear one voice from each corner before the reveal.
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 |
|---|---|
| Add resistances directly only in series; in parallel you must add the reciprocals, and the result is always less than the smallest branch. | In a parallel pair the voltage is shared (the same across each), while in series the current is shared (the same through each). When applying the second law, track the signs of e.m.f.s and p.d.s consistently around the loop. |
Kirchhoff laws: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
State Kirchhoff first and second laws.
Resistors of 2.0 Ω, 3.0 Ω and 5.0 Ω are connected in series. Find the combined resistance.
Two 4.0 Ω resistors are connected in parallel. Find the combined resistance.
A 12 V supply is connected across a 3.0 Ω and a 6.0 Ω resistor in series. Find the current and the p.d. across the 6.0 Ω resistor.
At a junction a current of 5.0 A flows in. It divides into two branches, one carrying 2.0 A. Find the current in the other branch.
A 6.0 Ω resistor is in series with a parallel pair of 4.0 Ω resistors, across a 16 V supply. Find the total resistance and the current from the supply.
Answer key and marking notes
- First law: the total current into a junction equals the total current out (conservation of charge) ✓
- Second law: around any closed loop, the sum of the e.m.f.s equals the sum of the p.d.s (conservation of energy) ✓
- In series the resistances add ✓
- R = 2.0 + 3.0 + 5.0 = 10 Ω ✓
- 1/R = 1/4.0 + 1/4.0 = 1/2.0 ✓
- R = 2.0 Ω ✓
- Total R = 9.0 Ω; I = 12/9.0 = 1.33 A ✓
- V = IR = 1.33 × 6.0 = 8.0 V ✓
- By the first law the currents out must total 5.0 A ✓
- Other branch = 5.0 − 2.0 = 3.0 A ✓
- Parallel pair: 1/R = 1/4.0 + 1/4.0 = 1/2.0, so 2.0 Ω ✓
- Total = 6.0 + 2.0 = 8.0 Ω; I = 16/8.0 = 2.0 A ✓
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.
Kirchhoff first law expresses conservation of:
- Aenergy
- Bcharge
- Cmomentum
- Dpower
Kirchhoff second law expresses conservation of:
- Acharge
- Bmass
- Ccurrent
- Denergy
Resistors of 2.0, 4.0 and 6.0 Ω in series have a combined resistance of:
- A12 Ω
- B1.1 Ω
- C48 Ω
- D4.0 Ω
A 6.0 Ω and a 3.0 Ω resistor in parallel give:
- A9.0 Ω
- B4.5 Ω
- C2.0 Ω
- D18 Ω
Currents of 4.0 A and 1.0 A flow into a junction; one current flows out. It is:
- A3.0 A
- B5.0 A
- C4.0 A
- D2.5 A
In a series circuit, the quantity that is the same through every component is the:
- Apotential difference
- Bpower
- Cresistance
- Dcurrent
In a parallel circuit, the quantity that is the same across every branch is the:
- Apotential difference
- Bcurrent
- Cpower
- Dcharge
The combined resistance of resistors in parallel is always:
- Agreater than the largest
- Bequal to the sum
- Cless than the smallest individual resistance
- Dequal to the average
Three 3.0 Ω resistors in parallel give:
- A9.0 Ω
- B1.0 Ω
- C3.0 Ω
- D6.0 Ω
A 12 V supply is across a 4.0 Ω and an 8.0 Ω resistor in series. The p.d. across the 4.0 Ω resistor is:
- A8.0 V
- B12 V
- C6.0 V
- D4.0 V
Two 10 Ω resistors in parallel give:
- A5.0 Ω
- B20 Ω
- C10 Ω
- D0.2 Ω
Resistances add directly when the resistors are connected in:
- Aparallel
- Ba loop
- Cseries
- Da junction
Applying Kirchhoff second law around a loop, the sum of the e.m.f.s equals the sum of the:
- Acurrents
- Bp.d.s (IR terms)
- Ccharges
- Dresistances
A 4.0 Ω and a 12 Ω resistor in parallel give:
- A16 Ω
- B8.0 Ω
- C6.0 Ω
- D3.0 Ω
Quiz key
Current in = current out: charge is conserved.
ΣE = ΣIR around a loop: energy is conserved.
Series: 2.0 + 4.0 + 6.0 = 12 Ω.
1/R = 1/6.0 + 1/3.0 = 1/2.0, so R = 2.0 Ω.
Current out = 4.0 + 1.0 = 5.0 A.
Series components carry the same current.
Parallel branches share the same p.d.
Adding parallel paths lowers the resistance below the smallest branch.
1/R = 3 × (1/3.0) = 1, so R = 1.0 Ω.
I = 12/12 = 1.0 A; V = IR = 1.0 × 4.0 = 4.0 V.
1/R = 1/10 + 1/10 = 1/5.0, so R = 5.0 Ω.
In series, R = R₁ + R₂ + ...
ΣE = ΣIR around any closed loop.
1/R = 1/4.0 + 1/12 = 4/12, so R = 3.0 Ω.
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.