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Quick subject briefing
Before this lesson: Consolidation after pupils have met circuit symbols, current, voltage, resistance and V = IR. Pupils use calculators; no circuit-building is required.
Current is charge flowing each second. One series loop carries the same current throughout; its resistances add and its voltages share the supply. Parallel branches each connect across the full supply voltage, and their currents add. An open switch stops current only in paths it breaks.
Words and symbols you will need
- Current, I (A)
- Charge flowing per second, measured in amperes. Charge is conserved.
- Voltage, V (V)
- Also called potential difference: energy transferred per unit charge.
- Resistance, R (Ω)
- For these fixed resistors, R = V/I. At the same voltage, larger resistance gives less current.
- Series
- Components in one complete path.
- Parallel
- Separate complete branches between the same two supply terminals.
- Ideal supply
- Its stated voltage stays fixed when a branch opens in this lesson.
Help pupils choose a method
- Trace complete paths between supply terminals. Decide whether components share one loop or separate branches.
- Choose the rule first. For a resistor, use its own voltage and current in V = IR, I = V/R or R = V/I.
- Check the sums. After a switch opens, trace which paths remain complete before recalculating.
Use the supplied model — pupil page 2
Pupil page 2 has the series model. Add 4 Ω and 8 Ω before finding the common current. The two resistors have the same current, but their unequal resistances give unequal voltages.
Follow the revision note’s series example: an ideal 12 V supply feeds 4 Ω and 8 Ω fixed resistors in one loop. Find the total resistance, current and potential difference across each resistor.
Worked answer: Rtotal = 4 + 8 = 12 Ω. I = 12/12 = 1 A through both resistors. Their potential differences are 1 × 4 = 4 V and 1 × 8 = 8 V. The check is 4 + 8 = 12 V: equal current does not mean equal potential difference.
If pupils are still stuck
Use the relevant coaching and named retry within its replacement time. Trace one complete path together or write the correct equation with units. If the retry remains insecure, record whether paths, equation choice or arithmetic caused the problem for the usual teacher.
KEEP THIS OPEN DURING THE LESSON
Timed teacher run sheet
- Download and print one complete 7-page pupil handout per pupil, in order. Double-sided printing is fine; staple each booklet.
- Read the subject briefing on teacher page 1. Use the timed run sheet on page 2 and the numbered answers. Each check has Say / Ask / Expect coaching and a fresh retry.
- Pupils need a pen, a different-coloured pen and a calculator. All reading and questions are included; no pupil accounts, devices or apparatus are needed.
0–5 minPupil page / Q
1 / 1–3
Read aloud / check: “Start on page 1. Work from memory.” At 5 minutes check the three answers.
Next: If insecure, Try again A on page 1 in the first 2 reading minutes; otherwise page 2.
Answers: 1, 2, 3.
Open check 1 and its coaching →
5–15 minPupil page / Q
2 / reading
Read aloud / check: “Read the rules and follow the models. You can use this page during practice.”
Next: Check Try again A, if used. Do not add it to the lesson time.
15–20 minPupil page / Q
3 / 4
Read aloud / check: “Choose an answer and write your reasons. Stop at the check.”
Next: Check B and both reasons. If insecure, Try again B on page 3 replaces 8–9.
Answers: 4.
Open check 2 and its coaching →
20–30 minPupil page / Q
3–4 / 5–9
Read aloud / check: “Work through questions 5–9. Show working and units.”
Next: If Try again B was used, leave 8–9 blank. Check the retry before continuing.
Answers: 5, 6, 7, 8, 9.
30–33 minPupil page / Q
4 / 10
Read aloud / check: “Pause the practice. Answer question 10 independently.”
Next: If insecure, Try again C on page 4 replaces 14–15. Otherwise continue.
Answers: 10.
Open check 3 and its coaching →
33–45 minPupil page / Q
5–6 / 11–18
Read aloud / check: “Continue the practice. Use page 2 if you need a rule.”
Next: Check support responses promptly. If finished early, improve the explanation in question 18.
Answers: 11, 12, 13, 14, 15, 16, 17, 18.
45–55 minPupil page / Q
6 / 19
Read aloud / check: “Check your practice in a different colour, then explain one correction.”
Next: Read the practice answers from this guide. Keep the exit answer hidden.
Answers: 19.
55–60 minPupil page / Q
7 / 20
Read aloud / check: “Cover the earlier pages. Complete the final question on your own.”
Next: Collect every booklet. Check the exit afterwards; record any Try again D follow-up.
Answers: 20.
Open check 4 and its coaching →
The run sheet totals 60 minutes. Try again A uses reading time; B and C replace named practice questions; D is next lesson only. Timings are estimates, not classroom-trial results.
Read the check answers at each marked pause. At 45 minutes share only the practice answers, then ask for the explained correction. Keep the exit and next-lesson answers hidden until pupils have submitted.
Download teacher guide & answers ↓If several pupils are stuck
Use the coaching for the shared error, then the named independent retry within its replacement time. If the retry remains insecure, use the “If pupils are still stuck” guidance on teacher page 1 and record the specific difficulty for the usual teacher.
The run sheet totals 60 minutes. Try again A uses reading time; B and C replace named practice questions; D is next lesson only. Timings are estimates, not classroom-trial results.
ENTRY CHECK
Check 1: Question 1, Question 2, Question 3
Look for: Question 1: 0.30 A with conserved charge. Question 2: 30 Ω from V/I. Question 3: parallel with separate complete paths between the same terminals.
If secure: Turn to the reading and models on pupil page 2.
If: question 1 reduces current after the lamp, or question 3 misidentifies the arrangement or gives no path reason.Open coaching
Likely issue: Charge conservation or the distinction between one path and separate branches is insecure.
- Say
- Charge is conserved: a lamp transfers energy but does not use up charge. Current is the same through one unbranched loop. Separate complete branches between the same terminals are parallel.
- Ask
- What does the ammeter after the lamp read in question 1? What arrangement is described in question 3?
- Expect
- 0.30 A after the lamp; parallel in question 3, because there are separate complete paths between the same terminals.
Next: Use pupil page 2, Series: one complete path and Parallel: separate complete branches. Then ask pupils to attempt Try again A independently on pupil page 1. Use the first 2 reading minutes. Check the fresh response against the numbered answer key.
pupil page 2, Series: one complete path and Parallel: separate complete branches; Try again box on pupil page 1; online #concept-1.
If: question 2 has a wrong equation, calculation or unit.Open coaching
Likely issue: Identify whether equation choice, substitution, arithmetic or units caused the error.
- Say
- Use quantities for the same resistor. Rearrange V = IR to find resistance: divide the voltage by the current. Volts per ampere give ohms.
- Ask
- For question 2, which division uses 6.0 V and 0.20 A?
- Expect
- R = 6.0 ÷ 0.20 = 30 Ω. A correct equation with an arithmetic slip needs a calculator check.
Next: Use pupil page 2, equation box: R = V ÷ I. Then ask pupils to attempt Try again A independently on pupil page 1. Use the first 2 reading minutes. Check the fresh response against the numbered answer key.
pupil page 2, equation box: R = V ÷ I; Try again box on pupil page 1; online #concept-1.
HINGE CHECK
Check 2: Question 4
Look for: B, with both reasons: each branch spans 12 V and the branch currents add to 0.70 A. A correct letter alone is insufficient.
If secure: Start questions 5–9. Require the explanation as well as the correct choice.
If: A, or a reason that divides 12 V between the parallel branches.Open coaching
Likely issue: The series sharing rule has been applied to parallel branches.
- Say
- Each parallel branch joins the same two supply terminals, so each has the full supply voltage. Sharing the supply voltage applies to components in series.
- Ask
- In question 4, branch P spans which two terminals? What voltage is across it?
- Expect
- The same two supply terminals; the full 12 V is across P.
Next: Use pupil page 2, Parallel: separate complete branches. Then ask pupils to attempt Try again B independently on pupil page 3. This replaces questions 8–9. Then continue with questions 5–7. Check the fresh response against the numbered answer key.
pupil page 2, Parallel: separate complete branches; Try again box on pupil page 3; online #concept-2.
If: C uses one branch as the total; D subtracts two branch currents as though one were the total.Open coaching
Likely issue: The given branch currents have not been distinguished from the current entering the junction.
- Say
- At a junction, the current entering equals the sum leaving. Here both given currents belong to branches, so add them to find the supply current.
- Ask
- Are 0.20 A and 0.50 A in question 4 branch currents or the total? What must the total be?
- Expect
- They are branch currents; supply current = 0.20 + 0.50 = 0.70 A. Subtract only if the total and one branch are given.
Next: Use pupil page 2, Worked model — junction. Then ask pupils to attempt Try again B independently on pupil page 3. This replaces questions 8–9. Then continue with questions 5–7. Check the fresh response against the numbered answer key.
pupil page 2, Worked model — junction; Try again box on pupil page 3; online #concept-2.
If: B without both requested reasons.Open coaching
Likely issue: The correct choice may be a guess; understanding has not yet been demonstrated.
- Say
- The correct letter needs both rules: a parallel branch has the full supply voltage, and the supply current equals the sum of branch currents.
- Ask
- Explain the voltage and current values in your choice for question 4.
- Expect
- B: 12 V because P spans the supply terminals; 0.70 A because 0.20 + 0.50 = 0.70 A.
Next: Use pupil page 2, Parallel rule and Worked model — junction. Then ask pupils to attempt Try again B independently on pupil page 3. This replaces questions 8–9. Then continue with questions 5–7. Check the fresh response against the numbered answer key.
pupil page 2, Parallel rule and Worked model — junction; Try again box on pupil page 3; online #concept-2.
MID PRACTICE CHECK
Check 3: Question 10
Look for: 0.60 A through 10 Ω; 0.20 A through 30 Ω; 0.80 A total. Both branches have 6.0 V, so the larger resistance gives less current.
If secure: Continue with questions 11–18 on pages 5–6.
If: Adds 10 Ω and 30 Ω first, shares 6 V between branches, or makes the branch currents equal.Open coaching
Likely issue: A series/equal-sharing model is being used for unequal parallel resistors.
- Say
- These resistors are on separate branches, each across 6.0 V. Find each branch current using its own resistance, then add the two currents.
- Ask
- For question 10, divide 6.0 V by each branch resistance. What are the currents and their sum?
- Expect
- Through 10 Ω: 0.60 A. Through 30 Ω: 0.20 A. Total: 0.80 A. Do not add the resistances as if they were in series.
Next: Use pupil page 2, Parallel: separate complete branches. Then ask pupils to attempt Try again C independently on pupil page 4. This replaces questions 14–15. Continue with the other practice questions. Check the fresh response against the numbered answer key.
pupil page 2, Parallel: separate complete branches; Try again box on pupil page 4; online #concept-2.
If: Gives each branch 6.0 V but has a calculation/sum/unit error or no reason for the unequal currents.Open coaching
Likely issue: The path model may be correct; identify the method error or missing fixed-voltage/resistance link.
- Say
- Keep 6.0 V across each branch. At the same voltage, a larger resistance gives less current because I = V/R. Add the branch currents for the supply current.
- Ask
- Which branch in question 10 carries less current, and why? Check the two divisions and sum.
- Expect
- The 30 Ω branch: 0.20 A rather than 0.60 A through 10 Ω, because its resistance is larger. Total = 0.80 A.
Next: Use pupil page 2, equation box: I = V ÷ R, then add the branch currents. Then ask pupils to attempt Try again C independently on pupil page 4. This replaces questions 14–15. Continue with the other practice questions. Check the fresh response against the numbered answer key.
pupil page 2, equation box: I = V ÷ R, then add the branch currents; Try again box on pupil page 4; online #concept-2.
EXIT CHECK
Check 4: Question 20
Look for: Initially 0.60 A total; after Q opens, P and the supply each carry 0.40 A. The reason identifies P’s complete path, unchanged 8.0 V and fixed resistance.
If secure: Record the objectives demonstrated. If insecure, use Try again D next lesson after revisiting the named reading.
If: After Q opens, P becomes zero, rises to the old total, or receives a different voltage.Open coaching
Likely issue: Opening one parallel branch has been treated like breaking the only loop or redistributing a fixed total current.
- Say
- Opening Q breaks only Q’s branch. P still has a complete path across the unchanged 8.0 V supply, so its own current stays the same.
- Ask
- In question 20, is P’s path still complete after Q opens? What voltage and current does P have?
- Expect
- Yes: 8.0 V and 8.0 ÷ 20 = 0.40 A. With only P conducting, the supply current is also 0.40 A.
Next: Next lesson, first use pupil page 2, What happens when a switch opens? and the parallel rule. Then ask pupils to attempt Try again D independently on pupil page 7. This is outside today’s 60 minutes. Check the fresh response against the numbered answer key.
pupil page 2, What happens when a switch opens? and the parallel rule; Try again box on pupil page 7; online #concept-3.
If: Predicts P is unchanged, but its value or the initial total is wrong, or gives no physical reason.Open coaching
Likely issue: The method, current sum or explanation needs confirming before the objective is secure.
- Say
- Find each initial branch current with I = V/R, then add. After Q opens, use P’s unchanged voltage and resistance; do not send the old total current through P.
- Ask
- Calculate the initial P and Q currents in question 20. What remains after Q opens?
- Expect
- Initially P = 0.40 A, Q = 0.20 A, total = 0.60 A. Afterwards P and supply = 0.40 A because P remains complete across 8.0 V.
Next: Next lesson, first use pupil page 2, equation box, parallel rule and complete-path explanation. Then ask pupils to attempt Try again D independently on pupil page 7. This is outside today’s 60 minutes. Check the fresh response against the numbered answer key.
pupil page 2, equation box, parallel rule and complete-path explanation; Try again box on pupil page 7; online #concept-2.
Answers keyed to the pupil handout
- Question 1
Pupil page 1 - 0.30 A. Charge is conserved and does not get used up in the lamp, so the steady current is the same at both points in this unbranched loop.
- Question 2
Pupil page 1 - R = V/I = 6.0/0.20 = 30 Ω. Divide potential difference by current.
- Question 3
Pupil page 1 - Parallel. There is a separate complete path through each component between the same two terminals.
- Question 4
Pupil page 3 - B: 12 V and 0.70 A. P connects across the same supply terminals, so it has the full 12 V. The supply current is the sum of the branch currents: 0.20 + 0.50 = 0.70 A.
- Question 5
Pupil page 3 - A series circuit has one complete path. A parallel circuit has separate complete branches between the same two supply terminals.
- Question 6
Pupil page 3 - The current is the same through both series resistors. Charge is conserved and is not used up; energy is transferred in the resistors.
- Question 7
Pupil page 4 - R = 10 + 20 = 30 Ω. I = 6.0 / 30 = 0.20 A.
- Question 8
Pupil page 4 - R = 15 + 25 = 40 Ω. I = 12 / 40 = 0.30 A.
- Question 9
Pupil page 4 - 15 Ω resistor: V = 0.30 × 15 = 4.5 V. 25 Ω resistor: V = 0.30 × 25 = 7.5 V. The values add to 12 V.
- Question 10
Pupil page 4 - Each branch has 6.0 V. Through 10 Ω: I = 6.0/10 = 0.60 A. Through 30 Ω: I = 6.0/30 = 0.20 A. The total is 0.60 + 0.20 = 0.80 A. At the same potential difference, the larger resistance gives the smaller current.
- Question 11
Pupil page 5 - 0.20 A: the same current flows everywhere in this single loop.
- Question 12
Pupil page 5 - R₁: V = 0.20 × 10 = 2.0 V. R₂: V = 0.20 × 20 = 4.0 V.
- Question 13
Pupil page 5 - 6.0 V across each resistor: both branches connect across the supply.
- Question 14
Pupil page 5 - R₁: I = 6.0 / 10 = 0.60 A. R₂: I = 6.0 / 20 = 0.30 A.
- Question 15
Pupil page 5 - I = 0.60 + 0.30 = 0.90 A.
- Question 16
Pupil page 6 - 0.30 A. The R₁ branch is broken, but the R₂ branch remains complete across 6.0 V and carries 0.30 A.
- Question 17
Pupil page 6 - 0 A through both. Opening the switch breaks the only complete path.
- Question 18
Pupil page 6 - B. Each unit connects across the full 6.0 V supply, and each has a separate complete path, so opening one branch leaves the other operating. In A the supply potential difference is shared and breaking the loop stops both.
- Question 19
Pupil page 6 - Accept a corrected assigned answer with the applicable rule and a clear explanation of the original error. For the all-correct alternative: both branches have the supply potential difference, but I = V/R gives a smaller current in the larger resistance. Merely copying a final value is not an explained correction.
- Question 20
Pupil page 7 - Initially IP = 8.0/20 = 0.40 A and IQ = 8.0/40 = 0.20 A, so Isupply = 0.60 A. After Q opens, IP = 0.40 A and Isupply = 0.40 A. P still has a complete path across the unchanged 8.0 V; its resistance is unchanged. Q carries zero current.
- Try again A
- The second resistor carries 0.40 A because charge is conserved in the single complete loop. The separate sketch is parallel: the resistors lie on separate complete paths between the same terminals. In the final test, R = V/I = 3.6/0.30 = 12 Ω.
- Try again B
- Y has 9.0 V because it connects across the supply terminals. The total is 0.15 + 0.25 = 0.40 A because branch currents add at a junction.
- Try again C
- I30 = 12/30 = 0.40 A; I60 = 12/60 = 0.20 A; Isupply = 0.60 A. The 30 Ω branch carries more current because its resistance is lower at the same 12 V potential difference.
- Try again D — next lesson
- Initially I25 = 10/25 = 0.40 A and I50 = 10/50 = 0.20 A, giving 0.60 A total. After the 25 Ω branch opens, the 50 Ω branch and supply each carry 0.20 A: the remaining branch still has a complete path across 10 V. Opening the only series path gives 0 A everywhere.
Worksheet selection and marking
Use the numbered answer key in this guide, accepting equivalent correct wording and units. Credit sound methods where arithmetic slips occur. Mark only attempted questions after support substitutions. The provenance table links each adapted practice question to the maintained source and solutions.
Selection: Source PDF pages 1–3: Q1(a,b), Q2(a,b,c), Q3(a–e), Q4(b,c,e). Thirteen parts, 24 practice marks before support replacements.
Open the maintained worked solutions →Collect and follow up: Collect the whole named booklet. Check the independent exit against the key, record which learning objectives are secure, and tell the usual teacher which pupils need Try again D next lesson.
