Why does adding a parallel branch increase the source current? Try it in a virtual circuit, then connect what you see to the equations. Practise resistance, power and energy without mixing up quantities or units.
Official syllabus version 2 for examinations in 2026–2028 and links checked 2 October 2026. Original TutorMax questions and solutions.
EXPLORE BEFORE YOU CALCULATE
Change one value at a time. Predict what happens to the current, then check the model. Parallel calculations are Supplement skills.
One path for the current
Ideal source and fixed resistors. Values shown to at most two decimal places. This model excludes heating, wire resistance and internal resistance.
R = R₁ + R₂ = 6 + 3 = 9 Ω
The same current passes through both components. The supply voltage is shared between them.
Keep the voltage and both resistors fixed. What happens to the source current when you change from series to parallel?
Choose an answer to check your reasoning.
Use this virtual circuit to explore calculations. Do not recreate circuits using mains electricity.
Now explain it in a practice question →A resistor has 6.0 V across it and 0.25 A through it. Find its resistance, power and energy transferred in 60 seconds. Assume its resistance stays constant.
Use the voltage across and current through this same resistor.
A watt is one joule per second.
Multiply the transfer rate by the time in seconds.
For the 2026–2028 syllabus, Core uses Papers 1 and 3; Extended uses Papers 2 and 4. Both routes also take Paper 5 or Paper 6. The school confirms the entry. The syllabus's Core/Supplement labels matter: Q4 uses parallel-circuit calculations and Q5 uses I = Q/t, which are Supplement skills.
Swipe across the table to see all columns.
| Use here | Questions | Then return to |
|---|---|---|
| Core electricity skills | Q1, Q2, Q3, Q6 | School-confirmed Core paper and practical component |
| Supplement calculation skills | Q4, Q5 | School-confirmed Extended paper and practical component |
Core starters: Q1–Q3 and Q6. Supplement calculations: Q4–Q5. Start with Q1 and Q2 if time is short. Work on paper before revealing an answer; these calculations do not replace practical preparation.
Resistance uses division; power uses multiplication. Write the required unit first.
Use voltage divided by current.
Use current multiplied by voltage.
Answer: Resistance 30 Ω; power 4.8 W.
Before moving on: can you explain why your method works, as well as give the answer?
Convert minutes to seconds before using E = IVt.
Three minutes is 180 seconds.
Substitute the converted time.
The power is 1.0 W: one joule each second for 180 seconds.
Answer: 180 J.
Before moving on: can you explain why your method works, as well as give the answer?
There is one current path. Series resistances add.
Both components are on the same path.
Adding positive resistance in series increases the total.
Answer: 20 Ω initially; 25 Ω after adding the third resistor.
Before moving on: can you explain why your method works, as well as give the answer?
Each branch has 12 V across it. Find the two currents separately, then add them at the source.
The smaller resistance carries the larger current.
The branches split the current, which recombines at the source.
It should be below the smaller branch resistance, 3 Ω.
Answer: Branch currents 2 A and 4 A; source current 6 A; equivalent resistance 2 Ω.
Before moving on: can you explain why your method works, as well as give the answer?
Use I = Q/t, with t in seconds.
Two minutes is 120 seconds.
The result is an average over this interval.
Answer: Average current 1.5 A.
Before moving on: can you explain why your method works, as well as give the answer?
Calculate V/I for each pair. Limit your conclusion to the measured conditions.
Each measured pair gives the same resistance.
The data are consistent with constant resistance over this measured range. They do not establish what happens at every temperature or outside these conditions.
Answer: 15 Ω for every pair; consistent with constant resistance over this range.
Before moving on: can you explain why your method works, as well as give the answer?
Attempt these after a break without copying the earlier solution. Use the units and physical checks to justify the result.
Write R = V/I and P = IV separately.
Divide voltage by current.
Multiply current by voltage.
Answer: 30 Ω and 2.7 W.
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
Both branches have 6 V. Source current is the sum of branch currents.
Use the resistance of each branch.
The equivalent is smaller than the smallest branch resistance.
Answer: 1.5 A and 0.5 A; source current 2 A; equivalent 3 Ω.
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
Return to official available paper materials through Cambridge, using the component confirmed by your school. Practise a missed calculation with changed numbers before another full timed attempt. For help, bring your attempted work and entry details; the teacher needs to see the first line that breaks down.
Q1–Q3 and Q6 practise selected Core skills. Q4–Q5 and retest R2 use Supplement calculations intended for Extended preparation. Always confirm your school's entry and current syllabus.
No. It is the Alternative to Practical assessment route and still assesses experimental skills. This calculation set does not prepare all those skills.
No. They are original TutorMax questions with worked solutions. The page links to official Cambridge sources separately.
No. Those are separate syllabus entries. Check the actual syllabus code and specification before treating a 0625 task as required content.
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