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S2 Science: Electric Circuits, Current, Voltage and Resistance (Paper 3: challenge)
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S2 Science: Electric Circuits, Current, Voltage and Resistance (Paper 3: challenge)
The challenge here is the mixed network: reduce the parallel part first, or every later step goes wrong. Another common slip is treating watt hours as kilowatt hours, which inflates the bill a thousand times.
Direct answers Answers in order: full main current passes through, ammeter in series and voltmeter in parallel, 6 ohms, 2 A, 8 V and 4 V, 1 kWh, 1.1 dollars, 3 ohms, 20 per cent, no.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S2 Science: Electric Circuits, Current, Voltage and Resistance |
| Level | S2 Science |
02 Questions (10 items)
- FoundationWhy must a fuse be connected in series?
- FoundationHow should an ammeter and a voltmeter be connected?
- IntermediateA 6 ohm and a 3 ohm resistor in parallel are then joined in series with a 4 ohm resistor. Find the total resistance.
- IntermediateThe same network is connected to 12 V. What is the main current?
- IntermediateFor the same network, what are the voltages across the 4 ohm resistor and across the parallel part?
- ChallengeA 100 W lamp is on for 10 hours a day. How many units does it use in a day?
- ChallengeAt 1.1 dollars per unit, what is the daily cost of that lamp?
- ChallengeA 12 ohm and a 4 ohm resistor are in parallel. What is the total resistance?
- FoundationA lamp takes in 20 W and gives out 4 W of light, the rest becoming heat. What is its efficiency?
- Error redoA student says current is used up by a resistor so the later lamp is dimmer. Is that right?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | So that the whole circuit current passes through it and melts it when the current is too large | A fuse melts from its own heating, so it has to carry the full current. |
| 2 | The ammeter in series with the branch and the voltmeter in parallel across the component | An ammeter has tiny resistance so it goes in series, a voltmeter has huge resistance so it goes in parallel. |
| 3 | 6 ohms | The parallel part is two ohms, plus four gives six ohms. |
| 4 | 2 A | Twelve divided by six is two. |
| 5 | 8 V across the 4 ohm resistor and 4 V across the parallel part | Two times four is eight volts, and twelve minus eight leaves four volts. |
| 6 | 1 kWh | One hundred times ten is a thousand watt hours, one kilowatt hour. |
| 7 | 1.1 dollars | One unit times 1.1 is 1.1 dollars. |
| 8 | 3 ohms | One over one twelfth plus one quarter is three ohms. |
| 9 | 20 per cent | Four over twenty times one hundred per cent is twenty per cent. |
| 10 | No | Series current is the same everywhere; brightness differs because voltage and power are shared, not because charge is consumed. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Reduce the parallel block first, then add the series part. For cost, convert watt hours to kilowatt hours before multiplying by the tariff.Tutor Registration →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.