中文（繁體）
中文（简体）
English
S4 Physics: Circuits in Series and Parallel – Challenge (10 Questions)
S4 Physics: Circuits in Series and Parallel – Challenge (10 Questions)
This paper is multi-step calculation and judgement: a two-stage network, an unknown resistance from the equivalent, total current for many parallel lamps, fuse selection, maximum power transfer, a load pulling down a divider, the same supply reconnected, internal resistance from two readings, two strings of three lamps at their rating, and a final item sorting three statements on totals, division and brightness.
Quick answer The 4 ohm series with the pair gives 6 ohm, 3 A and 6 V; 12 with 6 ohm gives 4 ohm; five 60 W lamps take 300 W and 1.36 A; 1 700 W needs a 10 A fuse; a 1 ohm internal resistance peaks at 9 W; the load pulls the divider from 6 V to 4.8 V; 20 W in series becomes 83.3 W in parallel; two readings give 1 ohm with terminal voltages of 11 V and 10 V; two strings of three lamps take 2 A each, 4 A in total, all at rating; and of three statements one is wrong and two are right.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 30 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S4 Physics | Circuits in Series and Parallel: Resistance, Current, Voltage and Power |
| Level | S4 |
02 Questions (10 items)
- FoundationA 4 ohm resistor is in series with the parallel pair of 6 ohm and 3 ohm across 18 V. Find the total resistance, the current and the voltage across the pair.
- FoundationA 12 ohm resistor is in parallel with an unknown resistance R and the equivalent is 4 ohm. Find R.
- FoundationFive 60 W lamps are in parallel across 220 V. Find the total power and the total current.
- FoundationA 220 V socket feeds appliances of 1 000 W, 500 W and 200 W. Find the total current and say whether a 5 A or a 10 A fuse should be used.
- IntermediateA battery of emf 6 V and internal resistance 1 ohm drives a variable resistor. At what load is the power transferred greatest and what is that power?
- IntermediateA 12 V supply feeds a divider of 10 ohm and 10 ohm, with the output across the second resistor. What is the output before and after a 20 ohm load is connected across it?
- IntermediateA 2 ohm and a 3 ohm resistor are in series across 10 V. Find each power and the total. What is the total if they are reconnected in parallel to the same 10 V supply?
- ChallengeA battery drives 1 A through 11 ohm and 2 A through 5 ohm. Find its emf and internal resistance, and the terminal voltage in each case.
- ChallengeSix lamps marked 12 V, 24 W are arranged as two parallel strings of three in series, across 36 V. Find the current per string, the total current and state whether each lamp runs at its rating.
- Error redoThree students write that adding a parallel branch lowers the total resistance and raises the battery current, that in series the larger resistor takes the larger voltage, and that a lamp brightness depends only on the current through it. Correct or confirm each.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | The pair is 2 ohm, the total resistance 6 ohm, the current 3 A and the pair takes 6 V. | Reduce the pair to 2 ohm, add the series 4 ohm, then apply Ohm law to the whole loop. |
| 2 | R is 6 ohm. | The reciprocal of the equivalent equals the sum of the branch reciprocals, so subtract the known one. |
| 3 | The total power is 300 W and the total current about 1.36 A. | Parallel branches share the same voltage, so the total power is the sum of the individual powers. |
| 4 | The total power is 1 700 W and the current about 7.73 A, so a 10 A fuse is needed. | The fuse rating must sit just above the working current, and a 5 A fuse would blow at once. |
| 5 | The power peaks when the load equals 1 ohm, with a current of 3 A and a maximum of 9 W. | The load takes the most power when it matches the internal resistance, which a quadratic or direct substitution confirms. |
| 6 | The output is 6 V unloaded and falls to 4.8 V with the load, since the lower arm becomes 6.67 ohm. | The load sits in parallel with the lower arm and lowers that arm, which shifts the divider ratio. |
| 7 | In series the current is 2 A, giving 8 W and 12 W for 20 W in total; in parallel they take 50 W and 33.3 W, about 83.3 W in total. | In parallel the total resistance is smaller and the current larger, so the total power rises sharply on the same supply. |
| 8 | The emf is 12 V and the internal resistance 1 ohm, giving terminal voltages of 11 V and 10 V. | Write Ohm law for both cases and subtract to remove the emf and leave the internal resistance. |
| 9 | Each lamp is 6 ohm, so a string is 18 ohm, the string current 2 A, the total 4 A, and each lamp takes 24 W at its rating. | The series divider gives each lamp 12 V, exactly its rated voltage. |
| 10 | The first two are correct, since a parallel branch lowers the total and series voltage rises with resistance; the third is wrong because brightness follows power, the square of the current times resistance. | Decide each case by asking which quantity is common, current in series and voltage in parallel. |
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.
Find a tutor freeTutor 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.