These eleven questions cover R = V ÷ I, series and parallel circuits, current-voltage data, Ohm’s law and meter placement from potential difference, resistance and circuits. All data is invented for practice and is not taken from any exam paper. They go from easier to harder.
Write your answer with its unit on paper first, then open the answer. Questions 6 and 7 use the combined resistance of parallel resistors, which depends on your syllabus route, so check the Cambridge page for your exam year. You can test a prediction in the circuit reasoning simulator.
Questions
1. A resistor has 6.0 V across it and carries 0.30 A. Find its resistance.
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R = V ÷ I = 6.0 ÷ 0.30 = 20 Ω. Check: 0.30 × 20 = 6.0 V.
2. A 15 Ω resistor carries a current of 0.40 A. What is the p.d. across it?
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V = I × R = 0.40 × 15 = 6.0 V. Check: 6.0 ÷ 15 = 0.40 A.
3. A 240 Ω resistor is connected across a 12 V supply. Find the current in amperes and in milliamperes.
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I = V ÷ R = 12 ÷ 240 = 0.050 A. In milliamperes, 0.050 × 1000 = 50 mA. Check: 0.050 × 240 = 12 V.
4. A component has 7.0 V across it and carries 35 mA. Find its resistance.
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Convert: I = 35 ÷ 1000 = 0.035 A. R = 7.0 ÷ 0.035 = 200 Ω. Check: 0.035 × 200 = 7.0 V.
5. Resistors of 6.0 Ω, 10 Ω and 14 Ω are in series across a 15 V supply. Find the current and the p.d. across the 10 Ω resistor.
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Total R = 6 + 10 + 14 = 30 Ω. I = 15 ÷ 30 = 0.50 A. The p.d. across the 10 Ω resistor is 0.50 × 10 = 5.0 V. Check: the three p.d.s are 3.0 + 5.0 + 7.0 = 15 V.
6. Resistors of 12 Ω and 24 Ω are in parallel across a 6.0 V supply. Find the current in each, the total current and the combined resistance.
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I₁ = 6.0 ÷ 12 = 0.50 A. I₂ = 6.0 ÷ 24 = 0.25 A. Total = 0.75 A. R = 6.0 ÷ 0.75 = 8.0 Ω. Check: 1/12 + 1/24 = 2/24 + 1/24 = 3/24 = 1/8, so R = 8.0 Ω, smaller than 12 Ω.
7. A 10 Ω resistor is in series with two parallel 20 Ω resistors, all across a 12 V supply. Find the supply current and the p.d. across the parallel pair.
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Two identical 20 Ω resistors in parallel give 20 ÷ 2 = 10 Ω. Total R = 10 + 10 = 20 Ω. I = 12 ÷ 20 = 0.60 A. The p.d. across the parallel pair is 0.60 × 10 = 6.0 V. Check: 6.0 V across the series resistor plus 6.0 V across the pair makes 12 V. Each 20 Ω branch carries 6.0 ÷ 20 = 0.30 A, and 0.30 + 0.30 = 0.60 A.
8. Component Y gives these readings. Is it ohmic? Give the resistance at each reading.
| V (V) | 2.0 | 4.0 | 6.0 |
|---|---|---|---|
| I (A) | 0.10 | 0.16 | 0.20 |
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R = 2.0 ÷ 0.10 = 20 Ω. R = 4.0 ÷ 0.16 = 25 Ω. R = 6.0 ÷ 0.20 = 30 Ω. The resistance is not constant, so Y is non-ohmic. Its resistance increases with p.d., which is consistent with a heating filament.
9. Two lamps are in parallel with a 6.0 V supply. An ammeter in the main wire reads 0.50 A, and an ammeter in the branch of lamp 1 reads 0.20 A. Find the current in lamp 2 and the resistance of each lamp.
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Branch currents add, so I₂ = 0.50 − 0.20 = 0.30 A. R₁ = 6.0 ÷ 0.20 = 30 Ω. R₂ = 6.0 ÷ 0.30 = 20 Ω. Check: 1/30 + 1/20 = 2/60 + 3/60 = 5/60 = 1/12, and 6.0 ÷ 0.50 = 12 Ω.
10. A student connects a voltmeter in series with a lamp and a cell. Describe and explain what is seen.
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The lamp does not light. The voltmeter has a very high resistance, so the current in the loop is almost zero. Nearly all the p.d. of the cell is across the voltmeter, so it reads close to the cell’s p.d. The correct connection is across the lamp, in parallel.
11. A student says: “The filament lamp breaks V = I × R because its resistance is not constant.” Using 4.0 V, 0.80 A and 6.0 V, 1.00 A, comment on this statement.
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At 4.0 V, R = 4.0 ÷ 0.80 = 5.0 Ω. At 6.0 V, R = 6.0 ÷ 1.00 = 6.0 Ω. V = I × R holds at each point, because it defines the resistance there. What the lamp does not obey is Ohm’s law, since V ÷ I is not constant. The resistance rises because the filament is hotter.
If you got these wrong
- Answer 1000 times too large or small, or a current that looks huge: a unit slip with milliamperes. Revisit using a resistance relationship with units (questions 1 to 4).
- A combined resistance larger than the smallest branch, or a p.d. that does not add up: review current in series and parallel branches (questions 5 to 7 and 9).
- A resistance worked out from a difference instead of from V ÷ I at a point: review interpreting a current-voltage graph (question 8).
- A statement about “breaking” V = I × R: review component law versus universal rule (questions 8 and 11).
- Meters in the wrong place, or a reading that does not match the circuit: review meter placement on a diagram (questions 9 and 10).
Log any repeated error in the mistake log and retest queue, then retest it a few days later with a fresh question. If the same mistake returns, a teacher watching your working live can find the cause faster than another set of questions. That is what our online one-to-one Physics tuition is designed to offer.