OxfordAQA International GCSE Physics: Series and parallel circuits and heating (9203) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards, section 3.5.1 (electrical circuits), statements p to t: cells in series, series and parallel circuits, mixed circuits and the heating effect of current.

Links: study guide, revision notes, course hub, printable checklist and the 10-minute diagnostics. Use V = IR throughout. Ammeters have negligible resistance and meters do not change the circuit.

Questions

1. State how the combined resistance, the current and the potential difference behave for components connected in series. [3]

2. Rechargeable cells each have a voltage of 1.2 V. (a) Calculate the combined voltage of six of these cells joined in series. [1] (b) A radio needs 3.6 V. State how many cells in series are needed. [1]

3. A 24 V supply is connected in series to a 120 Ω resistor and an 80 Ω resistor. Calculate (a) the total resistance, (b) the current, and (c) the potential difference across each resistor. [5]

4. An 18 V supply is connected across a 90 Ω resistor and a 60 Ω resistor in parallel. Calculate (a) the current in each resistor, (b) the current drawn from the supply, and (c) the combined resistance of the pair. [5]

5. Two identical lamps are lit from a battery. A third identical lamp is added to the circuit. (a) The first two lamps were in parallel and the new lamp is added in parallel. Describe what happens to the brightness of the first two lamps and to the current from the battery. [2] (b) The first two lamps were in series and the new lamp is added in series. Describe and explain what happens to their brightness. [2]

6. A 14 V supply is connected to a 20 Ω resistor in series with a parallel block. The block is a 10 Ω resistor and a 40 Ω resistor. An ammeter in the lead from the supply reads 0.50 A. (a) Calculate the potential difference across the 20 Ω resistor. [2] (b) Calculate the potential difference across the parallel block. [1] (c) Calculate the current in each resistor in the block. [2] (d) Calculate the resistance of the block. [1] (e) Calculate the total resistance of the circuit. [1]

7. Explain why the lamps in a house are connected in parallel and not in series. [3]

8. (a) Explain, in terms of charges and atoms, why a wire becomes hot when there is a current in it. [3] (b) A heater element and its connecting leads carry the same current. Suggest why the element becomes very hot and the leads stay cool. [1]

9. A filament lamp is supplied with 48 J of electrical energy each second and transfers 2.4 J each second as light. A compact fluorescent lamp (CFL) gives the same light and is supplied with 9.6 J each second. (a) Calculate the efficiency of each lamp as a percentage. [3] (b) Calculate the energy wasted each second by each lamp. [2] (c) State which lamp you would choose to buy. [1]

10. A 12 V supply is connected to a 30 Ω resistor in series with two identical lamps, X and Y, which are in parallel with each other. An ammeter in the lead from the supply reads 0.20 A. (a) Calculate the potential difference across the 30 Ω resistor. [2] (b) Calculate the potential difference across each lamp. [1] (c) State the current in each lamp. [1] (d) Calculate the resistance of one lamp. [1] The filament of lamp Y now breaks. Assume the resistance of lamp X does not change. (e) Calculate the new total resistance of the circuit. [1] (f) Calculate the new current from the supply. [1] (g) Calculate the new potential difference across lamp X, and state what happens to its brightness. [2]

11. A flat has six lamps. Each filament lamp is supplied with 40 J each second, and a CFL giving the same light is supplied with 8.0 J each second. The six lamps are on for 4.0 hours each day. (a) Calculate the energy saved each second by changing one filament lamp for a CFL. [1] (b) Calculate the energy saved in one day by changing all six lamps. [2] (c) Explain, in terms of energy transfer, why a room lit by filament lamps is warmer than one lit by CFLs. [1] (d) Give one reason, other than cost to the owner, for choosing the more efficient lamps. [1]

12. Explain why the combined resistance of two resistors in parallel is less than the resistance of either one. [3]

Answers

1. The combined resistance is the sum of the resistances of the components [1]. The current is the same in each component [1]. The supply potential difference is divided between the components [1]. [3] Examiner insight: Name each quantity and give its behaviour. Writing “they add up” without saying which quantity adds up leaves the answer ambiguous.

2. (a) 6 x 1.2 [1] = 7.2 V (the sum of the voltages). (b) 3.6 / 1.2 = 3 cells [1]. [2] Examiner insight: A one-mark part still needs a clear answer with the unit. Check that your count of cells is a whole number.

3. (a) R = 120 + 80 [1] = 200 Ω. (b) I = V/R = 24 / 200 [1] = 0.12 A. (c) 120 Ω: 0.12 x 120 = 14.4 V [1]. 80 Ω: 0.12 x 80 = 9.6 V [1]. Check: 14.4 + 9.6 = 24 V [1]. [5] Examiner insight: Quoting the equation and substituting makes your method visible. The check that the two pds add to the supply pd shows that the current is right and is worth including.

4. (a) Each resistor has 18 V across it. 90 Ω: I = 18 / 90 = 0.20 A [1]. 60 Ω: I = 18 / 60 = 0.30 A [1]. (b) 0.20 + 0.30 = 0.50 A [1]. (c) R = V/I = 18 / 0.50 [1] = 36 Ω [1]. [5] Examiner insight: Use the supply pd for each branch and the supply current for the combined resistance. A common slip is dividing by the wrong current. 36 Ω is below 60 Ω, which is a quick sense check.

5. (a) The first two lamps stay the same brightness, because each still has the full supply pd across it [1]. The current from the battery increases [1]. (b) The lamps become dimmer [1], because the total resistance is larger so the current is smaller, and the supply pd is shared between three lamps [1]. [4] Examiner insight: Give the effect and then the reason. “Dimmer” alone is a statement of what happens, with no explanation.

6. (a) The supply current passes through the 20 Ω resistor: V = IR = 0.50 x 20 [1] = 10 V [1]. (b) 14 - 10 = 4.0 V [1]. (c) 10 Ω: 4.0 / 10 = 0.40 A [1]. 40 Ω: 4.0 / 40 = 0.10 A [1]. (d) 4.0 / 0.50 = 8.0 Ω [1]. (e) 20 + 8.0 = 28 Ω [1] (check: 14 / 0.50 = 28 Ω). [7] Examiner insight: The pd across a parallel block is found by subtraction from the supply pd. Showing that 0.40 + 0.10 matches the ammeter reading of 0.50 A is a good check.

7. Each lamp has the full supply pd across it, so each works at its normal brightness [1]. Each lamp can be switched on and off on its own [1]. If one lamp fails, the others stay lit [1]. [3] Examiner insight: Three separate benefits earn three separate points. Two wordings of the same idea, such as “independent” and “one at a time”, count as one.

8. (a) The charges that move through the wire (electrons) collide with the stationary atoms [1]. Energy is transferred from the moving charges to the atoms, which vibrate more [1]. The temperature of the wire rises, and thermal energy is transferred to the surroundings [1]. (b) The element has a much higher resistance than the leads, so a larger share of the energy is transferred in the element [1]. [4] Examiner insight: Name the collision between charges and atoms. Saying that the wire “resists” the current repeats the question, and does not explain the heating.

9. (a) Efficiency = useful energy transferred / total energy supplied [1]. Filament: 2.4 / 48 = 0.050, so 5.0% [1]. CFL: 2.4 / 9.6 = 0.25, so 25% [1]. (b) Filament: 48 - 2.4 = 45.6 J [1]. CFL: 9.6 - 2.4 = 7.2 J [1]. (c) The CFL, since it wastes less energy for the same light [1]. [6] Examiner insight: Divide by the total energy supplied, not by the wasted energy. The choice in part (c) needs a reason from your results.

10. (a) V = IR = 0.20 x 30 [1] = 6.0 V [1]. (b) 12 - 6.0 = 6.0 V [1]. (c) The lamps are identical, so they share the 0.20 A equally: 0.10 A each [1]. (d) R = V/I = 6.0 / 0.10 = 60 Ω [1]. (e) Only X remains: 30 + 60 = 90 Ω [1]. (f) I = 12 / 90 = 0.133, so 0.13 A [1]. (g) V = 0.133 x 60 = 8.0 V [1]. Lamp X has more current and a larger pd than before, so it is brighter [1]. [9] Examiner insight: In a multi-part question, label each result clearly so that you can reuse it in the next part. In part (g), compare X with its old values of 0.10 A and 6.0 V, not just with Y.

11. (a) 40 - 8.0 = 32 J [1]. (b) Time = 4.0 x 3600 = 14 400 s [1]. Energy saved = 32 x 6 x 14 400 = 2 764 800 J, so 2.76 x 10^6 J [1]. (c) Filament lamps transfer more of the energy supplied as thermal energy to the surroundings [1]. (d) Less energy needs to be generated, which reduces the fuel used at power stations [1]. [5] Examiner insight: Convert hours to seconds before multiplying, because the power values are in joules per second. In part (d) keep to a reason that is not about money.

12. Each extra branch gives the charge another path [1]. For the same supply pd, more current flows from the supply than through either resistor alone [1]. Since resistance = V/I, a larger current for the same pd means a smaller combined resistance [1]. [3] Examiner insight: Link the extra path to the increased current and then to R = V/I. A statement of the rule without a reason does not answer an “explain” question.

Where marks are usually lost

• Giving each parallel branch a share of the supply pd instead of the full supply pd.
• Dividing the supply pd by one branch resistance to find the supply current.
• Forgetting that the series resistor in a mixed circuit carries the whole supply current.
• Adding branch resistances to find the combined resistance of a parallel block.
• Writing that current is “used up” in a series circuit.
• Giving heating as a result of “resistance” without naming collisions between charges and atoms.
• Calculating efficiency as useful energy divided by wasted energy.
• Leaving out units, or leaving energies in kilojoules and hours when the equation needs joules and seconds.

Next steps

• Revise with the revision notes.
• Re-read the study guide.
• Open the course hub and the printable checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class: /trial/.

Official syllabus

OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.5.1 (electrical circuits), statements p to t.