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S4 Physics: Circuits in Series and Parallel – Error Re-drill (10 Questions)

S4 Physics: Circuits in Series and Parallel – Error Re-drill (10 Questions)

  • 2026-11-05

S4 Physics: Circuits in Series and Parallel – Error Re-drill (10 Questions)

This drill targets the ten most frequent errors: adding parallel resistances directly, inverting the divider, making branch current rise with resistance, taking V over R as power, keeping a lamp rating at double voltage, ignoring the internal drop, thinking a parallel total exceeds the smallest branch, forgetting the kilo in kWh, putting an ammeter in parallel, and judging series brightness by rated wattage.

Quick answer Parallel 6 and 3 ohm give 2 ohm; 12 V across 4 and 8 ohm gives 4 V and 8 V at 1 A; 12 V across 6 and 3 ohm in parallel gives 2 A and 4 A; 12 V on 3 ohm is 48 W; the 6 V 3 W lamp takes 12 W at 12 V; 0.5 ohm internal resistance leaves 11 V and 91.7 percent efficiency; 2 and 8 ohm in parallel give 1.6 ohm; 2 000 W for half an hour is 1 kWh; an ammeter of 0.01 ohm across 12 V would take 1 200 A; and in series the 3 W lamp outshines the 6 W one, 1.33 W against 0.67 W.

01 Paper details (check answers afterwards)

Item Detail
Questions 10 questions
Suggested time about 25 minutes
Difficulty mix Error redo 10
Chapter Chapter: S4 Physics | Circuits in Series and Parallel: Resistance, Current, Voltage and Power
Level S4

02 Questions (10 items)

  1. Error redoA student puts 6 ohm and 3 ohm in parallel and writes a total of 9 ohm. What is the correct total and where is the mistake?
  2. Error redoA 12 V battery drives 4 ohm and 8 ohm in series. A student says the 4 ohm takes 8 V and the 8 ohm takes 4 V. What are the correct voltages?
  3. Error redoA 12 V supply feeds 6 ohm and 3 ohm in parallel. A student says 4 A through the 6 ohm and 2 A through the 3 ohm. What are the correct branch currents?
  4. Error redoA student applies 12 V across 3 ohm and uses P = V divided by R to get 4 W. What is the correct power and why is the method wrong?
  5. Error redoA lamp marked 6 V 3 W is connected to 12 V. One student says it still takes 3 W and another says 6 W. What power does it actually take?
  6. Error redoA cell of e.m.f. 12 V has an internal resistance of 0.5 ohm and delivers 2 A. A student says the terminal voltage is still 12 V. Find the terminal voltage and the efficiency.
  7. Error redoFor a 2 ohm and an 8 ohm resistor a student says the series total is 10 ohm and that any parallel total must exceed 2 ohm. Judge both statements.
  8. Error redoA 2 000 W kettle runs for 30 minutes. A student says it uses 60 kWh. What is the correct energy use?
  9. Error redoA student connects an ammeter straight across a 12 V battery, saying it will read the voltage. The ammeter has a resistance of 0.01 ohm. What happens?
  10. Error redoA 6 V supply runs a 6 W lamp of 6 ohm and a 3 W lamp of 12 ohm in series. A student says the 6 W lamp is brighter. Which one really is?

03 Answers and one-line notes

# Answer One-line note
1 The total is 2 ohm, because the reciprocals add: 1/6 + 1/3 = 1/2, so R = 2 ohm. The 9 ohm figure is the series answer for the same pair. Learn the rule with its condition: resistances add directly in series and add as reciprocals in parallel.
2 The 4 ohm takes 4 V and the 8 ohm takes 8 V, since the current is 12 divided by 12 ohm, that is 1 A. Voltage rises with resistance in series, so the larger resistor takes the larger share. Find the current first and multiply by each resistance; that prevents inverting the ratio.
3 The 6 ohm carries 2 A and the 3 ohm carries 4 A, giving 6 A in total from an equivalent resistance of 2 ohm. In parallel the branch currents are inversely proportional to the resistances. In parallel the voltage is common rather than the current, so the smaller resistance carries the larger current.
4 The correct power is 48 W from V squared over R, that is 144 divided by 3. Dividing V by R gives the current, 4 A, not the power. Pick the power form that matches what you know: with V and R use V squared over R.
5 The filament resistance is 36 divided by 3, that is 12 ohm, so at 12 V the power is 144 divided by 12, that is 12 W, four times the rating, enough to burn it out. Doubling the voltage quadruples the power, and a rating only holds at its rated voltage.
6 The terminal voltage is 11 V because the internal drop is 2 times 0.5, that is 1 V. The output is 22 W with 2 W lost inside, an efficiency of about 91.7 percent. The terminal voltage is the e.m.f. minus the internal drop, so it sags further as the current rises.
7 The series total of 10 ohm is right, but the parallel total is 1.6 ohm, below the smallest branch, so the second statement is wrong. A parallel total is always below the smallest resistance. Use one check: a parallel branch is an extra path, so the total can only fall.
8 It uses 1 kWh: convert 2 000 W to 2 kW and multiply by 0.5 hour, which is 3.6 million joules. The k in kWh stands for kilowatts, so divide the wattage by 1 000 first.
9 The current would be 12 divided by 0.01, that is 1 200 A, effectively shorting the battery and destroying the meter and the leads. Voltage is measured with a high resistance voltmeter in parallel. An ammeter belongs in series; putting it in parallel creates a short circuit, the most common laboratory accident.
10 The shared current is one third of an ampere, so the powers are 0.67 W in the 6 ohm and 1.33 W in the 12 ohm, making the 3 W lamp the brighter one. In series, compare resistances at the same current rather than comparing rated wattages.

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.

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