IGCSE Physics 4.3 Electric Circuits MCQs – 30 Practice Questions with Answers (Cambridge 0625)

Revise Cambridge IGCSE Physics (0625) Section 4.3, Electric Circuits, with 30 original multiple-choice questions on series and parallel circuits, circuit components, potential dividers, power and energy. Every question has a worked answer – just tap “Show Answer” to check yourself as you go.

Formula recap: V = IR • Series: R = R1 + R2 • Parallel: 1/R = 1/R1 + 1/R2 • P = IV • E = IVt. For the full list, see our IGCSE Physics formula sheet.

Series & Parallel Circuits

Q1. Two resistors of 4 Ω and 6 Ω are connected in series to a 20 V battery. What is the current in the circuit?

  • (A) 0.50 A
  • (B) 2.0 A
  • (C) 5.0 A
  • (D) 8.3 A
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Answer: B – In series the resistances add: R = 4 + 6 = 10 Ω. Then I = V / R = 20 / 10 = 2.0 A. (Using only the 4 Ω resistor gives the 5.0 A trap, and 8.3 A comes from wrongly combining the resistors as if they were in parallel.)

Q2. A 12 V battery supplies two lamps connected in series. The voltage across one lamp is 5 V. What is the voltage across the other lamp?

  • (A) 5 V
  • (B) 12 V
  • (C) 17 V
  • (D) 7 V
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Answer: D – In a series circuit the voltages across the components add up to the supply voltage, so the second lamp has 12 − 5 = 7 V across it.

Q3. Three resistors of 3 Ω, 5 Ω and 7 Ω are connected in series. What is their combined resistance?

  • (A) 15 Ω
  • (B) 1.5 Ω
  • (C) 35 Ω
  • (D) 105 Ω
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Answer: A – For resistors in series, simply add them: R = 3 + 5 + 7 = 15 Ω. (The 1.5 Ω value is what you would get if they were in parallel.)

Q4. A student wants to measure the current flowing through a resistor. How should the ammeter be connected?

  • (A) In parallel with the resistor
  • (B) In parallel with the battery
  • (C) In series with the resistor
  • (D) Across the resistor, in place of a voltmeter
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Answer: C – An ammeter measures current, so it is connected in series with the component so that all the current through the component also passes through the ammeter. It has a very low resistance so that it hardly affects the circuit.

Q5. A student wants to measure the potential difference across a lamp. How should the voltmeter be connected?

  • (A) In series with the lamp
  • (B) In series with the battery
  • (C) In place of the lamp
  • (D) In parallel with the lamp
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Answer: D – A voltmeter measures the potential difference between two points, so it is connected in parallel across the component. It has a very high resistance so that it draws almost no current from the circuit.

Q6. A 6 V battery is connected across two resistors joined in parallel. What is the voltage across each resistor?

  • (A) 6 V
  • (B) 3 V
  • (C) 12 V
  • (D) It depends on the resistance of each resistor
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Answer: A – In a parallel circuit every branch is connected directly across the supply, so each resistor has the full supply voltage of 6 V across it, whatever its resistance.

Q7. A 6 Ω resistor and a 3 Ω resistor are connected in parallel. What is their combined resistance?

  • (A) 9 Ω
  • (B) 4.5 Ω
  • (C) 2 Ω
  • (D) 18 Ω
Show Answer

Answer: C – For two resistors in parallel, 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, so R = 6/3 = 2 Ω. Note that the combined resistance is smaller than the smallest resistor, because the current now has two paths to flow through.

Q8. Two parallel branches of a circuit carry currents of 2 A and 3 A. What current is supplied by the battery?

  • (A) 1 A
  • (B) 5 A
  • (C) 6 A
  • (D) 2.5 A
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Answer: B – At a junction, the currents in the parallel branches add up to the total current from the supply: 2 + 3 = 5 A.

Q9. Three lamps are connected in series with a battery. The filament of one lamp breaks. What happens to the other two lamps?

  • (A) They become brighter
  • (B) They become dimmer
  • (C) They stay on as before
  • (D) They both go out
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Answer: D – A series circuit is a single loop. A break anywhere in the loop stops the current everywhere, so all the lamps go out.

Q10. Three lamps are connected in parallel with a battery. The filament of one lamp breaks. What happens to the other two lamps?

  • (A) They stay on with the same brightness
  • (B) They both go out
  • (C) They become dimmer
  • (D) They become brighter and burn out
Show Answer

Answer: A – In parallel, each lamp is on its own branch connected directly across the battery. A break in one branch does not affect the others, and each remaining lamp still has the full battery voltage across it, so its brightness is unchanged. This is why house lights are wired in parallel.

Components & Their Behaviour

Q11. In a circuit diagram, a symbol drawn as one long thin line next to one short thick line represents which component?

  • (A) A cell
  • (B) A fuse
  • (C) A switch
  • (D) A resistor
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Answer: A – The pair of unequal parallel lines is the symbol for a cell; the longer line is the positive terminal. Two or more cells joined together form a battery.

Q12. What happens to the resistance of a thermistor as its temperature increases?

  • (A) It increases
  • (B) It decreases
  • (C) It stays the same
  • (D) It becomes zero
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Answer: B – A thermistor is a temperature-dependent resistor whose resistance decreases as the temperature rises. This makes it useful as the sensor in thermostats and temperature alarms.

Q13. What happens to the resistance of a light-dependent resistor (LDR) as the light intensity falling on it increases?

  • (A) It increases
  • (B) It stays the same
  • (C) It decreases
  • (D) It doubles
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Answer: C – The resistance of an LDR decreases as the light intensity increases. In bright light its resistance is low, and in darkness it is high, which is why LDRs are used in automatic lighting circuits.

Q14. A diode is included in a circuit because it

  • (A) stores electrical charge
  • (B) increases the supply voltage
  • (C) resists the current equally in both directions
  • (D) allows current to flow in one direction only
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Answer: D – A diode conducts only when it is connected in the forward direction. Connected the other way round it blocks the current, so it can be used to protect components or to rectify an alternating supply.

Q15. Which statement about a light-emitting diode (LED) is correct?

  • (A) It emits light only when current flows through it in the forward direction
  • (B) It emits light whichever way the current flows through it
  • (C) It emits light when it is connected in the reverse direction
  • (D) It works as a resistor and never emits light
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Answer: A – An LED is a diode that gives out light, so it must be connected the correct way round: in the forward direction. Connected in reverse it does not conduct and stays dark.

Q16. What is the function of a relay in an electrical circuit?

  • (A) To measure a large current safely
  • (B) To store energy for later use
  • (C) To use a small current in one circuit to switch on a large current in a separate circuit
  • (D) To reduce the resistance of a circuit
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Answer: C – In a relay, a small current flows through a coil and makes it act as an electromagnet, which pulls switch contacts together and completes a second circuit. This lets a delicate low-current control circuit switch a powerful motor or heater on and off safely.

Q17. What is the purpose of a fuse in a circuit?

  • (A) To increase the current in the circuit
  • (B) To melt and break the circuit if the current becomes too large
  • (C) To store charge in case of a power cut
  • (D) To change alternating current into direct current
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Answer: B – A fuse contains a thin wire that heats up and melts when the current exceeds its rating. This breaks the circuit and stops the excessive current before the wiring overheats or a fire starts.

Q18. A circuit is needed that switches an electric heater on automatically when the temperature falls. Which component should be used as the temperature sensor?

  • (A) An LDR
  • (B) A diode
  • (C) A variable resistor
  • (D) A thermistor
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Answer: D – A thermistor changes its resistance as the temperature changes, so it can sense the temperature and trigger the switching circuit. An LDR senses light level instead, not temperature.

Q19. Street lamps are designed to switch on automatically at dusk. Which component is used to detect the falling light level?

  • (A) An LDR
  • (B) A thermistor
  • (C) A fuse
  • (D) An ammeter
Show Answer

Answer: A – An LDR (light-dependent resistor) has a high resistance in the dark and a low resistance in bright light, so it can detect dusk and switch the lamp on.

Q20. A variable resistor is connected in series with a lamp. Why is it included in the circuit?

  • (A) To measure the voltage across the lamp
  • (B) To protect the lamp like a fuse
  • (C) To adjust the current in the circuit and so change the brightness of the lamp
  • (D) To keep the resistance of the circuit fixed
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Answer: C – Sliding the contact on a variable resistor changes its resistance. A larger resistance means a smaller current, so the lamp becomes dimmer; a smaller resistance makes the lamp brighter.


Potential Dividers, Power & Energy

Q21. A potential divider consists of a 2 kΩ resistor R1 in series with a 4 kΩ resistor R2, connected across a 12 V supply. The output voltage is taken across R2. What is the output voltage? [Supplement]

  • (A) 4 V
  • (B) 6 V
  • (C) 2 V
  • (D) 8 V
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Answer: D – Vout = Vin × R2 / (R1 + R2) = 12 × 4 / (2 + 4) = 12 × 4/6 = 8 V. The 4 V trap is the voltage across R1 instead.

Q22. Two equal resistors are connected in series across a 10 V supply to form a potential divider. What is the output voltage taken across one of the resistors? [Supplement]

  • (A) 10 V
  • (B) 5 V
  • (C) 2.5 V
  • (D) 0 V
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Answer: B – Vout = Vin × R2 / (R1 + R2). With equal resistors this is 10 × R / (R + R) = 10 × 1/2 = 5 V: the supply voltage is shared equally.

Q23. In a potential divider, the output voltage is taken across a thermistor connected in series with a fixed resistor. The temperature of the thermistor increases. What happens to the output voltage? [Supplement]

  • (A) It increases
  • (B) It stays the same
  • (C) It decreases
  • (D) It becomes zero immediately
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Answer: C – As the temperature rises, the resistance of the thermistor falls. The output is the thermistor’s share of the supply voltage, Vout = Vin × Rthermistor / (Rfixed + Rthermistor), so a smaller thermistor resistance means a smaller output voltage.

Q24. In a potential divider, the output voltage is taken across an LDR connected in series with a fixed resistor. As the surroundings get darker, what happens to the output voltage? [Supplement]

  • (A) It increases
  • (B) It decreases
  • (C) It stays the same
  • (D) It falls to zero
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Answer: A – In the dark the resistance of the LDR rises, so it takes a larger share of the supply voltage and the output voltage across it increases. Circuits like this are used to switch on security lights at night.

Q25. A lamp is connected to a 12 V supply and draws a current of 0.50 A. What is the power of the lamp?

  • (A) 24 W
  • (B) 6.0 W
  • (C) 0.042 W
  • (D) 12 W
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Answer: B – P = IV = 12 × 0.50 = 6.0 W. (Dividing 0.50 by 12 gives the 0.042 W trap.)

Q26. A current of 2.0 A flows through a resistor of 5 Ω. What is the power dissipated in the resistor? [Supplement]

  • (A) 4.0 W
  • (B) 10 W
  • (C) 50 W
  • (D) 20 W
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Answer: D – Because the current through the resistor is known, use P = I2R = (2.0)2 × 5 = 4 × 5 = 20 W. (The 10 W trap is I × R, which is the voltage across the resistor, not the power.)

Q27. A potential difference of 12 V is applied across a resistor of 6 Ω. What is the power dissipated in the resistor? [Supplement]

  • (A) 24 W
  • (B) 2.0 W
  • (C) 12 W
  • (D) 72 W
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Answer: A – Because the voltage across the resistor is known, use P = V2 / R = 122 / 6 = 144 / 6 = 24 W. (The 2.0 W trap is actually the current, I = V / R = 2.0 A, not a power.)

Q28. A device is connected to a 6.0 V supply and draws a current of 0.50 A for 200 s. How much electrical energy is transferred?

  • (A) 3.0 J
  • (B) 300 J
  • (C) 600 J
  • (D) 6000 J
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Answer: C – E = IVt = 6.0 × 0.50 × 200 = 600 J. (The 3.0 J trap is just the power, P = IV = 3.0 W, without multiplying by the time.)

Q29. A 2 kW electric heater is switched on for 3 hours. How much energy, in kilowatt-hours (units), does it use?

  • (A) 0.67 kWh
  • (B) 6 kWh
  • (C) 1.5 kWh
  • (D) 6000 kWh
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Answer: B – Energy in kWh = power in kW × time in hours = 2 × 3 = 6 kWh. This is the “unit” shown on an electricity bill.

Q30. Two identical lamps can be connected to a battery either in series or in parallel. In which arrangement is each lamp brighter, and why?

  • (A) In series, because each lamp receives the full battery voltage
  • (B) In series, because the current through each lamp is larger
  • (C) In parallel, because the current through each lamp is smaller
  • (D) In parallel, because each lamp has the full battery voltage across it and so dissipates more power
Show Answer

Answer: D – In parallel, each lamp has the full battery voltage across it, so from P = IV each lamp dissipates its full power and shines brightly. In series the voltage is shared between the lamps, so each receives only part of the voltage and is dimmer.

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