IGCSE Physics 4.2 Electrical Quantities MCQs – 30 Practice Questions with Answers (Cambridge 0625)

Free practice questions for Cambridge IGCSE Physics (0625) Section 4.2, Electrical Quantities, covering current and charge, voltage and EMF, and resistance and Ohm’s law for both Core and Supplement students. Every question is an original practice question with a fully worked answer – tap “Show Answer” to check your working.

Formula recap: I = Q / t  |  V = E / Q  |  E = QV  |  R = V / I  |  R = ρL / A (Supplement)

Revise all the equations in one place on our IGCSE Physics Formula Sheet.

Current & Charge

Q1. Which statement correctly defines electric current?

  • (A) The energy transferred by each unit of charge
  • (B) The rate at which charge flows past a point
  • (C) The total charge stored in a battery
  • (D) The speed of the electrons in a wire
Show Answer

Answer: B – Current is the rate of flow of charge, I = Q / t, measured in amperes (1 A = 1 C per second). Energy per unit charge is the definition of voltage, not current.

Q2. A charge of 480 C flows past a point in a wire in 4.0 minutes. What is the current in the wire?

  • (A) 120 A
  • (B) 1920 A
  • (C) 2.0 A
  • (D) 0.50 A
Show Answer

Answer: C – First convert the time: 4.0 min = 240 s. I = Q / t = 480 / 240 = 2.0 A. (Option A is the trap answer from forgetting to convert minutes to seconds.)

Q3. A current of 0.50 A flows in a circuit for 2.0 minutes. How much charge passes?

  • (A) 60 C
  • (B) 1.0 C
  • (C) 120 C
  • (D) 0.25 C
Show Answer

Answer: A – Time = 2.0 min = 120 s. Q = I × t = 0.50 × 120 = 60 C.

Q4. A metal wire is connected in a circuit with a cell. Which statement about the directions of current and electron flow in the external circuit is correct?

  • (A) Electrons and conventional current both flow from the positive terminal to the negative terminal
  • (B) Conventional current is the direction in which the electrons actually move
  • (C) Electrons flow from the positive terminal to the negative terminal, in the same direction as conventional current
  • (D) Conventional current flows from the positive terminal to the negative terminal, while the electrons flow from the negative terminal to the positive terminal
Show Answer

Answer: D – Conventional current is defined as flowing from + to − around the external circuit (the direction positive charge would move). Electrons are negatively charged, so in a metal they drift the opposite way, from − to +.

Q5. An ammeter is connected in series in a circuit. Why must an ammeter have a very low resistance?

  • (A) So that it does not significantly reduce the current it is measuring
  • (B) So that it increases the current in the circuit
  • (C) So that it can also measure potential difference
  • (D) So that no current passes through the rest of the circuit
Show Answer

Answer: A – In series, the ammeter carries the full circuit current. If its resistance were large, it would reduce the current and give a false (too low) reading, so it is made with almost zero resistance.

Q6. A current is measured as 350 mA. What is this current in amperes?

  • (A) 0.035 A
  • (B) 0.35 A
  • (C) 3.5 A
  • (D) 350 A
Show Answer

Answer: B – 1 mA = 10-3 A, so 350 mA = 350 × 10-3 = 0.35 A.

Q7. A small current is measured as 2.4 × 10-3 A. What is this current in microamperes (µA)?

  • (A) 2.4 µA
  • (B) 240 µA
  • (C) 2400 µA
  • (D) 24 000 µA
Show Answer

Answer: C – 1 µA = 10-6 A, so the current = 2.4 × 10-3 / 10-6 = 2.4 × 103 = 2400 µA.

Q8. Which row correctly identifies the charge carriers in a metal and in an electrolyte (such as an aqueous solution)?

  • (A) Metal: free electrons; Electrolyte: positive and negative ions
  • (B) Metal: protons; Electrolyte: free electrons
  • (C) Metal: positive ions; Electrolyte: free electrons
  • (D) Metal: free electrons; Electrolyte: free electrons only
Show Answer

Answer: A – In metals the moving charges are free (delocalised) electrons; the positive ions stay fixed in the lattice. In an electrolyte there are no free electrons – the current is carried by positive ions moving one way and negative ions moving the other way.

Q9. A current of 3.0 A flows in a lamp. How long does it take for 900 C of charge to pass through the lamp?

  • (A) 2700 s
  • (B) 300 s
  • (C) 0.0033 s
  • (D) 30 s
Show Answer

Answer: B – t = Q / I = 900 / 3.0 = 300 s (5.0 minutes).

Q10. A current of 5.0 mA flows for 100 s. How much charge passes?

  • (A) 500 C
  • (B) 50 C
  • (C) 0.50 C
  • (D) 5.0 C
Show Answer

Answer: C – Convert first: 5.0 mA = 0.0050 A. Q = I × t = 0.0050 × 100 = 0.50 C. (Option A is the trap from not converting mA to A.)

Voltage, EMF & Energy

Q11. The potential difference (p.d.) between two points in a circuit is defined as…

  • (A) The charge flowing per unit time between the points
  • (B) The energy transferred per unit charge between the points
  • (C) The force acting on each unit of charge
  • (D) The power developed per unit of charge
Show Answer

Answer: B – V = E / Q. A p.d. of 1 volt means 1 joule of energy is transferred for each coulomb of charge that passes.

Q12. A charge of 30 C passes through a component and transfers 240 J of energy. What is the p.d. across the component?

  • (A) 7200 V
  • (B) 0.125 V
  • (C) 8.0 V
  • (D) 270 V
Show Answer

Answer: C – V = E / Q = 240 / 30 = 8.0 V.

Q13. Which statement about the electromotive force (EMF) of a cell is correct?

  • (A) It is measured in newtons
  • (B) It is the energy supplied by the cell to each unit of charge that passes through it
  • (C) It is the energy used by the external components per unit charge
  • (D) It is exactly the same quantity as the p.d. across a resistor
Show Answer

Answer: B – EMF (in volts) is the total energy the source supplies per unit charge, converting chemical energy into electrical energy. A p.d., by contrast, is the energy transferred to a component (or part of the circuit) per unit charge.

Q14. A voltmeter is used to measure the p.d. across a lamp. How is it connected, and what resistance should it have?

  • (A) In series with the lamp; very low resistance
  • (B) In parallel with the lamp; very low resistance
  • (C) In parallel with the lamp; very high resistance
  • (D) In series with the lamp; very high resistance
Show Answer

Answer: C – A voltmeter is always connected in parallel across the component. Its resistance is made very high so that it draws almost no current and does not change the p.d. it is measuring.

Q15. A 12 V battery supplies a circuit. A charge of 200 C passes through the battery. How much energy does the battery transfer?

  • (A) 16.7 J
  • (B) 2400 J
  • (C) 212 J
  • (D) 0.060 J
Show Answer

Answer: B – E = QV = 200 × 12 = 2400 J.

Q16. Three cells, each of EMF 1.5 V, are connected in series. What is the total EMF of the combination?

  • (A) 1.5 V
  • (B) 0.50 V
  • (C) 3.0 V
  • (D) 4.5 V
Show Answer

Answer: D – When identical cells are connected in series (positive to negative), their EMFs add up: 1.5 + 1.5 + 1.5 = 4.5 V.

Q17. A device connected to a 6.0 V supply transfers 720 J of electrical energy. How much charge passes through the device?

  • (A) 4320 C
  • (B) 120 C
  • (C) 0.0083 C
  • (D) 726 C
Show Answer

Answer: B – From E = QV, Q = E / V = 720 / 6.0 = 120 C.

Q18. A charge of 15 C flows through a lamp and transfers 90 J of energy to the lamp. What is the p.d. across the lamp?

  • (A) 1350 V
  • (B) 0.17 V
  • (C) 105 V
  • (D) 6.0 V
Show Answer

Answer: D – V = E / Q = 90 / 15 = 6.0 V.

Q19. A student wants to measure both the current through a resistor and the p.d. across it. Which arrangement is correct?

  • (A) Ammeter in series with the resistor; voltmeter in parallel with the resistor
  • (B) Ammeter in parallel with the resistor; voltmeter in series with the resistor
  • (C) Both meters in series with the resistor
  • (D) Both meters in parallel with the resistor
Show Answer

Answer: A – The ammeter must be in series so the full current passes through it, and the voltmeter must be in parallel across the resistor so it measures the p.d. between its two ends.

Q20. A charge of 40 C flows from a 230 V mains supply through a heater. How much electrical energy is transferred?

  • (A) 9200 J
  • (B) 5.75 J
  • (C) 270 J
  • (D) 0.17 J
Show Answer

Answer: A – E = QV = 40 × 230 = 9200 J (9.2 kJ).


Resistance & Ohm’s Law

Q21. A p.d. of 12 V across a resistor produces a current of 3.0 A. What is the resistance of the resistor?

  • (A) 36 Ω
  • (B) 0.25 Ω
  • (C) 4.0 Ω
  • (D) 15 Ω
Show Answer

Answer: C – R = V / I = 12 / 3.0 = 4.0 Ω.

Q22. One ohm (1 Ω) is defined as the resistance of a conductor in which…

  • (A) a p.d. of 1 V produces a current of 1 A
  • (B) a charge of 1 C flows in 1 s
  • (C) an energy of 1 J is transferred per coulomb
  • (D) a current of 1 A transfers 1 J in total
Show Answer

Answer: A – From R = V / I, a resistance of 1 Ω means 1 V is needed to drive a current of 1 A. (Option B defines the ampere, and option C defines the volt.)

Q23. A metal resistor is kept at constant temperature and the current through it is measured for different values of p.d. Which graph is obtained?

  • (A) A curve that bends towards the p.d. axis
  • (B) A straight line that does not pass through the origin
  • (C) A horizontal straight line
  • (D) A straight line through the origin
Show Answer

Answer: D – For an ohmic conductor at constant temperature, I is directly proportional to V (Ohm’s law), so the I-V graph is a straight line through the origin and its gradient is constant.

Q24. Why does the resistance of a filament lamp increase as the current through it increases?

  • (A) The filament gets hotter as the current increases, and the resistance of the hot filament is greater
  • (B) The electrons in the filament are gradually used up
  • (C) The p.d. across the filament decreases
  • (D) The filament becomes thinner because of the larger current
Show Answer

Answer: A – A larger current heats the filament to a higher temperature. The resistance of a metal increases with temperature, so the I-V graph for a filament lamp curves instead of being a straight line.

Q25. A uniform wire of length 0.80 m has a resistance of 6.0 Ω. A second wire of the same material and same cross-sectional area has a length of 1.2 m. What is its resistance?

  • (A) 4.0 Ω
  • (B) 3.0 Ω
  • (C) 12 Ω
  • (D) 9.0 Ω
Show Answer

Answer: D – Resistance is directly proportional to length. The length ratio is 1.2 / 0.80 = 1.5, so R = 6.0 × 1.5 = 9.0 Ω.

Q26. A wire has a resistance of 8.0 Ω. It is replaced by a wire of the same material and the same length, but with double the cross-sectional area. What is the new resistance?

  • (A) 16 Ω
  • (B) 4.0 Ω
  • (C) 8.0 Ω
  • (D) 2.0 Ω
Show Answer

Answer: B – Resistance is inversely proportional to cross-sectional area. Doubling the area halves the resistance: 8.0 / 2 = 4.0 Ω.

Q27. A copper wire of length 2.0 m has a cross-sectional area of 0.50 mm2. The resistivity of copper is 1.7 × 10-8 Ω m. What is the resistance of the wire? [Supplement]

  • (A) 6.8 Ω
  • (B) 0.68 Ω
  • (C) 68 Ω
  • (D) 0.068 Ω
Show Answer

Answer: D – Convert the area: 0.50 mm2 = 0.50 × 10-6 m2. R = ρL / A = (1.7 × 10-8 × 2.0) / (0.50 × 10-6) = 3.4 × 10-8 / 5.0 × 10-7 = 0.068 Ω.

Q28. A wire of length 1.0 m and cross-sectional area 2.0 × 10-7 m2 has a resistance of 5.0 Ω. What is the resistivity of the wire material? [Supplement]

  • (A) 1.0 × 10-6 Ω m
  • (B) 2.5 × 107 Ω m
  • (C) 1.0 × 10-5 Ω m
  • (D) 4.0 × 10-7 Ω m
Show Answer

Answer: A – From R = ρL / A, ρ = RA / L = (5.0 × 2.0 × 10-7) / 1.0 = 1.0 × 10-6 Ω m.

Q29. Four wires are available. Which wire has the least resistance?

  • (A) A long, thin copper wire
  • (B) A long, thick wire made of a high-resistivity alloy
  • (C) A short, thick copper wire
  • (D) A short, thin wire made of a high-resistivity alloy
Show Answer

Answer: C – Resistance decreases with shorter length, larger cross-sectional area, and lower resistivity. Copper has a low resistivity, so a short, thick copper wire has the least resistance.

Q30. When a metal wire is heated, its resistance increases. What is the reason for this increase?

  • (A) The number of free electrons in the wire falls to almost zero
  • (B) The wire melts and becomes a liquid
  • (C) The current through the wire becomes zero
  • (D) The metal ions vibrate with larger amplitude, making it harder for the electrons to pass through
Show Answer

Answer: D – At a higher temperature the lattice ions vibrate more vigorously and obstruct the drift of the free electrons, so the resistance of a metal rises with temperature.

Keep practising: Explore more boards and topics on our Physics MCQs hub page, and keep the IGCSE Physics Formula Sheet open while you revise – every equation for Section 4.2 is listed there.

All questions are original practice questions written for QuaidianPoint – not Cambridge past-paper questions. QuaidianPoint is not affiliated with Cambridge Assessment International Education.