A Level Physics 9.1 Electric Current MCQs – 30 Practice Questions with Answers (Cambridge 9702)
This page has 30 original multiple-choice questions on Section 9.1 Electric Current of the Cambridge AS & A Level Physics (9702) syllabus. Work through charge and current, Kirchhoff’s first law and charge carriers, then test yourself on the drift-speed equation I = nAvq.
Formula recap: I = Q / t · Q = Ne, where e = 1.6 × 10−19 C · I = nAvq. For the full list of equations, see the A Level Physics Formula Sheet.
Charge & Current
Q1. Electric current in a wire is best defined as
- (A) the total charge stored in the wire
- (B) the energy transferred per unit charge
- (C) the rate of flow of electric charge
- (D) the number of electrons in the conductor
Show Answer
Answer: C – Current is the rate at which charge flows past a point: I = Q / t. It is measured in amperes, where 1 A = 1 C s−1.
Q2. A steady current of 2.0 A flows in a circuit for 5.0 minutes. How much charge passes a point in the circuit?
- (A) 10 C
- (B) 600 C
- (C) 60 C
- (D) 300 C
Show Answer
Answer: B – t = 5.0 × 60 = 300 s. Q = It = 2.0 × 300 = 600 C.
Q3. A current of 3.0 mA flows through a sensor for 2.0 hours. What total charge flows?
- (A) 21.6 C
- (B) 6.0 C
- (C) 216 C
- (D) 2.16 C
Show Answer
Answer: A – I = 3.0 × 10−3 A and t = 2.0 × 3600 = 7200 s. Q = It = 3.0 × 10−3 × 7200 = 21.6 C.
Q4. What is the magnitude of the charge on a single electron?
- (A) 1.6 × 10−16 C
- (B) 1.6 × 10−23 C
- (C) 6.2 × 1018 C
- (D) 1.6 × 10−19 C
Show Answer
Answer: D – The elementary charge is e = 1.6 × 10−19 C. An electron carries charge −e.
Q5. A charge of 3.2 C flows through a lamp. How many electrons pass through the lamp?
- (A) 2.0 × 1018
- (B) 2.0 × 1019
- (C) 5.0 × 1018
- (D) 2.0 × 1020
Show Answer
Answer: B – Q = Ne, so N = Q / e = 3.2 / (1.6 × 10−19) = 2.0 × 1019 electrons.
Q6. In a electrolysis experiment, 5.0 × 1020 singly-charged ions reach an electrode. What charge do they deliver?
- (A) 8.0 C
- (B) 32 C
- (C) 80 C
- (D) 800 C
Show Answer
Answer: C – Each ion carries charge e, so Q = Ne = 5.0 × 1020 × 1.6 × 10−19 = 80 C.
Q7. Which statement about conventional current and electron flow in a metal wire connected to a battery is correct?
- (A) Electrons flow from the negative terminal to the positive terminal through the external circuit, opposite to the conventional current
- (B) Electrons flow in the same direction as the conventional current
- (C) The conventional current in the wire is carried by protons
- (D) Electrons flow from the positive terminal to the negative terminal through the external circuit
Show Answer
Answer: A – Conventional current is defined as the direction of flow of positive charge, from + to − outside the battery. Electrons, being negative, actually drift the other way, from − to +.
Q8. A current of 250 μA flows for 40 s. What charge passes?
- (A) 10 C
- (B) 0.10 C
- (C) 1.0 C
- (D) 0.010 C
Show Answer
Answer: D – I = 250 × 10−6 A. Q = It = 250 × 10−6 × 40 = 1.0 × 10−2 C = 0.010 C.
Q9. A battery delivers a charge of 48 C while supplying a steady current of 4.0 A. For how long does it supply this current?
- (A) 192 s
- (B) 0.083 s
- (C) 12 s
- (D) 44 s
Show Answer
Answer: C – t = Q / I = 48 / 4.0 = 12 s.
Q10. The coulomb, the unit of charge, is equivalent to
- (A) 1 A s−1
- (B) 1 A s
- (C) 1 V s
- (D) 1 W V−1
Show Answer
Answer: B – From Q = It, charge = current × time, so 1 C = 1 A s. (1 W V−1 is 1 A, not a charge.)
Current in Circuits & Charge Carriers
Q11. In a metal wire, the current is carried by
- (A) free (delocalised) electrons
- (B) positive metal ions
- (C) protons
- (D) neutral atoms
Show Answer
Answer: A – The positive ions stay fixed in the metal lattice. The free electrons drift through the lattice and carry the charge.
Q12. In an electrolyte such as copper sulfate solution, the current inside the liquid is carried by
- (A) electrons only
- (B) positive ions only
- (C) negative ions only
- (D) positive and negative ions moving in opposite directions
Show Answer
Answer: D – Positive ions drift in the direction of the conventional current and negative ions drift the opposite way. Both movements transfer charge in the same sense, so both contribute to the current.
Q13. Ammeters are placed at several different positions in a series circuit containing a battery and two resistors. Which statement is correct?
- (A) The current is largest nearest the positive terminal
- (B) The current gets smaller each time it passes through a resistor
- (C) The current is the same at every position in the circuit
- (D) The current is zero inside a resistor
Show Answer
Answer: C – Charge is not used up in components; the same charge per second passes every point of a series circuit. Resistors transfer energy from the charge, but they do not consume the charge itself.
Q14. At a junction in a circuit, a current of 5.0 A flows in. One branch carries 2.0 A away from the junction. What current flows away along the only other branch?
- (A) 7.0 A
- (B) 3.0 A
- (C) 2.0 A
- (D) 10 A
Show Answer
Answer: B – By Kirchhoff’s first law, total current in = total current out: 5.0 = 2.0 + I, so I = 3.0 A.
Q15. Kirchhoff’s first law (the sum of the currents entering a junction equals the sum leaving it) is a consequence of
- (A) conservation of energy
- (B) conservation of momentum
- (C) conservation of mass
- (D) conservation of charge
Show Answer
Answer: D – Charge cannot be created, destroyed or stored up at a junction, so the charge flowing in each second must equal the charge flowing out each second.
Q16. How is an ammeter connected in a circuit, and what resistance does an ideal ammeter have?
- (A) In series with the component; zero resistance
- (B) In parallel with the component; zero resistance
- (C) In series with the component; very large resistance
- (D) In parallel with the component; very large resistance
Show Answer
Answer: A – An ammeter measures the current through a component, so it is placed in series with it. An ideal ammeter has zero resistance so that it does not reduce the current it is measuring.
Q17. Currents of 2.5 A and 1.5 A flow into a junction. One wire carries 0.80 A away from the junction. What current flows away along the remaining wire?
- (A) 4.8 A
- (B) 0.20 A
- (C) 3.2 A
- (D) 4.0 A
Show Answer
Answer: C – Total current in = 2.5 + 1.5 = 4.0 A. Total current out = 0.80 + I = 4.0 A, so I = 3.2 A.
Q18. A real ammeter is made with a very low resistance so that
- (A) it increases the current to a value that is easier to measure
- (B) connecting it hardly changes the current in the circuit
- (C) it does not get hot when current passes through it
- (D) it can also be used to measure potential difference
Show Answer
Answer: B – The ammeter is in series with the circuit, so any resistance it adds would reduce the current. Keeping its resistance very small keeps this effect negligible.
Q19. A current of 0.50 A flows in a circuit. How much charge passes any given point in the circuit in 1.0 minute?
- (A) 0.50 C
- (B) 120 C
- (C) 3.0 C
- (D) 30 C
Show Answer
Answer: D – Q = It = 0.50 × 60 = 30 C. The same charge passes every point of the series circuit in that time.
Q20. Conventional current flows from left to right through an electrolyte. Which statement about the ions is correct?
- (A) Positive ions move from left to right and negative ions move from right to left
- (B) Both positive and negative ions move from left to right
- (C) Neither type of ion moves; only electrons carry the current
- (D) Negative ions stay still and positive ions move from right to left
Show Answer
Answer: A – Positive ions drift in the direction of the conventional current (left to right) and negative ions drift in the opposite direction. The two movements add together to give the current.
Drift Speed
Q21. In the equation I = nAvq for a current-carrying wire, the symbol n represents
- (A) the total number of charge carriers in the wire
- (B) the number of charge carriers per unit volume of the conductor
- (C) the charge carried by each charge carrier
- (D) the cross-sectional area of the wire
Show Answer
Answer: B – n is the number density of charge carriers, in m−3. In the equation, A is the cross-sectional area, v is the mean drift speed of the carriers and q is the charge on each carrier (q = e for electrons).
Q22. A copper wire of cross-sectional area 2.0 × 10−6 m2 carries a current of 1.6 A. The number density of free electrons in copper is 8.0 × 1028 m−3. What is the mean drift speed of the electrons?
- (A) 6.3 × 10−4 m s−1
- (B) 6.3 × 10−3 m s−1
- (C) 6.3 × 10−5 m s−1
- (D) 6.3 × 10−6 m s−1
Show Answer
Answer: C – v = I / (nAq) = 1.6 / (8.0 × 1028 × 2.0 × 10−6 × 1.6 × 10−19) = 1.6 / (2.56 × 104) = 6.3 × 10−5 m s−1. This is only about 0.2 m per hour – drift speeds are extremely small.
Q23. A wire of diameter 1.0 mm carries a current of 2.0 A. The conductor has a free-electron number density of 8.0 × 1028 m−3. What is the mean drift speed of the electrons?
- (A) 2.0 × 10−5 m s−1
- (B) 5.0 × 10−5 m s−1
- (C) 1.0 × 10−4 m s−1
- (D) 2.0 × 10−4 m s−1
Show Answer
Answer: D – First find the area from the radius, r = 0.50 mm = 5.0 × 10−4 m: A = πr2 = π × (5.0 × 10−4)2 = 7.9 × 10−7 m2. Then v = I / (nAq) = 2.0 / (8.0 × 1028 × 7.9 × 10−7 × 1.6 × 10−19) ≈ 2.0 × 10−4 m s−1. (Using the diameter as the radius makes the area four times too large and gives 5.0 × 10−5 m s−1.)
Q24. The current in a wire is doubled, while the wire itself is unchanged. What happens to the mean drift speed of the electrons?
- (A) It doubles
- (B) It halves
- (C) It stays the same
- (D) It becomes four times as large
Show Answer
Answer: A – From I = nAvq, with n, A and q fixed, v is proportional to I. Doubling the current means the electrons drift twice as fast.
Q25. A wire is replaced by a wire of the same metal but twice the cross-sectional area, carrying the same current. The mean drift speed of the electrons is now
- (A) twice as large
- (B) unchanged
- (C) half as large
- (D) one quarter as large
Show Answer
Answer: C – v = I / (nAq). With I, n and q fixed, v is inversely proportional to A, so doubling A halves the drift speed: there are twice as many electrons passing each second, so each one needs to move only half as fast.
Q26. A lamp lights almost instantly when the switch is closed, even though the electrons in the wires drift at only about 10−4 m s−1. Why?
- (A) Electrons travel from the switch to the lamp at the speed of light
- (B) An electric field is set up through the whole circuit almost instantaneously, so the free electrons everywhere begin drifting at the same time
- (C) The electrons in the wire are already moving towards the lamp at high speed
- (D) The battery ejects electrons at the speed of light
Show Answer
Answer: B – The signal (the electric field) spreads through the circuit at a speed close to the speed of light. The wire is already full of free electrons along its whole length, so electrons near the lamp start drifting through the lamp immediately – no electron has to travel from the switch.
Q27. In a wire of cross-sectional area 1.0 × 10−6 m2, the free-electron number density is 5.0 × 1028 m−3 and the electrons drift at 2.0 × 10−4 m s−1. What is the current in the wire?
- (A) 0.16 A
- (B) 16 A
- (C) 0.80 A
- (D) 1.6 A
Show Answer
Answer: D – I = nAvq = 5.0 × 1028 × 1.0 × 10−6 × 2.0 × 10−4 × 1.6 × 10−19 = 1.6 A.
Q28. Two wires P and Q of the same metal carry the same current. Wire P has half the diameter of wire Q. The mean drift speed of the electrons in P, compared with that in Q, is
- (A) four times as large
- (B) twice as large
- (C) the same
- (D) one quarter as large
Show Answer
Answer: A – Area is proportional to diameter squared, so halving the diameter makes the area one quarter as large. Since v = I / (nAq) with I, n and q the same, the drift speed in P is four times that in Q.
Q29. A wire is replaced by one of a different metal that has twice the number density of charge carriers. The new wire has the same cross-sectional area and carries the same current. The mean drift speed is now
- (A) twice as large
- (B) half as large
- (C) unchanged
- (D) four times as large
Show Answer
Answer: B – v = I / (nAq). With I, A and q fixed, v is inversely proportional to n, so doubling the number density halves the drift speed.
Q30. A wire of cross-sectional area 3.0 × 10−6 m2 has a free-electron number density of 1.0 × 1029 m−3 and carries a current of 4.8 A. What is the mean drift speed of the electrons?
- (A) 1.0 × 10−3 m s−1
- (B) 1.0 × 10−5 m s−1
- (C) 1.0 × 10−4 m s−1
- (D) 3.2 × 10−4 m s−1
Show Answer
Answer: C – v = I / (nAq) = 4.8 / (1.0 × 1029 × 3.0 × 10−6 × 1.6 × 10−19) = 4.8 / (4.8 × 104) = 1.0 × 10−4 m s−1, a typical drift speed in a metal.
Keep practising: find more topic-by-topic questions in the Physics MCQs section, and revise every equation for this topic on the A Level Physics Formula Sheet.
These are original practice questions written for revision; they are not Cambridge past-paper questions, and this page is not affiliated with or endorsed by Cambridge Assessment International Education.