These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data and assumptions for every question: e = 1.60 × 10⁻¹⁹ C; electrons have charge −e; the ions named carry charge +e or −e; “current” means conventional current, the direction positive charge would move. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
A steady current of 0.25 A flows through a heating element for 2.0 minutes. How much charge passes through the element?
- (A) 30 C
- (B) 0.50 C
- (C) 480 C
- (D) 2.1 × 10⁻³ C
Answer and explanation
Answer: (A). Convert the time first: 2.0 min = 120 s. Then ΔQ = IΔt = (0.25 A)(120 s) = 30 C.
- (B) uses 2.0 as the time in seconds without converting minutes.
- (C) divides the time by the current (120 ÷ 0.25). The units of that are s/A, not coulombs.
- (D) divides the current by the time (0.25 ÷ 120). That would be a rate of change of current, not a charge.
Question 2 (multiple choice · core)
A copper wire lies on a bench, not connected to anything. There is no current in it. Which statement about the free electrons in the wire is correct?
- (A) They move rapidly in random directions, so equal numbers cross any cross-section each way on average.
- (B) They are at rest, because there is no potential difference to push them.
- (C) They all drift slowly in the same direction, but too slowly to count as a current.
- (D) They have left the wire, because copper only has free electrons when it is connected to a battery.
Answer and explanation
Answer: (A). The free electrons in a metal always move randomly at very high speeds. Zero current means the net charge passing a cross-section is zero, not that the carriers are still.
- (B) confuses zero net motion with zero motion. A potential difference is needed for a drift, not for motion itself.
- (C) is wrong because any steady drift of charge in one direction is a current, however slow. With no potential difference there is no drift.
- (D) is wrong because the free electrons belong to the metal. A battery does not supply the carriers.
Question 3 (multiple choice · core)
Inside an evacuated glass tube, a beam of electrons travels from an electrode at the west end to a screen at the east end. Which statement describes the current carried by the beam?
- (A) The current is directed from east to west.
- (B) The current is directed from west to east, because the electrons carry the current.
- (C) There is no current, because there is no wire.
- (D) The current is directed from east to west only if positive ions also move in the tube.
Answer and explanation
Answer: (A). Conventional current is the direction positive charge would move. Negative charges moving east are equivalent to positive charge moving west, so the current is from east to west.
- (B) gives the direction the electrons move. Conventional current is opposite to the motion of negative carriers.
- (C) is wrong because current is any net flow of charge through a cross-section. A beam in a vacuum is a current.
- (D) is wrong because the electrons alone give a current from east to west. Positive ions, if present, would add to it, not create it.
Question 4 (multiple choice · core)
The current in a small sensor is 4.0 mA. How many electrons pass through the sensor each second?
- (A) 2.5 × 10¹⁶
- (B) 2.5 × 10¹⁹
- (C) 6.4 × 10⁻²²
- (D) 4.0 × 10⁻¹⁷
Answer and explanation
Answer: (A). In 1 s, ΔQ = IΔt = 4.0 × 10⁻³ C. The number of electrons is N = ΔQ / e = (4.0 × 10⁻³ C) ÷ (1.60 × 10⁻¹⁹ C) = 2.5 × 10¹⁶.
- (B) uses 4.0 A instead of 4.0 mA: the milli prefix (10⁻³) was dropped.
- (C) multiplies the current by e instead of dividing. A number of electrons cannot be smaller than 1 here.
- (D) divides e by the current, which is upside down.
Question 5 (calculation · core)
A battery-powered drill is switched on. The current in its motor is constant at 0.40 A from t = 0 to t = 3.0 s. The drill is then released and the current falls in a straight line to zero at t = 5.0 s.
(a) Sketch the current–time graph, with labelled axes and values. (b) Calculate the total charge that passes through the motor. (c) Calculate the average current over the 5.0 s. (d) Calculate the number of electrons that pass through the motor.
Worked solution and suggested Marlbridge rubric
(a) Axes: current I (A) vertically, time t (s) horizontally. A horizontal line at 0.40 A from 0 to 3.0 s, then a straight line sloping down to 0 A at 5.0 s.
(b) Area under the graph:
- rectangle: (0.40 A)(3.0 s) = 1.2 C
- triangle: ½(2.0 s)(0.40 A) = 0.40 C
- total ΔQ = 1.6 C.
(c) I_average = 1.6 C ÷ 5.0 s = 0.32 A.
(d) N = 1.6 C ÷ (1.60 × 10⁻¹⁹ C) = 1.0 × 10¹⁹ electrons.
| Point | What earns it |
|---|---|
| 1 | Graph with labelled axes and units, flat at 0.40 A to 3.0 s, then a straight fall to zero at 5.0 s |
| 1 | Charge found as the area under the graph, including the triangle: 1.6 C |
| 1 | Average current 0.32 A, from total charge divided by total time |
| 1 | Number of electrons 1.0 × 10¹⁹, consistent with the charge in (b) |
Common error: multiplying the peak current by the whole time, (0.40 A)(5.0 s) = 2.0 C. That ignores the fall in current after 3.0 s. Allow carried-forward answers in (c) and (d) from a wrong value in (b).
Question 6 (constructed response · core)
A glowing gas tube has an electrode at each end. Each second, 3.0 × 10¹⁷ electrons move through the middle of the tube towards the right-hand electrode, and 1.0 × 10¹⁷ positive ions, each of charge +e, move towards the left-hand electrode.
(a) Which electrode is at the higher electric potential? Justify your answer. (b) Calculate the size of the current in the tube and state its direction. (c) A student says: “The electrons and ions move in opposite directions, so their currents partly cancel. The current is (3.0 × 10¹⁷ − 1.0 × 10¹⁷) × e = 0.032 A.” Evaluate this claim. (d) How much charge passes through the middle of the tube in 1.0 minute?
Worked solution and suggested Marlbridge rubric
(a) The right-hand electrode. Negative charges are pushed towards higher potential, and the electrons move right. This agrees with the positive ions, which are pushed towards lower potential and move left.
(b) Electrons moving right carry negative charge to the right, which is the same as positive charge moving left. The ions carry positive charge to the left. Both contributions point left, so they add: I = (3.0 × 10¹⁷ + 1.0 × 10¹⁷)(1.60 × 10⁻¹⁹ C) per second = 0.064 A, directed to the left (towards the left-hand electrode).
(c) The claim is wrong. The two kinds of carrier do move in opposite directions, but they have opposite charges. A negative charge moving right changes the charge on each side in the same way as a positive charge moving left. So the contributions add, giving 0.064 A, not 0.032 A.
(d) ΔQ = IΔt = (0.064 A)(60 s) = 3.8 C (3.84 C).
| Point | What earns it |
|---|---|
| 1 | Right-hand electrode, justified by electrons moving towards higher potential (or ions towards lower potential) |
| 1 | Both contributions identified as being in the same direction (to the left) |
| 1 | I = 0.064 A, directed to the left |
| 1 | Rejects the claim because opposite charges moving in opposite directions give currents in the same direction |
| 1 | ΔQ = 3.8 C, with time converted to 60 s |
Allow a carried-forward answer in (d) from (b). Do not award the (c) point for “the student is wrong” with no reason.
Question 7 (constructed response · stretch)
A lamp is connected to a battery by two wires, each 1.5 m long. Point P is on the wire between the positive terminal and the lamp. Point Q is on the wire between the lamp and the negative terminal. When the circuit is complete, the lamp lights within a tiny fraction of a second. The drift speed of the electrons in the wires is about 0.10 mm/s.
(a) Compare the current at P with the current at Q. Justify your answer. (b) At point P, which way do the electrons drift: towards the lamp or away from it? Explain. (c) A student says: “The lamp lights at once because electrons leave the negative terminal and race through the wire to the lamp.” Use the drift speed to evaluate this claim, then explain why the lamp does light quickly. (d) The battery is disconnected. Describe the motion of the free electrons in the wire at P afterwards, compared with their motion while the lamp was lit.
Worked solution and suggested Marlbridge rubric
(a) The currents are equal. Charge is conserved and does not build up in the lamp, so every coulomb per second that enters the lamp from P leaves it towards Q. The lamp transfers energy, not charge.
(b) Away from the lamp, towards the positive terminal. Conventional current leaves the positive terminal and passes P on its way to the lamp. Electrons are negative, so they drift the opposite way.
(c) At 0.10 mm/s, an electron would need t = 1.5 m ÷ (1.0 × 10⁻⁴ m/s) = 1.5 × 10⁴ s, about 4.2 hours, to drift 1.5 m. The lamp lights in far less time, so the claim is wrong. The lamp lights quickly because the wires and filament are already full of free electrons. When the circuit is completed, an electric field is set up along the whole loop almost at once, and it pushes on the carriers everywhere, including those already in the filament.
(d) The free electrons keep moving rapidly in random directions, just as before. The only change is that the slow drift stops, so the net motion through a cross-section, and therefore the current, becomes zero.
| Point | What earns it |
|---|---|
| 1 | Currents at P and Q equal |
| 1 | Justified by conservation of charge: no charge builds up in, or is used up by, the lamp |
| 1 | Electrons at P drift away from the lamp (towards the + terminal), opposite to conventional current |
| 1 | Time to drift 1.5 m calculated: about 1.5 × 10⁴ s (about 4 h), so the claim is rejected |
| 1 | Explains fast lighting: carriers already present everywhere, and the push (field) acts along the whole loop almost at once |
| 1 | After disconnection: random motion continues, drift and net flow stop, current zero |
Do not award the (a) justification for “current is the same everywhere” without a reason. Accept “the potential difference is set up across the whole circuit almost at once” in place of “field” in (c).
How did you do?
- Q1, Q4 or Q5 wrong: re-read “What current measures” and “Current–time graphs” in the study guide. Check units and prefixes first.
- Q2 or Q7(c)–(d) wrong: revisit “Random motion and drift”.
- Q3, Q6 or Q7(b) wrong: revisit “Which way is the current?” and Worked example 3.
- Q7(a) wrong: remember that charge is conserved, so current is not used up.
Then tick off the topic checklist.