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: batteries, wires and closed switches are ideal, so there is no change in potential along a wire or a closed switch; ammeters and voltmeters are ideal; a bulb carries current only if it is in a closed loop with the battery and has a potential difference across it; “current” means conventional current. A scientific calculator is assumed.
Question 1 (multiple choice · foundation)
A battery is connected to two bulbs in a single loop. Which change creates a short circuit across one of the bulbs?
- (A) Connecting a plain wire from one end of the first bulb to its other end
- (B) Opening a switch placed between the two bulbs
- (C) Connecting a voltmeter from one end of the first bulb to its other end
- (D) Breaking the loop and inserting an ammeter between the two bulbs
Answer and explanation
Answer: (A). A plain wire joins the two ends of the bulb with a path that has no change in potential. The bulb then has zero potential difference across it and carries no current: it is short-circuited.
- (B) makes an open circuit, not a short circuit. No charge can flow anywhere in the loop.
- (C) is how a voltmeter is meant to be connected. An ideal voltmeter lets no charge through, so it does not change the circuit.
- (D) is the correct way to connect an ammeter. It carries the loop current and does not bypass any element.
Question 2 (multiple choice · foundation)
A schematic shows a zigzag line with a diagonal arrow drawn through it. What does this symbol represent?
- (A) A variable resistor
- (B) A resistor with conventional current in the direction of the arrow
- (C) A variable capacitor
- (D) A resistor that has been short-circuited
Answer and explanation
Answer: (A). The zigzag is a resistor. A diagonal arrow drawn through any symbol marks a variable element, so this is a resistor whose value can be adjusted.
- (B) is wrong because the diagonal arrow does not show current. Current directions are shown by arrows drawn along the wires.
- (C) is wrong because a capacitor is two parallel lines, not a zigzag.
- (D) is wrong because a short circuit is shown by a wire connected across an element, not by an arrow through it.
Question 3 (multiple choice · core)
A battery, bulb X, an open switch and bulb Y are connected one after the other in a single loop, in that order. Which statement is correct?
- (A) Neither bulb carries a current.
- (B) Bulb X carries a current, but bulb Y does not, because the current stops at the switch.
- (C) Both bulbs carry a small current, because charge can jump across the switch.
- (D) Bulb Y carries a current, but bulb X does not, because current flows from the negative terminal.
Answer and explanation
Answer: (A). The open switch breaks the only loop, so this is an open circuit. With no closed loop, there is no current anywhere in it, on either side of the switch.
- (B) pictures current travelling up to the gap and stopping. There is no steady flow in any part of an open loop.
- (C) is wrong: an ideal open switch is a gap that charges cannot cross.
- (D) uses the direction of electron flow, and still wrongly lets one side of an open loop carry a current.
Question 4 (multiple choice · core)
A battery and three bulbs, X, Y and Z, are all connected between the same two junctions, each in its own branch. In how many different closed loops is bulb X included?
- (A) 1
- (B) 2
- (C) 3
- (D) 6
Answer and explanation
Answer: (C). A loop through X must return through one of the other three branches: the battery, Y or Z. So X is in three loops: battery–X, X–Y and X–Z.
- (A) counts only the loop that contains the battery.
- (B) counts only the loops made of bulbs and forgets the loop with the battery.
- (D) is the total number of loops in the whole circuit (every pair of the four branches), including the three that do not contain X.
Question 5 (constructed response · core)
A torch contains two cells joined end to end (the positive terminal of one touching the negative terminal of the other), a push switch and a bulb, all in one loop.
(a) Draw a schematic of the torch. Add an ammeter that measures the current in the bulb and a voltmeter that measures the potential difference across the pair of cells. (b) With the switch closed, the ammeter reads 0.30 A. Calculate the charge that passes through the bulb in 5.0 minutes. (c) A student moves the ammeter to the other side of the bulb. Predict the new reading and justify your prediction.
Worked solution and suggested Marlbridge rubric
(a) One loop with two cell symbols in a row, the long line of one next to the short line of the other, plus a switch and a bulb. The ammeter is drawn in the loop. The voltmeter is connected from the free terminal of the first cell to the free terminal of the second cell, across both cells, without breaking the loop.
(b) Δt = 5.0 × 60 s = 300 s. ΔQ = IΔt = (0.30 A)(300 s) = 90 C.
(c) The reading is still 0.30 A. There is only one loop, so there is only one path for charge. Charge is conserved and does not build up in the bulb, so the same current passes every point in the loop.
| Point | What earns it |
|---|---|
| 1 | Correct symbols for two cells (joined + to −), switch and bulb in a single closed loop |
| 1 | Ammeter in the loop and voltmeter across both cells, not breaking the loop |
| 1 | ΔQ = 90 C, with time converted to 300 s |
| 1 | Same reading, 0.30 A, justified by a single path and conservation of charge |
Common error: using 5.0 s instead of 300 s, giving 1.5 C. Any order of elements around the loop is acceptable.
Question 6 (constructed response · core)
A circuit has a battery and bulb A in the main line. After bulb A, the circuit splits at a junction into two branches that rejoin before returning to the battery. One branch contains switch S₁ and bulb B. The other contains switch S₂ and bulb C.
(a) Copy and complete a table showing which bulbs carry current for each of the four combinations of S₁ and S₂ (both open, only S₁ closed, only S₂ closed, both closed). (b) Explain why bulb A can never be the only bulb lit. (c) Both switches are closed. A plain wire is then connected from one end of bulb A to the other. State which bulbs carry current, and justify your answer.
Worked solution and suggested Marlbridge rubric
(a)
| S₁ | S₂ | Bulbs carrying current |
|---|---|---|
| open | open | none |
| closed | open | A and B |
| open | closed | A and C |
| closed | closed | A, B and C |
(b) Every closed loop containing the battery and A must return through one of the branches, so it must also pass through B or C. If both branches are open, A is in no closed loop. If a branch is closed, its bulb is in the same loop as A, so it carries current too.
(c) B and C carry current; A does not. The wire joins the two ends of A with no change in potential, so the potential difference across A is zero and A is short-circuited. B and C are still in closed loops with the battery (battery → wire → branch → battery), so they still carry current.
| Point | What earns it |
|---|---|
| 1 | Both-open row: no bulbs |
| 1 | The two one-switch rows: A with B, and A with C |
| 1 | Both-closed row: all three bulbs |
| 1 | (b) Any loop with A and the battery must pass through B or C |
| 1 | (c) A carries no current because the wire leaves zero potential difference across it |
| 1 | (c) B and C still in closed loops with the battery, so they carry current |
Do not award the last point for “B and C get A’s current” with no reference to closed loops.
Question 7 (constructed response · stretch)
A battery of emf 6.0 V, a bulb and a switch are connected in a single loop. An ideal voltmeter is connected across the switch.
(a) The switch is open. State the voltmeter reading and justify it. (b) The switch is closed. State the voltmeter reading and justify it. (c) A student says: “When the switch is open there is no current, so there can be no potential difference anywhere. The voltmeter across the switch must read zero.” Evaluate this claim. (d) With the switch still open, the voltmeter is moved so that it is across the bulb instead. State the new reading.
Worked solution and suggested Marlbridge rubric
(a) 6.0 V. With the switch open there is no current, so no charge flows through the bulb. A conductor with no charge flowing through it has the same potential at both ends (as in Unit 10, a conductor in equilibrium). So one contact of the switch is at the potential of the positive terminal (joined by an ideal wire) and the other is at the potential of the negative terminal (joined through the bulb and an ideal wire). The voltmeter reads the full potential difference of the battery.
(b) 0 V. A closed ideal switch is a path along which charges can flow with no change in potential, just like a plain wire. Its two contacts are at the same potential.
(c) The claim is wrong. Zero current means no flow of charge, but the battery still keeps its terminals at different potentials. The whole of that potential difference appears across the gap in the open switch, because nothing else in the loop has a potential difference across it. The claim confuses “no current” with “no potential difference”.
(d) 0 V. The bulb carries no current, so its two ends are at the same potential.
| Point | What earns it |
|---|---|
| 1 | (a) 6.0 V |
| 1 | (a) Justified: no current, so the bulb and wires have no potential difference across them, leaving the whole emf across the gap |
| 1 | (b) 0 V, because a closed switch has no change in potential across it |
| 1 | (c) Rejects the claim: the battery maintains its potential difference even when there is no current |
| 1 | (d) 0 V across the bulb with the switch open |
Accept equivalent wording in (a), for example: “the voltmeter is connected to the battery’s terminals through elements that have no potential difference across them”.
How did you do?
- Q1 or Q7(b) wrong: re-read “Closed, open and short circuits” in the study guide and Worked example 2.
- Q2 or Q5(a) wrong: revisit Figure 1 and the rules for reading and drawing schematics.
- Q3 or Q7(c) wrong: remember that an open loop has no current anywhere, but the battery still keeps its potential difference.
- Q4 or Q6 wrong: work through Worked example 1 again, listing every loop.
Then tick off the topic checklist.