Question 1

In the circuit below, all the bulbs, B1, B2, B3 and B4, are lit. What is the highest number of bulbs that would remain lit when one of the bulbs is fused?

Accepted Answer

(4) 3. The circuit has a series-parallel arrangement. B1 is in series with the parallel combination of B2 and B3, which is then in series with B4. 
- If B1 or B4 fuses, the entire circuit breaks, and 0 bulbs light up.
- If B2 fuses, the current can still flow through B1, B3, and B4, leaving 3 bulbs lit.
- If B3 fuses, the current can still flow through B1, B2, and B4, leaving 3 bulbs lit.
Therefore, the highest number of bulbs that would remain lit is 3.

Question 2

Clara used a circuit tester to test a circuit card. Her results are recorded in the table below. Which of the following represents the circuit card Clara tested?

Accepted Answer

(4). By systematically checking the connectivity between points for each diagram against the provided table:
- Diagram (1) shows A-C-E and A-D-E paths, so A and E would be connected, but the table says 'No'.
- Diagram (2) shows A-C-D-E path, so A and E would be connected, but the table says 'No'.
- Diagram (3) shows A-C-B-E path, so A and E would be connected, but the table says 'No'.
- Diagram (4) correctly matches all conditions from the table:
    - A and C: Yes (direct connection)
    - A and D: Yes (direct connection)
    - A and E: No (no path)
    - B and C: No (no path)
    - B and D: No (no path)
    - B and E: Yes (direct connection)
    - D and E: No (no path)
All connections match for diagram (4).

Question 3

Wei Qing had three rods, X, Y and Z, made of unknown materials. He placed them at various positions, A, B and C, in the circuit below. The results are shown in the table below. When any of the lamps, L1, L2, L3 or L4, lit up, a tick (✔) was placed in the box. Which of the following shows what the result will be when the rods, X, Y and Z, are placed at the different positions?

Accepted Answer

(4). The circuit diagram shows the bulbs arranged as follows: Battery(+) -> A -> [ (B -> L1 -> L2) || L3 ] -> L4 -> C -> Battery(-).

From the solution key, the correct option is (4).
Let's analyze the results of option (4): A: Y, B: X, C: Z. Lamps: L1(X), L2(X), L3(X), L4(✔).
1. For L4 to light up, the main circuit path through A(Y), L4, and C(Z) must be complete. Thus, Y must be a Conductor and Z must be a Conductor.
2. For L1 and L2 not to light up, the branch B(X)-L1-L2 must be open. Since Y(A) is a conductor and L1, L2 are assumed to be working bulbs, then X (at B) must be an Insulator.
3. The fact that L3 does not light up (L3 X) is contradictory with the derived properties. If Y(A) and Z(C) are conductors, then the path A-L3-L4-C is complete, and L3 should light up. This suggests a flaw in the question's data or an implicit assumption that L3 is faulty or not a functional bulb.

Assuming the deduction for X, Y, Z from L1, L2, L4 is correct (Y=Conductor, Z=Conductor, X=Insulator):
- Let's check the initial table given in the question: A: X(Insulator), B: Y(Conductor), C: Z(Conductor).
- If A is an insulator, the entire circuit is open, so all bulbs (L1, L2, L3, L4) should be OFF.
- However, the initial table states L1(✔) and L4(✔). This is a direct contradiction to the properties deduced from the answer option.

Due to these contradictions, the question appears flawed. However, following the provided solution, we deduce Y=Conductor, X=Insulator, Z=Conductor. Option (4) is the only one showing L4 lit and L1, L2 off, which is consistent with X being an insulator at B, and Y, Z being conductors at A, C respectively (assuming L3 is faulty or non-functional in all cases as implied by option 4's data).

Question 4

The diagram below shows a circuit made up of identical bulbs and batteries. Initially, bulb A is unlit while bulb B is lit. Which of the following will be observed when the metal rod is pressed at point X until it hits point Y?

Accepted Answer

(1) Only bulb A will light up.. Initially, bulb B is lit, meaning there's a closed circuit path for bulb B. Bulb A is unlit, meaning its circuit path is open.
When the metal rod is pressed at point X until it hits point Y:
- The circuit for bulb B will be short-circuited by the metal rod (path Y to the right). Current will bypass bulb B through the rod, causing bulb B to go out.
- The circuit for bulb A will be completed. The metal rod at Y completes the path for bulb A.
Therefore, only bulb A will light up.

Question 5

Farah set up a circuit with the switches, P, Q, R and S, as shown below. All five bulbs were lit when all four switches were closed. Farah wanted the fewest number of bulbs to be lit by opening only one switch. Which switch should she open?

Accepted Answer

(2) Q. Analyze the circuit: There are two parallel branches from the battery. The top branch has two bulbs in series with switch P. The bottom branch has three bulbs. Switch Q is in series with the entire bottom branch. Switch R is in parallel with one bulb in the bottom branch. Switch S is in series with another bulb in the bottom branch. When all switches are closed, all 5 bulbs are lit.

Let's evaluate opening each switch:
- If Q is opened: This switch is in series with the entire bottom branch (3 bulbs). Opening Q will turn off all 3 bulbs in the bottom branch. The top branch (2 bulbs) will remain lit. (2 bulbs lit)
- If P is opened: This switch is in series with the top branch (2 bulbs). Opening P will turn off the 2 bulbs in the top branch. The bottom branch (3 bulbs) will remain lit. (3 bulbs lit)
- If R is opened: R is in parallel with one bulb in the bottom branch. Opening R will force current through that bulb (increasing resistance for that path), so all 3 bulbs in the bottom branch remain lit (possibly dimmer), and 2 bulbs in the top branch remain lit. (5 bulbs lit)
- If S is opened: S is in series with one bulb in the bottom branch. Opening S will turn off only that one bulb and the bulb in series with it in that part of the parallel branch, making two bulbs go off in the bottom branch (if the connection is after the parallel section) or just one. Looking closely, S is in series with the leftmost bulb of the bottom branch. If S opens, that specific bulb goes out, and the rest of the bottom branch (two bulbs) and the top branch (two bulbs) remain lit. (4 bulbs lit)

To have the fewest number of bulbs lit, opening switch Q (turning off 3 bulbs, leaving 2 lit) is the best option.

Question 6

The diagram below shows an electromagnet and a toy car. When the switch is closed, the toy car moves up the ramp towards the electromagnet. Based on the above information, which of the following statement(s) is/ are true?
A: The toy car is made of magnetic material.
B: As the number of batteries increases, the toy car will move up the ramp faster.
C: As the number of coils around the electromagnet decreases, the toy car will move up the ramp faster.

Accepted Answer

(1) A and B only. A: The toy car moves towards the electromagnet when the switch is closed, indicating it is attracted by the magnetic field. Therefore, the toy car must be made of a magnetic material. (True)
B: Increasing the number of batteries increases the current flowing through the electromagnet, which strengthens the magnetic field. A stronger magnetic field will exert a greater attractive force on the toy car, causing it to move up the ramp faster. (True)
C: Decreasing the number of coils around the electromagnet weakens the magnetic field (fewer turns means weaker electromagnet for the same current). A weaker magnetic field will result in a weaker attractive force, causing the toy car to move slower, not faster. (False)
Thus, only statements A and B are true.

P5 Science CA2 2025 — Tao Nan

Related papers