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CIE iGCSE Co-Ordinated Science P4.3.1 Circuit diagrams and circuit components Exam Style Questions Paper 4

CIE iGCSE Co-Ordinated Science Paper 4 - All Chapters

Question

A student investigates an NTC thermistor.
(a) The student connects the thermistor in series with a cell and an ammeter.
The student also connects a voltmeter to measure the potential difference across the thermistor.
Fig. 3.1 shows an incomplete circuit diagram of the circuit used by the student.
(i) Complete Fig. 3.1.
(ii) When the thermistor is at room temperature, the ammeter reads \(3.0\,\text{A}\).
Calculate the charge that flows through the thermistor in \(60\,\text{s}\).
State the unit for your answer.
(b) The student places the thermistor into hot water.
The student records the values shown by the ammeter and the voltmeter as the temperature of the thermistor increases.
Describe how the power output of the thermistor changes as the temperature of the thermistor increases.
(c) Fig. 3.2 shows the current–voltage characteristic for an ohmic resistor.
Explain the shape of the graph shown in Fig. 3.2.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic P4.3.1 — Circuit diagrams and circuit components (Part (a)(i))
• Topic P4.2.2 — Electric current (Part (a)(ii))
• Topic P4.2.5 — Electrical energy and electrical power (Part (b))
• Topic P4.2.4 — Resistance (Part (c))

▶️ Answer/Explanation

(a)(i) Thermistor symbol added in series; ammeter and voltmeter correctly placed

The thermistor symbol must be drawn in the main series loop with the cell and ammeter.
The voltmeter must be connected in parallel across the thermistor only.

(a)(ii) \(180\,\text{C}\)

Charge is calculated using \(Q = It\).
\(Q = 3.0 \times 60 = 180\,\text{C}\) (coulombs).

(b) Power increases as temperature increases

As temperature rises, the resistance of the NTC thermistor decreases.
A lower resistance at constant voltage causes the current through the thermistor to increase.
Since \(P = IV\), an increase in current (with p.d. roughly maintained) means the power output increases.

(c) Straight line through the origin

The graph is a straight line passing through the origin, showing current is directly proportional to voltage.
This means the resistance of the component stays constant, which is the definition of an ohmic conductor obeying Ohm’s law.

Question

A student is investigating electrical circuits.
(a) Fig. 12.1 shows a circuit made by the student.
(i) The ammeter in Fig. 12.1 reads 0.50 A.
The voltmeter in Fig. 12.1 reads 2.0 V.
Calculate the resistance of the resistor labelled \(R\) in Fig. 12.1.
(ii) The student notices that resistor \(R\) gets hot if the circuit is left connected for too long.
Describe, in terms of current, how the student prevents resistor \(R\) from overheating using the circuit shown in Fig. 12.1.
(b) The student replaces the 6.0 V battery with a small solar cell.
The solar cell has an efficiency of 16%.
Calculate the power input to the solar cell when the solar cell provides 8.0 W of power to the circuit.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P4.3.1 — Circuit diagrams and circuit components (Part (a))
• Topic P4.2.1 — Electrical charge (Part (a)(i))
• Topic P1.6.1 — Energy (Part (b))

▶️ Answer/Explanation

(a)(i) 8.0 Ω

Since the battery, variable resistor and \(R\) are in series, the voltage across \(R\) \(= 6.0 – 2.0 = 4.0\) V.
Using \(R = \dfrac{V}{I} = \dfrac{4.0}{0.50} = 8.0\) Ω.

(a)(ii) Increase the resistance of the variable resistor to decrease the current flowing through R

Since the variable resistor is in series with \(R\), increasing its resistance reduces the total current in the circuit.
A lower current through \(R\) means less heat is generated, preventing it from overheating.

(b) 50 W

Efficiency \(= \dfrac{\text{power output}}{\text{power input}} \times 100\%\).
Power input \(= \dfrac{\text{power output}}{\text{efficiency}} = \dfrac{8.0}{0.16} = 50\) W.

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