CIE iGCSE Co-Ordinated Science P4.5.2 The a.c. generator Exam Style Questions Paper 1
▶️ Answer/Explanation
Using the formula \(R = \frac{V}{I}\), the resistance for each lamp is calculated: A = \(\frac{2.0}{0.40} = 5.0 \, \Omega\), B = \(\frac{4.0}{0.50} = 8.0 \, \Omega\), C = \(\frac{6.0}{2.0} = 3.0 \, \Omega\), D = \(\frac{12}{3.0} = 4.0 \, \Omega\). Lamp B has the highest resistance of 8.0 Ω.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
Electric current is defined as the rate of flow of electric charge, and in metallic conductors, this charge is carried by delocalised electrons moving through the material. Power is the rate of energy transfer, resistance opposes current, and voltage is the driving force that causes current to flow.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
Using Ohm’s law: \(R = \frac{V}{I} = \frac{8.0}{4.0} = 2.0 \ \Omega\). The resistance is the ratio of voltage to current.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
Increasing the number of cells increases the voltage (potential difference) across the motor, which increases the current flowing through it. Decreasing the resistance of the variable resistor also allows more current to flow. Since the speed of a motor increases with more current, both actions together will increase the motor’s speed. Therefore, option C is correct.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
To determine the resistance of a fixed resistor using \(R = \frac{V}{I}\), the voltmeter must be connected in parallel with the resistor to measure the potential difference across it, and the ammeter must be connected in series with the resistor to measure the current flowing through it. Option D correctly shows the voltmeter in parallel and the ammeter in series with the resistor. Options A, B, and C have incorrect meter connections that would give inaccurate readings or damage the meters.
✅ Answer: (D)
✅ Answer: (D)