CIE iGCSE Co-Ordinated Science P4.2.4 Resistance Exam Style Questions Paper 1
▶️ Answer/Explanation
Ohm’s law states that the current flowing through a conductor is directly proportional to the voltage across it, provided the temperature remains constant.
The relationship between voltage (\( V \)), current (\( I \)), and resistance (\( R \)) is given by the equation:
\[ V = I \times R \]
To find the equation for resistance, we rearrange this to make \( R \) the subject:
\[ R = \frac{V}{I} \]
This means resistance is equal to the voltage divided by the current. The other options are incorrect rearrangements or incorrect relationships.
✅ 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)
▶️ 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
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
Resistance is calculated using \(R = \dfrac{V}{I}\).
The potential difference is 12 V and the current is 4.0 A.
So, \(R = \dfrac{12}{4.0} = 3.0\ \Omega\).
✅ Answer: (B)
The potential difference is 12 V and the current is 4.0 A.
So, \(R = \dfrac{12}{4.0} = 3.0\ \Omega\).
✅ Answer: (B)