Q1. [3] deep thorough-understanding
A voltmeter connected across a resistor in a circuit reads 9 V, while an ammeter in series reads 3 A. (a) Calculate (i) the resistance of the resistor, and (ii) the work done in moving 6 C of charge through it. (b) A student claims that the potential difference across a resistor is a fixed property of the resistor alone, and will remain 9 V regardless of what else changes in the circuit. Evaluate this claim, explaining what factors in the circuit actually determine the potential difference across a given resistor.
Generated by claude-sonnet-4-6 · 2026-06-26 01:09 · grounding rag
Model Answer
(a)(i) Using Ohm's law: $R = V/I = 9\text{ V} / 3\text{ A} = \mathbf{3\ \Omega}$
(a)(ii) Work done = Potential difference × Charge
$$W = V \times Q = 9\text{ V} \times 6\text{ C} = \mathbf{54\ J}$$
(b) The student's claim is incorrect. Potential difference across a resistor is not a fixed property of the resistor alone. By Ohm's law, $V = IR$ — it depends on both the resistance and the current through it. The current itself depends on the total resistance of the circuit and the EMF of the source. For example, adding resistors in series reduces the current, which reduces the potential difference across any individual resistor, even though its resistance stays the same.
Source: Chapter 11, Section 11.4 (Ohm's Law) and Section 11.6.1 (Resistors in Series)
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Explanation
- Part (a)(i): Straightforward application of $R = V/I$. Always write the formula first.
- Part (a)(ii): Work done $W = VQ$ (since $V = W/Q$ by definition). Don't confuse this with $W = VIt$; here charge $Q$ is directly given.
- Part (b): Examiners want you to clearly say the claim is wrong, state that $V = IR$ means V depends on current, and link current to circuit conditions. One concrete example (series circuit changing current) earns full credit. Avoid vague answers like "it depends on other things."