Read the following information and answer the questions that follow:
A technician tests various conductors in a laboratory at constant temperature. She records that nichrome has a resistivity of 100 × 10⁻⁶ Ω m, while silver has a resistivity of 1.60 × 10⁻⁸ Ω m. She uses a nichrome wire of length 1 m and cross-sectional area 1 × 10⁻⁶ m² as a heating element.
(a) The technician observes that for the conductor at constant temperature, the current is directly proportional to the potential difference applied. Which law does this represent? State the law. [1 mark]
(b) Which material — nichrome or silver — is a better conductor of electricity? Justify your answer using the given data. [1 mark]
(c) Calculate the resistance of the nichrome heating element. [1 mark]
(d) If the cross-sectional area of the nichrome wire is doubled while its length remains the same, what will be the new resistance? [1 mark]
Generated by claude-sonnet-4-6 · 2026-06-26 01:10 · grounding rag
Model Answer
(a) This represents Ohm's Law. It states that the potential difference (V) across the ends of a conductor is directly proportional to the current (I) flowing through it, provided its temperature remains constant. i.e., V ∝ I, or V = IR.
(b) Silver is a better conductor. A lower resistivity means less opposition to current flow. Silver has resistivity $1.60 \times 10^{-8}$ Ω m, which is much lower than nichrome's $100 \times 10^{-6}$ Ω m.
(c) Using $R = \rho \dfrac{l}{A}$:
$$R = \frac{100 \times 10^{-6} \times 1}{1 \times 10^{-6}} = \frac{10^{-4}}{10^{-6}} = \textbf{100 Ω}$$
(d) Since $R \propto \dfrac{1}{A}$, doubling the area halves the resistance:
$$R_{\text{new}} = \frac{100}{2} = \textbf{50 Ω}$$
Source: Chapter 11 – Electricity, Sections 11.4 (Ohm's Law) and 11.5 (Factors on which resistance depends)
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Explanation
- (a) Always state the law in full — "provided temperature remains constant" is essential for full marks.
- (b) The key principle: lower resistivity = better conductor. Justify by comparing the actual values.
- (c) Substitute directly in $R = \rho l/A$. Show the formula, substitution, and final answer with units.
- (d) No re-calculation needed — just apply the inverse proportionality. Since area doubles, resistance halves. This tests conceptual understanding, not arithmetic.