Wire X is made of copper and wire Y is made of manganin. Both wires have identical dimensions. In which wire will electrons experience greater opposition to their flow? Given that the resistivity of manganin is approximately 2700 times that of copper, by what factor is the resistance of wire Y greater than that of wire X?
Generated by claude-sonnet-4-6 · 2026-06-26 01:10 · grounding rag
Model Answer
Wire Y (manganin) offers greater opposition to electron flow.
This is because resistance depends on the material's resistivity ($R = \rho \dfrac{l}{A}$). Since both wires have identical dimensions (same $l$ and $A$), the resistance depends only on resistivity.
Manganin has resistivity approximately 2700 times that of copper. Therefore:
$$\frac{R_Y}{R_X} = \frac{\rho_{\text{manganin}}}{\rho_{\text{copper}}} = 2700$$
The resistance of wire Y is 2700 times greater than that of wire X.
Source: Chapter 11, Section 11.5 – Factors on which the resistance of a conductor depends
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Explanation
- The key formula examiners expect is $R = \rho \dfrac{l}{A}$. Since $l$ and $A$ are identical, the ratio of resistances equals the ratio of resistivities directly.
- State which wire clearly first, then justify with the formula — this earns the first mark.
- The factor calculation is straightforward substitution — show the ratio setup to secure the final mark.
- Manganin is an alloy; alloys generally have much higher resistivity than pure metals, which is why they are used in heating devices.