A metallic wire has a resistance of 10.0\,\Omega at 0^\circ\text{C} and a temperature coef… — JAMB UTME

A metallic wire has a resistance of $10.0\,\Omega$ at $0^\circ\text{C}$ and a temperature coefficient of resistance $\alpha = 4.0 \times 10^{-3}\,\text{K}^{-1}$ . The wire is uniformly stretched at constant temperature until its length is doubled while maintaining constant mass and volume. It is subsequently heated to $50^\circ\text{C}$ . What is the electric current, in Amperes, that flows through the wire when a potential difference of $120.0\,\text{V}$ is applied across its ends?

Answer: 2.5 A

Answer

The electric current passing through the heated, stretched wire is 2.5 A.

When a wire of initial resistance 10.0 ohms is stretched to twice its original length, conservation of volume requires its cross-sectional area to halve, which quadruples its resistance to 40.0 ohms at 0 °C. Heating the wire by 50 K increases its resistance by a factor of (1 + 0.004 * 50) = 1.2, producing a final resistance of 48.0 ohms. Applying a 120.0 V potential difference across 48.0 ohms results in a current of 2.5 A.

Step-by-Step Solution

Determine resistance change due to wire stretching

R_0' = 40.0 ohms at 0 °C

Uniform stretching conserves total volume (V = A * L). Doubling length halves area, making resistance increase by a factor of 2^2 = 4.

Calculate resistance at 50 °C using temperature coefficient

R(50 °C) = 48.0 ohms

Resistance increases linearly with temperature: R(T) = R_0'(1 + alpha * Delta T).

Apply Ohm's law to find current

I = 2.5 A

Current is given by potential difference divided by total resistance at the operational temperature.

Key Concept

Resistance variation with geometric stretching and temperature coefficient of resistance

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