A heater coil, designed to operate at 500 W across a 250 V supply, is fabricated from a nichrome wire having a resistivity of 1.0 × 10^ -6 m . If the wire has a cross-sectional area of 0.5 mm^2 , what is the required

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Quantrex Academy · Accepted Answer

Correct answer: A. 62.5 m

The resistance of the heater coil is given by the formula:

R = \dfrac{V^2}{P}

Substituting the given values

V = 250 \text{ V}

and

P = 500 \text{ W}

:

R = \dfrac{(250)^2}{500} = \dfrac{62500}{500} = 125 \Omega

The resistance of a wire is related to its length

L

, cross-sectional area

A

, and resistivity

\rho

by:

R = \rho \dfrac{L}{A}

Rearranging for length

L

:

L = \dfrac{R \cdot A}{\rho}

Given

\rho = 1.0 \times 10^{-6} \Omega \text{m}

and

A = 0.5 \text{ mm}^2 = 0.5 \times 10^{-6} \text{ m}^2

, substituting these values gives:

L = \dfrac{125 \times 0.5 \times 10^{-6}}{1.0 \times 10^{-6}}

L = 125 \times 0.5 = 62.5 \text{ m}

A heater coil, designed to operate at 500 W across a 250 V supply,...

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