Question

A student needs to make a 0.12 Ω resistor. She has some copper wire of 0.80 mm diameter. Resistivity of copper is 1.8 × 10^–8 Ωm.

1. Determine the cross-sectional area of the wire.
2. Calculate the length of wire required for the 0.12 Ω resistor.

Answer 1

i. Given, diameter of wire = 0.8 mm

= 0.8 × 10⁻³ m

The cross-sectional area, A = πr²

= `pi (d/2)^2`

= `pi d^2/4`

A = `pi((0.8 xx 10^-3)^2)/4`

= π (0.40 × 10⁻³)²

= π × (0.16 × 10⁻⁶) m²

≈ 3.14 × 0.16 × 10⁻⁶ m²

≈ 5.024 × 10⁻⁷ m²

= 5.03 × 10⁻⁷ m²

Thus the cross-sectional area A is approximately 5 × 10⁻⁷ m².

ii. To find the length of the wire, we can use the formula of resistance.

R = `rho l/A`

l = `(RA)/rho`

= `(0.12 xx 5 xx 10^-7)/(1.8 xx 10^-8)`

= `0.6/1.8 xx 10`

= 3.33 m

The student needs a length of approximately 3.33 m of given copper wire to make a 0.12 Ω resistor.