Question

Calculate the resistance of a copper wire of length 1 m and the area across
section 2 mm 2 . Resistivity of copper is 1.7x10 -8 Ωm.

Answer 1

ANSWER:

8.5 × 10^(-3) Ω

STEP BY STEP SOLUTION:

RESISTANCE (R) = RESISTIVITY (rho) × LENGTH (L) / AREA OF CROSS SECTION (A)

R = rho × L / A

GIVEN:

THERE IS A COPPER WIRE

ITS LENGTH = 1 m

AREA OF CROSS SECTION = 2 mm² = 2 × 10^(-6) m²

RESISTIVITY = 1.7 × 10^(-8) Ωm

HENCE THE REQUIRED RESISTANCE OF THE COPPER WIRE

= 1.7 × 10^(-8) Ωm × 1 m / 2 × 10^(-6) m²

= 0.85 × 10^(-2) Ωm²/m²

= 8.5 × 10^(-3) Ω

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Answer 2

$\bigstar$ Given

A copper wire of length 1 m and area of cross section 2 mm²

Resistivity of copper is 1.7 x 10 -8 Ω m

$\bigstar$ To Find

The resistance of a copper wire

$\bigstar$ Solution

We know that

$\sf \longrightarrow R=\dfrac{\rho \times L}{A}$

Here

• R = resistance
• ρ = resistivity
• L = length
• A = area of cross section

According to the question

We are asked to find the resistance of a copper wire

Therefore

We must find "R"

Given that

A copper wire of length 1 m and area of cross section 2 mm²

Resistivity of copper is 1.7 x 10 -8 Ω m

Hence

• L = 1 m
• A = 2 mm²
• ρ = 1.7 x 10 -8 Ω m

Substituting the values

We get

$\sf \longrightarrow R=\dfrac{1.7 \times 10^{-8} \times 1}{2 \times (10^{-3})^{2}}$

$\sf \longrightarrow R=\dfrac{1.7 \times 10^{-8}}{2 \times (10^{-3})^{2}}$

$\sf \longrightarrow R=\dfrac{1.7 \times 10^{-8}}{2 \times 10^{-6}}$

On further simplification

We get

$\sf \longrightarrow R=0.0085 \ \Omega$

Hence

$\sf \longrightarrow R=8.5 \times 10^{-3} \ \Omega$

Therefore

Resistance of copper wire = 8.5 x 10⁻³