Question

A copper wire has diameter 0.5 mm and resistivity of \(1.6×10^{–8} \ Ω m\). What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Answer 1

Resistance (R) of a copper wire of length l and cross-section A is given by the expression,
\(R = ρ\frac {l}{A}\)
Where,
ρ is resistivity of copper = 1.6 × 10^–8 Ω m
R = 10 Ω,
radius of wire r = \(\frac {0.5}{2}\)mm = 0.25 mm = 0.00025 m
𝐴 = 𝜋𝑟² 
A = 3.14×(0.00025)² 
A = 0.000000019625 m²
⟹𝑙 = \(\frac {𝑅𝐴}{ρ}\)

⟹𝑙 = \(\frac {10×0.000000019625}{1.6×10^{−8}}\)
⟹𝑙 = 122.72 𝑚
If the diameter (radius) is doubled, the new radius r = 0.5 mm = 0.0005 m
𝐴 = 𝜋𝑟² 
A = 3.14×(0.0005)²
A = 0.000000785 m² 
So, the new resistance will be
\(𝑅^′=ρ\frac {𝑙}{𝐴}\) 

\(R' = \frac {1.6×10^{−8} \times 122.72}{0.000000785 }\)
\(R'=2.5\  Ω\)
Now,
\( \frac {𝑅^′}{𝑅}=\frac {2.5}{10}\)

⟹ \( \frac {𝑅^′}{𝑅}=\frac {1}{4}\)

⟹ \(𝑅^′=\frac 14𝑅\)

Hence, the new resistance will become \(\frac 14\) times the original resistance.