Practice question
Question
A wire of length \( 4 \, \text{m} \) and resistance \( 10 \, \Omega \) is stretched to \( 8 \, \text{m}
\). What is the new resistance?
Explanation
**Resistivity** depends on material and temperature, not geometry. For metallic conductor, ρ ≈10⁻⁸ Ω·m for copper. Given ρ=4×10⁻⁸ Ω·m, l=2 m, A=π r², R calculation uses R=ρ l/A. Stretching wire conserves volume V = l A = l' A', so A' = A l/l', new R' = ρ l'²/V. Volume constant: l A = l' A' ⇒ A' = (A/2) . New resistance: R' = (rho l'/A') = (rho (2l)/(A/2)) = 4 (rho l/A) = 4R = 4 × 10 = 40 Ω . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V =
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