Q#21 (Past Exam Paper – March 2018 Paper 12 Q5)
A person calculates the potential difference across a wire by using the measurements shown.
Which measured quantity has the greatest contribution to the percentage uncertainty in the calculated potential difference?
quantity value uncertainty
A current / A 5.0 ± 0.5
B diameter of wire / mm 0.8 ± 0.1
C length of wire / m 150 ± 5
D resistivity of metal in wire / Ω m 1.6 × 10-8 ± 0.2 × 10-8
Solution:
Answer: B.
Ohm’s law: V = IR
Resistance of wire: R = ρL / A
Cross-sectional area A = πd2 / 4
Resistance of wire: R = 4ρL / πd2
p.d. V = I × (4ρL / πd2) = 4IρL / πd2
ΔV / V = ΔI/I + Δρ/ρ + ΔL/L + 2(Δd/d)
Considering the percentage uncertainties,
%V = %I + %ρ + %L + 2%d
%I = 0.5/5 ×100% = 10%
%ρ = 0.2/1.6 ×100% = 12.5%
%L = 5/150 ×100% = 3.33%
2%d = 2 × (0.1/0.8 ×100%) = 25% [greatest contribution to percentage uncertainty]
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