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Resistivity Question

Q1.

(a) A metal wire of length 1.4 m has a uniform cross-sectional area = 7.8 × 10–7 m2. Calculate the resistance, R, of the wire. resistivity of the metal = 1.7 × 10–8 Ωm

(2 marks)

R = ρ L/A

R = 1.7 × 10–8 x 1.4/ (7.8 × 10–7)

R = 0.0305

R = 0.031 Ω

(b) The wire is now stretched to twice its original length by a process that keeps its volume constant. If the resistivity of the metal of the wire remains constant, show that the resistance increases to 4R.

The examiners expect you to reason this.

If the volume remains constant doubling length with halve the cross sectional area'

R2= ρ 2L/0.5A = 4ρ L/A = 4R1

 

Or this can be done as a calculation with L2 = 2.8 m and A2 = 3.9 m2 giving R = 0.124 Ω which is 4 x 0.031 Ω

(2 marks)

(Total 4 marks)