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Physics

There are two copper wires of length ratio 1:2 that have their cross-sectional areas in the ratio 1:4. What will be the ratio of their:

(a) resistances?
(b) specific resistances?

Current Electricity

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Answer

(a) We use the formula for resistance of a wire: R=ρLAR = \rho \dfrac{L}{A}
Where : R = resistance, ρ\rho = specific resistance (same for copper), L = length, A = cross-sectional area.

Since ρ\rho is the same for both:

R1R2=L1A1L2A2=L1L2⋅A2⋅A1R1R2=1⋅42⋅1=42=2:1\dfrac{R1}{R2} = \dfrac{\dfrac{L1}{A1}}{\dfrac{L2}{A2}} = \dfrac{L1}{L2}\dfrac{\cdot A2}{\cdot A1} \\[1em] \dfrac{R1}{R2} = \dfrac{1 \cdot 4}{2 \cdot 1} = \dfrac{4}{2} = 2:1 \\[1em]

(b) Specific resistance (resistivity) depends only on the material, and both wires are made of copper. So, specific resistance is the same for both.
∴ρ1:ρ2=1:1\therefore \rho1 : \rho2 = 1 : 1

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