Two copper wires have the same length, but one has twice the diameter of the other. Compared to the one that has the smaller diameter, the one that has the larger diameter has a resistance that is

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Answer:

Ratio of resistance of wire with larger diameter to smaller diameter is 4.

Explanation:

Resistance is defined as the property of the material/wire to oppose the flow of current through it.

Resistance (R) of the wire is determine by the relation:

R = (рL)/A

Here р is resistivity of the wire and it depends upon the material of the wire, L is length of the wire and A is the area of the wire.

Area of wire, A = π (d/2)²

Here d is diameter of the wire.

According to the question, let R₁ and R₂ be the resistance of the two copper wires, d₁ and d₂ be there diameters respectively and L and р be the length and resistivity of the two wire respectively. Since, L and р are same foe two wires.

Hence, there ratio of resistance is given by:

[tex]\frac{R_{1} }{R_{2} }[/tex]  = [tex]\frac{A_{2} }{A_{1} }[/tex]  

[tex]\frac{R_{1} }{R_{2} }[/tex] = [tex]\frac{d_{2}^{2} }{d_{1}^{2} }[/tex]

We know that d₂ = 2d₁

[tex]\frac{R_{1} }{R_{2} }[/tex] =  [tex]\frac{(2d_{1})^{2} }{d_{1}^{2} }[/tex] = 4

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