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