Respuesta :
Answer:
First wire is 10 times larger in diameter than the second wire.
[tex]\frac{ d_{1} }{ d_{2} }[/tex] ≅ 10
Explanation:
Given :
Current in first wire [tex]I_{1} = 112 \times 10^{-12}[/tex] A
Current in second wire [tex]I_{2} = 0.044 \times 10^{-12}[/tex] A
Length of first wire = [tex]l_{1}[/tex]
Length of second wire = [tex]25 l_{1}[/tex]
Here potential across a resistor is same.
From ohm's law,
[tex]V =IR[/tex]
Where [tex]R =[/tex]resistance of the wire,
Resistor of wire is given by,
[tex]R = \frac{\rho l}{A}[/tex]
Where [tex]A =[/tex] cross section area of the wire = [tex]\pi r^{2}[/tex]
We write area in terms of diameter [tex]d = 2r[/tex]
[tex]A = \frac{\pi d^{2} }{4}[/tex]
Now we have to compare current in both wires,
[tex]I_{1} R_{1} = I_{2}R_{2}[/tex]
Put the value of resistor in terms of its diameter,
[tex]\frac{l_{1} d_{2}^{2} }{l_{2} d_{1}^{2} } = \frac{I_{2} }{I_{1} }[/tex]
[tex]\frac{l_{1} d_{2}^{2} } { 25l_{1} d_{1}^{2} } = \frac{112 \times 10^{-12} }{0.044 \times 10^{-12} }[/tex]
[tex]\frac{ d_{2}^{2} }{ d_{1}^{2} } = 0.0098[/tex]
[tex]\frac{ d_{2} }{ d_{1}} } = \sqrt{0.0098} = 0.099[/tex]
[tex]\frac{ d_{1} }{ d_{2} } = \frac{1}{0.099} = 10.10[/tex]
[tex]\frac{ d_{1} }{ d_{2} }[/tex] ≅ 10
