Respuesta :
By using Ohm's law, we can calculate the resistance of the wire. Ohm's law states that:
[tex]V=IR[/tex]
where V is the potential difference across the conductor, I is the current and R the resistance. Rearranging the equation, we get
[tex]R= \frac{V}{I}= \frac{15 V}{0.96 A}=15.6 \Omega [/tex]
Now we can use the following equation to calculate the length of the wire:
[tex]R= \frac{\rho L}{A} [/tex] (1)
where
[tex]\rho[/tex] is the resistivity of the material
L is the length of the conductor
A is its cross-sectional area
In this problem, we have a wire of copper, with resistivity [tex]\rho=1.68 \cdot 10^{-8} \Omega m[/tex]. The radius of the wire is half the diameter:
[tex]r= \frac{d}{2}= \frac{0.44 mm}{2}=0.22 mm=0.22 \cdot 10^{-3} m [/tex]
And the cross-sectional area is
[tex]A=\pi r^2=\pi (0.22 \cdot 10^{-3}m)^2=1.52 \cdot 10^{-7} m^2[/tex]
So now we can rearrange eq.(1) to calculate the length of the wire:
[tex]L= \frac{RA}{\rho}= \frac{(15.6 \Omega)(1.52 \cdot 10^{-7} m^2)}{1.68 \cdot 10^{-8} \Omega m}=141.1 m [/tex]
[tex]V=IR[/tex]
where V is the potential difference across the conductor, I is the current and R the resistance. Rearranging the equation, we get
[tex]R= \frac{V}{I}= \frac{15 V}{0.96 A}=15.6 \Omega [/tex]
Now we can use the following equation to calculate the length of the wire:
[tex]R= \frac{\rho L}{A} [/tex] (1)
where
[tex]\rho[/tex] is the resistivity of the material
L is the length of the conductor
A is its cross-sectional area
In this problem, we have a wire of copper, with resistivity [tex]\rho=1.68 \cdot 10^{-8} \Omega m[/tex]. The radius of the wire is half the diameter:
[tex]r= \frac{d}{2}= \frac{0.44 mm}{2}=0.22 mm=0.22 \cdot 10^{-3} m [/tex]
And the cross-sectional area is
[tex]A=\pi r^2=\pi (0.22 \cdot 10^{-3}m)^2=1.52 \cdot 10^{-7} m^2[/tex]
So now we can rearrange eq.(1) to calculate the length of the wire:
[tex]L= \frac{RA}{\rho}= \frac{(15.6 \Omega)(1.52 \cdot 10^{-7} m^2)}{1.68 \cdot 10^{-8} \Omega m}=141.1 m [/tex]
The length of the copper wire, which flows the current of 0.96 amps at a potential difference of 15 v is 141.1 meters.
What is Ohm's law?
Ohm's law states that for a flowing current, the potential difference of the circuit is directly proportional to the current flowing in it. Thus,
[tex]V\propto I[/tex]
Here, (V) is the potential difference, and (I)is the current.
It can be written as,
[tex]V=IR[/tex]
Here, (R) is the resistance of the circuit.
The current of 0.96 amps flows through a copper wire when connected to a potential difference of 15 v. Thus, by using the above formula of ohm's law, the resistance of the wire can be found out as,
[tex]15=0.96R\\R=15.6\rm ohm[/tex]
Now, the resistance of a metal wire can be given as,
[tex]R=\dfrac{\rho L}{A}[/tex]
Here, ([tex]\rho[/tex]) is the resistivity of a wire (L) is the length of a wire, and A is the cross-sectional area of it.
As the resistivity of a copper wire is [tex]1.68\times10^{-8}\rm ohm-m[/tex] and the diameter of the wire is 0.44 mm. Thus, put the values in the above formula as,
[tex]15.6=\dfrac{1.68\times10^{-8} L}{\pi \times\dfrac{(0.44)^2}{4}}\\L=141.1\rm m[/tex]
Thus, the length of the copper wire, which flows the current of 0.96 amps at a potential difference of 15 v is 141.1 meters.
Learn more about Ohm's law here;
https://brainly.com/question/25822112
