Physics, 17.07.2019 20:30 boneyke3720

Acurrent of 0.96 amps flows through a copper wire 0.44 mm in diameter when connected to a potential difference of 15 v. how long is the wire?

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Answer from: kawaiiblurainbow

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,

$V\propto I$

Here, (V) is the potential difference, and (I)is the current.

It can be written as,

V=IR

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,

$15=0.96R\\R=15.6\rm ohm$

Now, the resistance of a metal wire can be given as,

$R=\dfrac{\rho L}{A}$

Here, ( $\rho$ ) 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 $1.68\times10^{-8}\rm ohm-m$ and the diameter of the wire is 0.44 mm. Thus, put the values in the above formula as,

$15.6=\dfrac{1.68\times10^{-8} L}{\pi \times\dfrac{(0.44)^2}{4}}\\L=141.1\rm m$

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;

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Answer from: timothycarter342

By using Ohm's law, we can calculate the resistance of the wire. Ohm's law states that:
V=IR

where V is the potential difference across the conductor, I is the current and R the resistance. Rearranging the equation, we get
$R= \frac{V}{I}= \frac{15 V}{0.96 A}=15.6 \Omega$

Now we can use the following equation to calculate the length of the wire:
$R= \frac{\rho L}{A}$ (1)
where
$\rho$ 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 $\rho=1.68 \cdot 10^{-8} \Omega m$ . The radius of the wire is half the diameter:
$r= \frac{d}{2}= \frac{0.44 mm}{2}=0.22 mm=0.22 \cdot 10^{-3} m$
And the cross-sectional area is
$A=\pi r^2=\pi (0.22 \cdot 10^{-3}m)^2=1.52 \cdot 10^{-7} m^2$

So now we can rearrange eq.(1) to calculate the length of the wire:
$L= \frac{RA}{\rho}= \frac{(15.6 \Omega)(1.52 \cdot 10^{-7} m^2)}{1.68 \cdot 10^{-8} \Omega m}=141.1 m$

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Answer from: Quest

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