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2. When a potential difference of 12 V is applied to a wire 6.9 m long and 0.33 mm in diameter
the result is an electric current of 2.1 A. What is the resistivity of the wire?​

Respuesta :

rokettojanpu

Apply Ohm's law:

V = IR

V = potential difference, I = current, R = resistance

Resistance of a wire:

R = ρL/A

ρ = resistivity, L = length, A = cross-sectional area

Area of wire's cross-sectional area:

A = π(d/2)²

A = area, d = diameter

Substitute R:

V = IρL/A

Substitute A:

V = IρL/(π(d/2)²)

Given values:

V = 12V, I = 2.1A, L = 6.9m, d = 0.33×10⁻³m

Plug in these values and solve for ρ:

12 = 2.1ρ(6.9)/(π(0.33×10⁻³/2)²)

ρ = 7.1×10⁻⁸Ω·m

shubhamchouhanvrVT

The resistivity of the wire will be ρ = 7.1×10⁻⁸Ω·m

What will be the resistivity of the Wire?

It is given that

Voltage V= 12 v

Length of wire l=6.9m

Diameter of wire =0.33mm

Current flowing from a wire I=2.1 A

now applying Ohm's law:

V = IR

V = potential difference, I = current, R = resistance

Resistance of a wire:

[tex]R=\dfrac{\rho\times L}{A}[/tex]

ρ = resistivity, L = length, A = cross-sectional area

Area of cross-sectional of the wire

[tex]A= \dfrac{\pi\times d^{2} }{4}[/tex]

A = area, d = diameter

By putting the value of R:

[tex]V=\dfrac{I\rho L}{A}[/tex]

Now putting the value of A

[tex]V= \dfrac{I\rho L}{\dfrac{\pi}{4} d^{2} }[/tex]

Given values:

V = 12V, I = 2.1A, L = 6.9m, d = 0.33×10⁻³m

Put all values in the above formula we get

[tex]12=\dfrac{2.1\times\rho \times6.9}{\dfrac{\pi}{4} \times0.33\times10^{-3} }[/tex]

[tex]\rho=7.1\times10^{-8}[/tex]Ω·m

Thus the resistivity of the wire will be ρ = 7.1×10⁻⁸Ω·m

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