Question

How long must a 0.62-mm-diameter aluminum wire be to have a 0.40 A current when connected to the terminals of a 1.5 V flashlight battery? Express your answer to two significant figures and include the appropriate units.

Answer 1

Step-by-step explanation:

• Given the current and voltage, we can find the resistance of the wore via the formula

R = V / I

V = Voltage = 1.5 V

I = Current = 0.40 A

R = 3.75Ω

• Now, from he formula of Resistance in a wire, we know that

R = ρL / A

ρ = Density of Aluminium = 2.710
`g/m^(3)`

Radius of wire = 0.00031 m³, from the data given

A = Area of Wire = π×r²

A = 3.14159×0.00031

A = 0.00097 m²

• Now, having the complete data, we can find the length of the wire

L = (R*A)/ρ

L = (3.75 * 0.00097) / 2.710

L = 0.0013 meters

Answer 2

42 m

Step-by-step explanation:

From ohm's law,

V = IR............... Equation 1

Where V = Voltage, I = current, R = Resistance.

Make R the subject of the equation

R = V/I............. Equation 2

Given: V = 1.5 V, I = 0.40 A.

Substitute into equation 2

R = 1.5/0.4

R = 3.75 Ω.

But,

R = ρL/A..............Equation 2

Where ρ = resistivity of the aluminum wire, l = length of the wire, A = cross sectional area of the wire.

make L the subject of the equation,

L = RA/ρ..............Equation 3

Given: R = 3.75 Ω,

A = πd²/4, where d = diameter = 0.62 mm = 0.00062 m, π = 3.14

A = 3.14× 0.00062²/4 = 0.000000302 m²

Constant: ρ = 2.65 × 10⁻⁸ Ω·m

Substitute into equation 3

L = 3.75(0.000000302)/(2.65 × 10⁻⁸ )

L = 3.75(3.02×10⁻⁷)/(2.65 × 10⁻⁸)

L = 42.70 m

L = 42 m ( Two significant figures)