Question

How long must a 0.54-mm-diameter aluminum wire be to have a 0.42 A current when connected to the terminals of a 1.5 V flashlight battery

Answer 1

L = 30.85 m

Step-by-step explanation:

First, we calculate the resistance of the wire by using Ohm's Law:

V = IR

where,

V = Potential Difference = 1.5 V

I = Current = 0.42 A

R = Resistance of Wire = ?

Therefore,

`R = (1.5\ V)/(0.42\ A)\\\\R = 3.57\ Ohms`

Now, the cross-sectional area of wire will be:

`Area = A = (\pi d^(2))/(4)\\\\A = (\pi (0.00054\ m)^(2))/(4)\\\\A = 2.29\ x\ 10^(-7)\ m^(2)`

Now, the resistance of the wire is given as:

`R = (\rho L)/(A)\\\\L = (RA)/(\rho)`

where,

L = Length of Wire = ?

ρ = resistivity of aluminum = 2.65×10⁻⁸ Ohm.m

Therefore,

`L = ((3.57\ Ohms)(2.29\ x\ 10^(-7)\ m^(2)))/(2.65\ x\ 10^(-8)\ Ohm.m)`

L = 30.85 m