Question

 

1. The resistance of a very fine aluminum wire with a 10 μm × 10 μm square cross section is 1200 Ω . A 1200 Ω resistor is made by wrapping this wire in a spiral around a 3.4-mm-diameter glass core.

Part A: How many turns of wire are needed? Express your answer using two significant figures.

2. What is the resistance of A 50.0 cm long piece of carbon with a 0.900 mm ×× 0.900 mm square cross section?

3. How long must a 0.56-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

Sol::

Given data

1)
Area(A)= 10 μm × 10 μm
Resistance(R)= 1200 Ω
Diameter(D)= 3.4 mm
Radius (r) = 1.7 mm =1.7*10^-3 m
  
R = 1200 Ω
Resistivity of aluminum is
ρ = 2.65*10^-8 Ωm
A = (10 μm)^2
= (10*10^-6)^2
= 1*10^-10 m^2

L = (R*A) / ρ
L = (1200* 1*10^-10) / (2.65*10^-8)
L = 4.528 m

C = Circumference of wire
= 2π*r = 2π*(1.7*10^-3 )
= 10.679*10^-3 m

Turns = L / C
Turns = (4.528) / (10.679*10^-3 )
Turns = 423.99
.......................................
2)
Length (L) = 50.0 cm
Area(A)= 0.900 mm ×× 0.900 mm
Resistivity of carbon is
ρ = 3.6*10^-5 Ωm

So resistance will be
R= ρL/A
= (3.6*10^-5)*0.5/(0.9*0.9*10^-6)
= 22.22 Ω
....................................

3)
Diameter (D) = 0.56-mm
Radius (r) = 0.28*10^-3 m
Current(i)= 0.40 A
Voltage(V)= 1.5 V
We have resistivity of aluminum as
ρ = 2.65*10^-8 Ωm

Resistance(R)= V/i
= 1.5/0.4
= 3.75 Ω
Area(A)= πr²

So the length will be L = (R*A) / ρ

L = 3.75*(π*(0.28*10^-3)²)/2.65*10^-8
= 34.84 m