- #1
gamesandmore
- 32
- 0
A cylindrical aluminum pipe of length 1.36 m has an inner radius of 1.80 10-3 m and an outer radius of 3.00 10-3 m. The interior of the pipe is completely filled with copper. What is the resistance of this unit? (Hint: Imagine that the pipe is connected between the terminals of a battery and decide whether the aluminum and copper parts of the pipe are in series or in parallel.)
Here is what I did:
L = 1.36 m
ri = 1.80e-3m
ro = 3.00e-3 m
R = ? ohms
R = pL/A
pcopper = 1.72e-8 ohm*m
palum = 2.82e-8 ohm*m
In = copper
Out = aluminum
Ai = pi*ri^2 = 1.01788e-5 m^2
Ao= pi*ro^2 = 2.827e-5 m^2
Ri = pcopper * L/Ai = .002298 ohms
Ro = palum * L/Ao = .0013566 ohms
Then I did
1 / ( (1/.002298ohms) + (1/.0013566 ohms) )
and got 8.53e-4 Ohms, but it was wrong...
Here is what I did:
L = 1.36 m
ri = 1.80e-3m
ro = 3.00e-3 m
R = ? ohms
R = pL/A
pcopper = 1.72e-8 ohm*m
palum = 2.82e-8 ohm*m
In = copper
Out = aluminum
Ai = pi*ri^2 = 1.01788e-5 m^2
Ao= pi*ro^2 = 2.827e-5 m^2
Ri = pcopper * L/Ai = .002298 ohms
Ro = palum * L/Ao = .0013566 ohms
Then I did
1 / ( (1/.002298ohms) + (1/.0013566 ohms) )
and got 8.53e-4 Ohms, but it was wrong...