Three conducting straight wires (though insulated from each other) are each 2 m long. They are electrically connected to each other only at their ends points A and B. AB = 2 meter.

The wires are cylindrical. One is made of copper, one of aluminum and one of iron. Their radii are 2, 3 and 4 mm, respectively.

a) What is the ohmic resistance between point A and B? Express your answer in Ohms. (you might need to look up the resistivity of these elements).

unanswered

We now attach at B a straight copper wire which is in electrical contact with the 3 wires at B. This 2nd copper wire is identical to the one between A and B. It runs from B to C. The distance AC is 4 m.

(b) What is the ohmic resistance between A and C? Express your answer in Ohms.

unanswered

We now turn AC into a near perfect circle. A is very close to C but it is not touching C.

(c)

What is the ohmic resistance between A and C now? Express your answer in Ohms.

Please some one answer.

To find the ohmic resistance between points A and B, we need to calculate the resistance of each wire separately and then add them up.

a) Ohmic resistance between A and B:
To calculate the resistance of a wire, we can use the formula:

R = (ρ * L) / A

Where R is the resistance, ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.

The resistivity values for copper, aluminum, and iron can be obtained from reference tables or online sources. Let's assume the resistivity values are:

Copper (ρcopper) = 1.7 x 10^-8 Ω.m
Aluminum (ρaluminum) = 2.7 x 10^-8 Ω.m
Iron (ρiron) = 10 x 10^-8 Ω.m

The cross-sectional areas, A, of the wires can be calculated using the formula for the area of a circle:

A = π * r^2

Where r is the radius of the wire.

Given the radii of the wires are:
Copper wire radius (rcopper) = 2 mm = 0.002 m
Aluminum wire radius (raluminum) = 3 mm = 0.003 m
Iron wire radius (riron) = 4 mm = 0.004 m

Let's calculate the resistance of each wire:

Resistance of copper wire (Rcopper) = (ρcopper * length) / (π * rcopper^2)
= (1.7 x 10^-8 * 2) / (π * (0.002)^2)
= 0.0017 / (π * 0.000004)
≈ 0.135 Ohms

Resistance of aluminum wire (Raluminum) = (ρaluminum * length) / (π * raluminum^2)
= (2.7 x 10^-8 * 2) / (π * (0.003)^2)
= 0.0027 / (π * 0.000009)
≈ 0.096 Ohms

Resistance of iron wire (Riron) = (ρiron * length) / (π * riron^2)
= (10 x 10^-8 * 2) / (π * (0.004)^2)
= 0.002 / (π * 0.000016)
≈ 0.039 Ohms

Now, we can sum up the resistances of the three wires to find the total resistance between A and B:

Total resistance (Rtotal) = Rcopper + Raluminum + Riron
= 0.135 + 0.096 + 0.039
≈ 0.27 Ohms

Therefore, the ohmic resistance between point A and B is approximately 0.27 Ohms.

b) To find the ohmic resistance between points A and C after attaching the second copper wire:

Since we have connected a second copper wire running from B to C, we can consider it as a parallel connection with the existing copper wire between A and B.

In a parallel circuit, the total resistance (Rtotal) can be calculated using the formula:

1/Rtotal = 1/Rcopper1 + 1/Rcopper2

where Rcopper1 and Rcopper2 are the resistances of the two copper wires.

Since the second copper wire is identical to the first one, the resistance of both wires will be the same. Therefore, the equation becomes:

1/Rtotal = 1/Rcopper + 1/Rcopper

Simplifying:

1/Rtotal = 2/Rcopper

To find Rtotal, we need to find the resistance of one copper wire (Rcopper), which we calculated already in part (a):

Rcopper ≈ 0.135 Ohms

Substituting this value into the equation:

1/Rtotal = 2/0.135

Rearranging the equation:

Rtotal = 0.135/2

Therefore, the ohmic resistance between points A and C, after attaching the second copper wire, is approximately 0.0675 Ohms.

c) When we turn AC into a near perfect circle, the resistance between A and C is affected by the change in geometry. However, since the details of the new geometry are not provided, we cannot calculate the exact resistance without the specific dimensions of the circle.