Question

Calculate the resistance of a 100-m length of wire having a uniform cross-sectional area of 0.1 square millimeter if the wire is made of manganin having a resistivity of 50 x 10^-8 ohm-meter. If the wire is drawn out to four times its original length, by how many times would the resistance be increased?

Answer 1

Answer:

R = p(L/A) Ohms

Where:

R= Resistance in Ohms

p(rho) = Resistivity in (Ohm meters)

L = Conductor length in meters (m)

A = Conductor cross sectional area in meters squared (m^2)

So we have:

R = (50×10^-18 Ohm m) × 100m/(0.1mm^2 × (1m/1000mm)^2)

R = 5×10^-7 Ohms

This number seems way off, as in way too low! Please double check your Resistivity. I believe it is not correct and is too low by several orders of magnitude

Answer 2

The resistance of the wire would be increased by a factor of 4 when the wire is drawn out to four times its original length.

The following formula can be used to determine the wire's resistance:

R = ρL/A

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

Given that the length of the wire is$100 m$, the cross-sectional area is $0.1$square millimeters ($or 0.0001 square meters$), and the resistivity of manganin is $50\times10^{-8}$ ohm-meter, we can calculate the resistance as:

R =$(50 \times 10^{-8} ohm-meter) \times(100 m) / (0.0001 m^2)$

R =$5 ohms$

Therefore, the resistance of the wire is   $5 ohms$.

If the wire is drawn out to four times its original length, the new length would be $4 \times 100 m = 400 m$. The cross-sectional area of the wire remains the same. Using the same formula, the new resistance can be calculated as:

R' = ρL'/A

R' = $(50 \times 10^{-8} ohm-meter) \times (400 m) / (0.0001 m^2)$

R' = $20 ohms$

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