Question

a wire of resistivity 4.40 * 10 ^ - 5 ohms cm has a cross-sectional area of 7.50 * 10 ^ - 4 cm². calculate the length of the wire that will be required to make a 4.0 ohm resistor . change to meter

Answer 1

To determine the length of the wire needed to create a resistor with a resistance of 4.0 ohms, we can use the formula for resistance (\(R\)) of a uniform wire:

\[
R = \rho \frac{L}{A}
\]

where:
- \(R\) is the resistance,
- \(\rho\) is the resistivity,
- \(L\) is the length of the wire,
- \(A\) is the cross-sectional area.

Given:
- \(R = 4.0 \, \text{ohms}\)
- \(\rho = 4.40 \times 10^{-5} \, \Omega \cdot \text{cm}\)
- \(A = 7.50 \times 10^{-4} \, \text{cm}^2\)

First, we can rearrange the formula to solve for \(L\):

\[
L = \frac{R \cdot A}{\rho}
\]

Now, we can substitute in the values:

\[
L = \frac{4.0 \, \Omega \cdot (7.50 \times 10^{-4} \, \text{cm}^2)}{4.40 \times 10^{-5} \, \Omega \cdot \text{cm}}
\]

Calculating the numerator:

\[
4.0 \, \Omega \cdot (7.50 \times 10^{-4} \, \text{cm}^2) = 3.0 \times 10^{-3} \, \Omega \cdot \text{cm}^2
\]

Now, we perform the division:

\[
L = \frac{3.0 \times 10^{-3} \, \Omega \cdot \text{cm}^2}{4.40 \times 10^{-5} \, \Omega \cdot \text{cm}} = \frac{3.0 \times 10^{-3}}{4.40 \times 10^{-5}}
\]

Calculating \(L\):

\[
L \approx 68.18 \, \text{cm}
\]

To convert this length into meters:

\[
L \approx 68.18 \, \text{cm} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 0.6818 \, \text{m}
\]

Therefore, the length of the wire required to make a 4.0 ohm resistor is approximately:

\[
\boxed{0.6818 \, \text{m}}
\]