Question

a wire of resistivity 4.40 * 10 ^ -5 ohm cm has a cross-sectional area of 7.50 * 10 ^ -14 cm². calculate the length of the wire that will be required to make a 4.0 ohms resistor

Answer 1

To calculate the length of the wire required to make a resistor of a given resistance, we can use the formula for resistance in terms of resistivity:

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

Where:
- \( R \) is the resistance (in ohms),
- \( \rho \) is the resistivity (in ohm-cm),
- \( L \) is the length of the wire (in cm),
- \( A \) is the cross-sectional area (in cm²).

Given:
- \( R = 4.0 \, \text{ohms} \)
- \( \rho = 4.40 \times 10^{-5} \, \text{ohm cm} \)
- \( A = 7.50 \times 10^{-14} \, \text{cm}^2 \)

We need to solve for \( L \):

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

Now, substituting in the values:

\[
L = \frac{4.0 \, \text{ohms} \cdot 7.50 \times 10^{-14} \, \text{cm}^2}{4.40 \times 10^{-5} \, \text{ohm cm}}
\]

Calculating the numerator:

\[
4.0 \cdot 7.50 \times 10^{-14} = 30.0 \times 10^{-14} \, \text{ohm cm}^2
\]

Calculating the length \( L \):

\[
L = \frac{30.0 \times 10^{-14}}{4.40 \times 10^{-5}}
\]

Now, calculating the division:

\[
L = \frac{30.0}{4.40} \times 10^{-14 + 5} = \frac{30.0}{4.40} \times 10^{-9}
\]
\[
L \approx 6.81818 \times 10^{-9} \, \text{cm}
\]

Calculating \( 30.0 \div 4.40 \):

\[
L \approx 6.8182 \times 10^{-9} \, \text{cm} \approx 6.82 \times 10^{-9} \, \text{cm}
\]

Therefore the required length of the wire to make a \(4.0 \, \text{ohm}\) resistor is approximately:

\[
L \approx 6.82 \, \text{cm} \, (rounded)
\]