If a resistance of 3.5 Ohms was measured using the 4-pin Wenner method with a pin spacing of 2 meters, what is the resistivity?

If a resistance of 3.5 Ohms was measured using the 4-pin Wenner method with a pin spacing of 2 meters, what is the resistivity?

• A. 44 Ohm-cm
• B. 132 Ohms
• C. 132 Ohms-cm
• D. 4397 Ohm-cm

Answer: (C)

To determine the resistivity from a resistance measurement taken using the 4-pin Wenner method, the formula used is: \[ \rho = R \cdot \frac{2 \pi a}{\ln(2)} \] where: - \(\rho\) is the resistivity, - \(R\) is the resistance measured (3.5 Ohms in this case), - \(a\) is the distance between the pins (2 meters for this scenario). First, let's compute the value of \( \frac{2 \pi a}{\ln(2)} \). The value of \( \ln(2) \) is approximately 0.693. Therefore, \[ \frac{2 \pi (2)}{\ln(2)} \approx \frac{4 \pi}{0.693} \approx 18.099 \] Now, substituting the values into the resistivity formula gives: \[ \rho = 3.5 \cdot 18.099 \approx 63.35 \, \text{Ohm-m} \] To convert Ohm-meters to Ohm-centimeters, we can multiply by 100 (since there are 100 cm

To determine the resistivity from a resistance measurement taken using the 4-pin Wenner method, the formula used is:

[

\rho = R \cdot \frac{2 \pi a}{\ln(2)}

]

where:

• (\rho) is the resistivity,

• (R) is the resistance measured (3.5 Ohms in this case),

• (a) is the distance between the pins (2 meters for this scenario).

First, let's compute the value of ( \frac{2 \pi a}{\ln(2)} ). The value of ( \ln(2) ) is approximately 0.693. Therefore,

[

\frac{2 \pi (2)}{\ln(2)} \approx \frac{4 \pi}{0.693} \approx 18.099

]

Now, substituting the values into the resistivity formula gives:

[

\rho = 3.5 \cdot 18.099 \approx 63.35 , \text{Ohm-m}

]

To convert Ohm-meters to Ohm-centimeters, we can multiply by 100 (since there are 100 cm