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To determine the conductance for a parallel-wire line, the formula for the conductance per unit length \( G \) of two parallel wires is given by:

\[
G = \frac{\pi \sigma}{\ln\left(\frac{d}{r}\right)}
\]

Where:
- \( \sigma \) is the conductivity of the material (in S/m),
- \( d \) is the distance between the wires (in meters),
- \( r \) is the radius of the conductor (in meters).

### Given dataTo determine the conductance for a parallel-wire line, the formula for the conductance per unit length \( G \) of two parallel wires is given by:

\[
G = \frac{\pi \sigma}{\ln\left(\frac{d}{r}\right)}
\]

Where:
- \( \sigma \) is the conductivity of the material (in S/m),
- \( d \) is the distance between the wires (in meters),
- \( r \) is the radius of the conductor (in meters).

### Given 
- Conductivity \( \sigma = 200 \times 10^{-6} \, \text{S/m} \),
- Conductor diameter = 12 mm, so the radius \( r = \frac{12}{2} = 6 \, \text{mm} = 6 \times 10^{-3} \, \text{m} \),
- Conductor spacing \( d = 10 \, \text{mm} = 10 \times 10^{-3} \, \text{m} \).

### Step-by-step calculation:

1. First, calculate \( \ln\left(\frac{d}{r}\right) \):
\[
\ln\left(\frac{d}{r}\right) = \ln\left(\frac{10 \times 10^{-3}}{6 \times 10^{-3}}\right) = \ln\left(\frac{10}{6}\right) = \ln\left(1.6667\right)
\]
\[
\ln\left(1.6667\right) \approx 0.5108
\]

2. Now, substitute the values into the conductance formula:
\[
G = \frac{\pi \times 200 \times 10^{-6}}{0.5108} = \frac{628.318 \times 10^{-6}}{0.5108}
\]
\[
G \approx 1.23 \, \text{mS/m}
\]

### Final answer:
The conductance is approximately \( 1.23 \, \text{mS/m} \), which corresponds to option B.

Would you like more details, or do you have any questions?

### Related questions:
1. How does the spacing between conductors affect the conductance?
2. What would happen if the conductivity of the material were increased?
3. How do different wire diameters impact the conductance calculation?
4. What is the relationship between conductance and resistance in a parallel-wire line?
5. How can you reduce losses in a parallel-wire transmission line?

### Tip:
Increasing the spacing between conductors reduces the conductance but increases the impedance of the line.

Determine the conductance for a parallel-wire line if the conductivity is 200 x 10^-6 S/m and the conductor diameter is 12 mm. The conductor spacing is 10 mm.

Learn how to calculate the conductance of a parallel-wire line using the formula G = (π * σ) / ln(d / r). Given the conductivity of 200 x 10^-6 S/m, a conductor diameter of 12 mm, and conductor spacing of 10 mm, the conductance is calculated as approximately 1.23 mS/m.

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