Question

For either single-phase two-wire line or balanced three-phase three-wire line with equal phase spacing and with conductor radius, the capacitance (line-to-neutral) in$\frac{\mathbf{F}}{\mathbf{m}}$ is given by${\mathit{C}}_{\mathbf{a}\mathbf{n}}\mathbf{=}\mathbf{\_}\mathbf{\_}\mathbf{\_}\mathbf{\_}$

Short answer

Therefore, the line to neutral capacitance is $\frac{2\pi \epsilon }{\ln \left(\frac{D}{r}\right)} \text{F/m}$.

Step-by-step solution

Determine the formulas to find the expression for the line-to-line capacitance.

The expression for single phase two wire line-to-line voltage is given by,

$\mathit{V}\mathbf{=}\frac{\mathbf{q}}{\mathbf{\pi }\mathbf{\epsilon }}\mathit{l}\mathit{n}\left(\frac{D}{r}\right)\mathbf{ }\mathit{V}$…… (1)

Here, is the charge of the conductor,is the distance between the conductors, are the radius of the conductors.

Determine the expression for the capacitance.

The expression for the capacitance is given by,

$C=\frac{q}{V}$

Substitute for $\frac{q}{\pi \epsilon }\ln \left(\frac{D}{r}\right)$in V the above expression.

$\begin{array}{rcl}C& =& \frac{q}{\frac{q}{\pi \epsilon }\ln \left(\frac{D}{r}\right)}\\ & =& \frac{1}{\frac{1}{\pi \epsilon }\ln \left(\frac{D}{r}\right)}\\ & =& \frac{\pi \epsilon }{\ln \left(\frac{D}{r}\right)}\\ & & \end{array}$

The relationship between the line to neutral and line-to-line capacitance is given by,

$C_{an}=2C$

Substitute$\frac{\pi \epsilon }{\ln \left(\frac{D}{r}\right)}$ for C in the above expression.

$\begin{array}{rcl}{C}_{an}& =& 2\times \frac{\pi \epsilon }{\ln \left(\frac{D}{r}\right)}\\ & =& \frac{2\pi \epsilon }{\ln \left(\frac{D}{r}\right)} \frac{\text{F}}{\text{m}}\\ & & \end{array}$

Hence the line to neutral capacitance is $\frac{2\pi \epsilon }{\ln \left(\frac{D}{r}\right)}\frac{\text{F}}{\text{m}}$.