Question

For a three-phase three-wire line consisting of three solid cylindrical conductors, each with radius $\mathbf{r}$and with equal phase spacing$\mathbf{D}$between any two conductors, the inductance in$\frac{\mathbf{H}}{\mathbf{m}}$per phase is given by

(a)$\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{ln}\left(\frac{D}{r^{\text{'}}}\right)$

(b)$\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{ln}\left(\frac{D}{r^{\text{'}}}\right)$

(c) $\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{ln}\left(\frac{D}{r^{\text{'}}}\right)$

Where ${\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{e}}^{\frac{\mathbf{-}\mathbf{1}}{\mathbf{4}}}\mathbf{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{778}\mathbf{r}$

Short answer

Therefore, the correct option is (a) ${\mathrm{L}}_{\mathrm{a}}=2\times {10}^{-7}\ln \left(\frac{\mathrm{D}}{{\mathrm{r}}^{\text{'}}}\right)$.

Step-by-step solution

Step 1:  Determine the formula of total inductance for first phase.

In a three-phase, three wire line consisting of three conductors, let a, b and c represents the balance three-phase.

Write the formula of total inductance for a-phase conductor.

${\mathbf{L}}_{\mathbf{a}}\mathbf{=}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{ln}\mathbf{\left(}\frac{\mathbf{D}}{{{\mathbf{r}}_{\mathbf{x}}}^{\mathbf{\text{'}}}}\mathbf{\right)}\frac{\mathbf{H}}{\mathbf{m}}$

Due to the symmetry, the total inductance of phases b and c will be same as phase a.

That is L_a=L_b=L_c .

Determine the total inductance.

In a balanced three-[phase line, the flux linkages has the same magnitude and ${120}^0$phase displacement.

So, only single phase is considered for balanced three phase condition.

Hence, correct option is (a) ${\mathrm{L}}_{\mathrm{a}}=2\times {10}^{-7}\ln \left(\frac{\mathrm{D}}{{\mathrm{r}}^{\text{'}}}\right)$.