SSE JE Electrical Engineering 24 March 2021 Shift 1 Paper

Loop inductance of a single-phase two-wire line
Considered a single-phase line consisting of two conductors (phase and neutral) a and b of equal radius r. They are situated at a distance D meters. The cross sections of conductors are shown in the diagram below.Let the current flow in the conductors are opposite in direction so that one becomes the return path for the other.
The flux linkages of conductor ‘a’ is given by the formula:
${\mathrm{\lambda }}_{\mathrm{a}}=2\times {10}^{-7}\left[{\mathrm{I}}_{\mathrm{a}}\ln \frac{1}{{\mathrm{D}}_{\mathrm{aa}}}+{\mathrm{I}}_{\mathrm{b}}\ln \frac{\mathrm{I}}{{\mathrm{D}}_{\mathrm{ab}}}\right]$
${\mathrm{I}}_{\mathrm{a}}=+\mathrm{I},{\mathrm{I}}_{\mathrm{b}}=-\mathrm{I},{\mathrm{D}}_{\mathrm{aa}}=\mathrm{r}\prime ,{\mathrm{D}}_{\mathrm{ab}}=\mathrm{D}$
${\mathrm{\lambda }}_{\mathrm{a}}=2\times {10}^{-7}\left[\mathrm{I}\left[\ln \frac{1}{{\mathrm{D}}_{\mathrm{aa}}}\right]-\mathrm{I}\left[\ln \frac{1}{{\mathrm{D}}_{\mathrm{ab}}}\right]\right]$ $=2\times {10}^{-7}\mathrm{I}\left[\ln \frac{{\mathrm{D}}_{\mathrm{ab}}}{{\mathrm{D}}_{\mathrm{aa}}}\right]$
${\mathrm{L}}_{\mathit{a}}={\mathrm{\lambda }}_{\mathrm{a}}\mathrm{I}$ $=2\times {10}^{-7}\left[\ln \frac{\mathrm{D}}{\mathrm{r}\prime }\right]$ $={\mathrm{L}}_{\mathrm{b}}={\mathrm{L}}_{\mathrm{c}}$
Loop inductance $={\mathrm{L}}_{\mathrm{a}}+{\mathrm{L}}_{\mathrm{b}}=4\times {10}^{-7}\left[\ln \frac{\mathrm{D}}{\mathrm{r}\prime }\right]$ ........(1)
Inductance per conductor in a three-phase line (symmetrical)
Let the spacing between the conductors be D and the radius of each conductor, r. The flux linkages of conductor a is given by the equation:
The flux linkages of the conductor ‘a’ is given by the formula:${\mathrm{\lambda }}_{\mathit{a}}=2\times {10}^{-7}\left[{\mathrm{I}}_{\mathrm{a}}\ln \frac{1}{{\mathrm{D}}_{\mathrm{aa}}}+{\mathrm{I}}_{\mathrm{b}}\ln \frac{1}{{\mathrm{D}}_{\mathrm{ab}}}+{\mathrm{I}}_{\mathrm{c}}\ln \frac{1}{{\mathrm{D}}_{\mathrm{ac}}}\right]$
Where ${\mathrm{D}}_{\mathrm{aa}}=\mathit{r}\prime$, ${\mathrm{D}}_{\mathrm{ab}}={\mathrm{D}}_{\mathrm{bc}}={\mathrm{D}}_{\mathrm{ac}}=\mathrm{D}$
${\mathrm{\lambda }}_{\mathrm{a}}=2\times {10}^{-7}\left[{\mathrm{I}}_{\mathrm{a}}\ln \frac{1}{\mathrm{r}\prime }+{\mathrm{I}}_{\mathrm{b}}\ln \frac{1}{\mathrm{D}}+{\mathrm{I}}_{\mathrm{c}}\ln \frac{1}{\mathrm{D}}\right]$
Now from $3-\mathrm{\varphi },3$ wire system
${\mathrm{I}}_{\mathrm{a}}+{\mathrm{I}}_{\mathrm{b}}+{\mathrm{I}}_{\mathrm{c}}=0$
${\mathrm{I}}_{\mathrm{a}}=-\left({\mathrm{I}}_{\mathrm{b}}+{\mathrm{I}}_{\mathrm{c}}\right)$
${\mathrm{\lambda }}_{\mathit{a}}=2\times {10}^{-7}\left[{\mathrm{I}}_{\mathrm{a}}\ln \frac{1}{\mathrm{r}\prime }+\left({\mathrm{I}}_{\mathrm{b}}+{\mathrm{I}}_{\mathrm{c}}\right)\ln \frac{1}{\mathrm{D}}\right]$
${\mathrm{\lambda }}_{\mathit{a}}=2\times {10}^{-7}\left[{\mathrm{I}}_{\mathrm{a}}\ln \frac{1}{\mathrm{r}\prime }-\left({\mathrm{I}}_{\mathrm{a}}\right)\ln \frac{1}{\mathrm{D}}\right]$
${\mathrm{\lambda }}_{\mathrm{a}}=2\times {10}^{-7}\left[{\mathrm{I}}_{\mathrm{a}}\ln \frac{\mathrm{D}}{\mathrm{r}\prime }\right]$
${\mathrm{L}}_{\mathrm{a}}=\frac{{\mathrm{\lambda }}_{\mathrm{a}}}{\mathrm{I}}=2\times {10}^{-7}\left[\ln \frac{\mathrm{D}}{\mathrm{r}\prime }\right]={\mathrm{L}}_{\mathrm{b}}={\mathrm{L}}_{\mathrm{c}}$ ....(2)
Where ${\mathrm{L}}_{\mathrm{a}}$ is the inductance per conductor in a three-phase line
From equations (1) and (2),
Inductance per conductor in a 3-phase line = $\left(\frac{1}{2}\right)$ times loop inductance in 1 phase.