Question

The arrangement of conductors of single phase transmission line is shown below where in the forward circuit is composed of two solid wires of 3mm in radius and the return circuit of one wire of radius 4 mm placed symmetrically with respect to forward circuit. The inductance of forward circuit A is _____$mH/km$

• A. 0.0089

Answer 1

The correct option is A 0.0089
Self GMD of side A is

$D_{SA}={(D_{11}\times D_{12})}^{1/2}={(0.7788\times 4\times 3\times {10}^{-3})}^{1/2}=0.096m$

Where $D_{11}=0.7788\times r=0.7788\times 3\times {10}^{-3}m$

$D_{12}=4m$
$D_{23}=D_{13}=\sqrt{8^2+2^2}=8.246m$

Mutual GMD:
$D_m={\left[\right(D_{13}\left)\right(D_{23}\left)\right]}^{1/2}$

$=\sqrt{8.246\times 8.246}$

$D_m=8.26m$

$L_A=2\times {10}^{-7}\ell n\frac{GMD}{selfGMD}H/m$

$=2\times {10}^{-3}\ell n\left(\frac{8.246}{0.096}\right)mH/km=0.0089$

Inductance of line A $=0.0089mH/km$