Question

For a single-phase, two-wire line consisting of two solid cylindrical conductors of same radius r, the total circuit inductance, also called loop inductance, is given by (in H/m)

(a) 2 x 10^-7 In $\left(\frac{D}{r^{\text{'}}}\right)$

(b) 4 x 10^-7 In $\left(\frac{D}{r\text{'}}\right)$

Where r^'= $e^{\frac{-1}{4}}r=0.778r$

Short answer

The correct option is (b) L_P= 4 x 10⁷In$\left(\frac{D}{r^{\text{'}}}\right)$

Step-by-step solution

 Determine the formula of total inductance for first conductor.

In a single-phase, two wire line consisting of two conductors, let x is first conductor and y is second conductor.

Write the formula of total inductance for first conductor.

L_x = 2 x 10^-7 In $\left(\frac{D}{r^{\text{'}}}\right)$$\frac{H}{m}$

Write the formula of total inductance for second conductor.

L_x = 2 x 10^-7 In$\left(\frac{D}{{r_y}^{\text{'}}}\right)\frac{H}{m}$

Determine the total inductance.

Write the formula of total inductance of the two wire line system.

L_p= L_x = L_y

Substitute the value and solve as,

Assume that, radius r^'_x and r^'_y for conductors x and y are same, therefore, substitute r^' for r^'_x and r^'_yin equation (1).

Hence, correct option is (b).