Question

Consider the equivalent circuit of figure$\mathbf{3}\mathbf{.}\mathbf{10}\left(c\right)$in the text. After neglecting the winding resistances and exciting current, could ${\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{,}{\mathbf{x}}_{\mathbf{3}}$becomes negative, even though the leakage reactance are always positive.

(a) Yes

(b) No

Short answer

Therefore the correct option is (a): Yes.

Step-by-step solution

Determine the leakage reactance for the exciting currents.

Consider the circuit diagram given in the figure$3.20\left(c\right)$.

Consider$X_{12},X_{13},X_{23}$,to be the per-unit leakage reactance between the winding$1,2$and $3$respectively.

The per unit reactance are given by,

$X_1=\frac{1}{2}\left(X_{12}+X_{13}-X_{23}\right)$…… (1)

$$\begin{gathered} X_{1}2=\frac{1}{2}\left(X_{12}+X_{23}-X_{13}\right) \\ X3=\frac{1}{2}\left(X_{13}+X_{23}-X_{12}\right) \end{gathered}$$

Explanation for correct answer.

The leakage reactance is always positive. From equation (1), if the leakage$\left(X_{23}\right)$reactance has a value greater than the sum of the other two leakage reactance$\left(X_{12}and{x}_{13}\right)$, then the per unit series reactance will be negative. Similarly, for the other two equations.

Since, after neglecting the winding resistances and exciting current, $X_1,X_2$, and $X_3$can become negative, even though the leakage reactances are always positive.

Thus (a) is the correct option.

As$X_1,X_2$and $X_3$can become negative even though the leakage reactances are always positive.

Thus the (b) is incorrect option.