Question

Consider an ideal single-phase-2 winding transformer of turns ratio$\raisebox{1ex}{${\mathbf{N}}_{\mathbf{1}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{N}}_{\mathbf{2}}$}\right.\mathbf{=}\mathit{a}$. If it is converted to an autotransformer arrangement with a transformation ratio of $\raisebox{1ex}{${\mathbf{V}}_{\mathbf{H}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{V}}_{\mathbf{x}}$}\right.\mathbf{=}\mathbf{1}\mathbf{+}\mathit{x}$, (the autotransformer rating /two-winding transformer rating) would then be

a)$\mathbf{1}\mathbf{+}\mathit{x}$

b) $\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{a}}$

c)$\mathit{a}$

Short answer

Therefore the correct option is (b):$1+\frac{1}{a}$

Step-by-step solution

Two winding transformer transformation ratio.

The KVA relation between two-winding transformer and auto transformer is given by,

${\left(\mathrm{KVA}\right)}_{\mathbf{Autotransformer}}\mathbf{=}{\left(\mathrm{KVA}\right)}_{\mathbf{2}\mathbf{}\mathbf{Winding}\mathbf{}\mathbf{transformer}}\mathbf{=}\left(\frac{1}{1-K}\right)$

Rearrange the above expression.

$\frac{{\left(KVA\right)}_{Autotransformer}}{{\left(KVA\right)}_{2windingtransformer}}=\left(\frac{1}{1-K}\right)$ ...... (1)

the value of K is,

$\mathbf{K}\mathbf{=}\frac{\mathbf{Low}\mathbf{}\mathbf{voltage}\mathbf{}\mathbf{side}\mathbf{}\left(V_X\right)}{\mathbf{High}\mathbf{}\mathbf{voltage}\mathbf{}\mathbf{side}\mathbf{}\left(V_H\right)}$ ...... (2)

Consider the expression for the transformation ratio.

$\frac{V_X}{V_H}=1+a$ ....... (3)

Using the equation (2) and (3), deduce the following expression.

$\mathbf{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}\mathbf{a}}$

Substitute $\frac{1}{1+a}$for $K$in equation (1).

$\begin{array}{rcl}\frac{{\left(KVA\right)}_{Autotransformer}}{{\left(KVA\right)}_{2windingtransformer}}& =& \left(\frac{1}{1-{\displaystyle \frac{1}{1+a}}}\right)\\ & =& \frac{1}{{\displaystyle \frac{1+a-1}{1+a}}}\\ & =& \frac{1+a}{a}\\ & =& 1+\frac{1}{a}\end{array}$

Explanation for option (a)

From equation (1) as the autotransformer rating to two-winding transformer is$1+\frac{1}{a}$ and not$1+a$ , the option (a) is incorrect.

Thus the (a) is incorrect option.

Explanation for option (b)

Rearrange the expression in equation (1).

${\left(KVA\right)}_{Autotransformer}=\left(\frac{1}{1+a}\right){\left(KVA\right)}_{2-windingtransformer}$

Hence, the KVA rating of autotransformer is $\left(\frac{1}{1+a}\right)$times the KVA rating of two-winding transformer.

Thus the correct option is (b).

Explanation for option (c)

From equation (1) as the autotransformer rating to two-winding transformer is$\left(1+\frac{1}{a}\right)$ and not , the option (a) is incorrect