Question

Consider a single-phase three-winding transformer with the primary excited winding of ${\mathit{N}}_{\mathbf{1}}$turns carrying a current ${\mathit{I}}_{\mathbf{1}}$and two secondary windings of ${\mathit{N}}_{\mathbf{2}}$and${\mathit{N}}_{\mathbf{3}}$turns, delivering currents of ${\mathit{I}}_{\mathbf{2}}$and ${\mathit{I}}_{\mathbf{3}}$respectively. For an ideal case, how are the ampere-turns balanced?

(a) ${\mathit{N}}_{\mathbf{1}}{\mathit{I}}_{\mathbf{1}}\mathbf{=}{\mathit{N}}_{\mathbf{2}}{\mathit{I}}_{\mathbf{2}}\mathbf{-}{\mathit{N}}_{\mathbf{3}}{\mathit{I}}_{\mathbf{3}}$

(b) ${\mathit{N}}_{\mathbf{1}}{\mathit{I}}_{\mathbf{1}}\mathbf{=}{\mathit{N}}_{\mathbf{2}}{\mathit{I}}_{\mathbf{2}}\mathbf{+}{\mathit{N}}_{\mathbf{3}}{\mathit{I}}_{\mathbf{3}}$

(c) ${\mathit{N}}_{\mathbf{1}}{\mathit{I}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\left(N_2{I}_2-N_3{I}_3\right)$

Short answer

Answer

Therefore, the correct option is (b) $N_1{I}_1=N_2{I}_2+N_3{I}_3$

Step-by-step solution

 Determine the turns ratio for two winding transformer.

The turns ratio of a two winding transformer is,

$$\begin{gathered} \frac{{\mathbf{N}}_{\mathbf{1}}}{{\mathbf{N}}_{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{I}}_{\mathbf{2}}}{{\mathbf{I}}_{\mathbf{1}}} \\ {\mathit{N}}_{\mathbf{1}}{\mathit{I}}_{\mathbf{1}}\mathbf{=}{\mathit{N}}_{\mathbf{2}}{\mathit{I}}_{\mathbf{2}} \end{gathered}$$…… (1)

Here, $N_1$and $N_2$are the number of turns of primary and secondary winding.

Also, $I_1$and $I_2$are the currents flowing in these respective windings.

determine the current relation in three winding transformer.

Now, a third winding is connected to the secondary winding.

So, $N_3$are the turns of tertiary winding and $I_3$is a current flowing through it.

Therefore, the equation (1) is modified as $N_1{I}_1=N_2{I}_2+N_3{I}_3$.

Hence, correct option is (b) $N_1{I}_1=N_2{I}_2+N_3{I}_3$.