Question

(a) An ideal single-phase two-winding transformer with turns ratio ${\mathit{a}}_{\mathbf{t}}\mathbf{=}\frac{{\mathbf{N}}_{\mathbf{1}}}{{\mathbf{N}}_{\mathbf{2}}}$is connected with a series impedance localid="1656741053389" ${\mathit{Z}}_{\mathbf{2}}$across winding 2. If one wants to replace localid="1656741059221" ${\mathit{Z}}_{\mathbf{2}}$with a series impedance localid="1656741063182" ${\mathit{Z}}_{\mathbf{1}}$across winding 1 and keep the terminal behavior of the two circuits to be identical, find localid="1656741066768" ${\mathit{Z}}_1$in terms of localid="1656741070118" ${\mathit{Z}}_{\mathbf{2}}$

(b) Would the above result be true if instead of a series impedance there is a shunt impedance?

(c) Can one refer a ladder network on the secondary (2) side to the primary (1) side simply by multiplying every impedance by localid="1656741074149" ${\mathit{a}}_{\mathbf{t}}^{\mathbf{2}}$?

Short answer

Answer

(a) The relation between $Z_1$and $Z_2$is $Z_1=a^2{Z}_2$
.

(b) The above result will be true as $Y_2=a^2{Y}_1$.

(c) The ladder network can be referred from the secondary to primary when impedance is multiplied by the $a_t^2$.

Step-by-step solution

Determine the formulas of primary and secondary impedances, transformation ratio.

Write the formula of primary impedance.

$$\begin{gathered} {\mathit{Z}}_{\mathbf{1}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{I}}_{\mathbf{1}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Y}}_{\mathbf{1}}} \end{gathered}$$ ……. (1)

Write the formula of secondary impedance.

$$\begin{gathered} {\mathit{Z}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{I}}_{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Y}}_{\mathbf{2}}} \end{gathered}$$ ……. (2)

Write the formula of transformer ratio

$$\begin{gathered} \frac{{\mathbf{N}}_{\mathbf{1}}}{{\mathbf{N}}_{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{V}}_{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{I}}_{\mathbf{2}}}{{\mathbf{I}}_{\mathbf{1}}} \end{gathered}$$ ……. (3)

Determine the relation between primary and secondary series impedances.

(a)

Take the ratio of primary and secondary impedance and solve.

$$\begin{gathered} \frac{Z_1}{Z_2}=\frac{{\displaystyle \frac{V_1}{I_1}}}{{\displaystyle \frac{V_2}{I_2}}} \\ =\frac{V_1}{V_2}\times \frac{I_2}{I_1} \\ =\frac{N_1}{N_2}\times \frac{N_1}{N_2} \\ =\left(\frac{N_1}{N_2}\right) \end{gathered}$$

Therefore, relation of primary and secondary impedance is,

$$\begin{gathered} \frac{Z_1}{Z_2}={\left(a\right)}^2 \\ {Z}_1=a^2{Z}_2 \end{gathered}$$ ……. (4)

Here, $a_t^2={\left(\frac{N_1}{N_2}\right)}^2$.

Determine the relation between primary and secondary shunt impedances.

(b)

Let $Y_1$and $Y_2$are the shunt impedances at primary and secondary side.

Take a ratio of primary and secondary shunt impedance.

$$\begin{gathered} \frac{Y_1}{Y_2}=\frac{{\displaystyle \frac{I_1}{V_1}}}{{\displaystyle \frac{I_2}{V_2}}} \\ =\frac{V_2}{V_1}\times \frac{I_1}{I_2} \\ =\frac{N_2}{N_1}\times \frac{N_2}{N_1} \\ ={\left(\frac{N_2}{N_1}\right)}^2 \end{gathered}$$

Therefore, relation of primary and secondary shunt impedance is,

$$\begin{gathered} \frac{Y_1}{Y_2}=\frac{1}{a_t^2} \\ {Y}_2=a_t^2{Y}_1 \end{gathered}$$

Therefore, the relation is true for shunt impedance.

Determine the condition for ladder network.

(c)

Using relation (4), every secondary impedance can be converted into primary impedance if turns ratio is known.

So, the ladder network connected in secondary can move to primary by multiplying every impedance by $a_t^2$.