Question

If positive-sequence voltages are assumed and the $\mathbf{Y}\mathbf{-}\mathbf{∆}$connection is considered,again with ideal transformers as in Problem 3.29, find the complexvoltage gain${\mathit{C}}_{\mathbf{3}}$.

(a) What would the gain be for a negative-sequence set?

(b) Comment on the complex power gain.

(c) When terminated in a symmetric Y-connected load, find the referred impedance${\mathbf{Z}}_{\mathbf{L}}\mathbf{\text{'}}$,the secondary impedance${\mathbf{Z}}_{\mathbf{L}}$referred to primary (i.e., the per-phase driving-point impedance on the primary side), in terms ofand the complex voltage gain$\mathit{C}$.

Short answer

Therefore, the voltage gain is${\mathrm{C}}_3=\frac{\mathrm{\eta }}{\sqrt{3}}{\mathrm{e}}^{\mathrm{j}30}$.

1. The voltage gain forthenegative sequence set is${\mathrm{C}}_3=\frac{\mathrm{\eta }}{\sqrt{3}}{\mathrm{e}}^{-\mathrm{j}30}={\mathrm{C}}_3^{*}$.
2. The complex power gain is 1.
3. The impedance gain is $\frac{1}{C_3^2}$.

Step-by-step solution

Determine the formulas of phase voltages in the star-delta connection

Draw the given diagram.

Using problem 3.29, write the formula of phase voltage at star connection with respect to line voltages at delta connection for positive sequence set.

${\mathbf{V}}_{\mathbf{a}\mathbf{\text{'}}\mathbf{n}\mathbf{\text{'}}}\mathbf{=}\sqrt{\mathbf{3}}{\mathbf{\eta e}}^{\mathbf{j}\mathbf{30}}{\mathbf{V}}_{\mathbf{an}}$ ……. (1)

Similarly, write the formula of phase voltage at delta connection with respect to line voltages at star connection for positive sequence set.

${\mathbf{V}}_{\mathbf{a}\mathbf{\text{'}}\mathbf{n}\mathbf{\text{'}}}\mathbf{=}\frac{\mathbf{\eta }}{\sqrt{\mathbf{3}}}{\mathbf{e}}^{\mathit{j}\mathbf{30}}{\mathbf{V}}_{\mathbf{an}}$ ……. (2)

Therefore, the voltage gain is ${\mathrm{C}}_3=\frac{\mathrm{\eta }}{\sqrt{3}}{\mathrm{e}}^{\mathrm{j}30}$.

Determine the voltage gain for the negative sequence

(a)

Equation (2) represents the phase voltage in the positive sequence set. Now, the phase voltage in the negative sequence set is given below.

$V_{a\text{'}n\text{'}}=\frac{\mathrm{\eta }}{\sqrt{3}}{\mathrm{e}}^{-\mathrm{j}30}{V}_{an}$ …… (3)

Therefore, the voltage gain is $\frac{\mathrm{\eta }}{\sqrt{3}}{\mathrm{e}}^{-\mathrm{j}30}={\mathrm{C}}_3^{*}$.

Determine the complex power gain

(b)

The line current in the negative sequence set is given below.

$$\begin{gathered} I_a=\frac{\mathrm{\eta }}{\sqrt{3}}{\mathrm{e}}^{\mathrm{j}30}{I}_a\text{'} \\ {I}_a\text{'}=\frac{I_a}{\frac{\mathrm{\eta }}{\sqrt{3}}{\mathrm{e}}^{\mathrm{j}30}} \end{gathered}$$ …… (4)

Therefore, the current gain is$\frac{\mathrm{\eta }}{\sqrt{3}}{\mathrm{e}}^{\mathrm{j}30}={\mathrm{C}}_3$.

The complex power gain is the product of voltage and current gain.

$$\begin{gathered} S\text{'}={\mathrm{V}}_{\mathrm{a}\text{'}\mathrm{n}\text{'}}{\left({\mathrm{I}}_{\mathrm{a}}\text{'}\right)}^{*} \\ S\text{'}=\left(\frac{\mathrm{\eta }}{\sqrt{3}}{\mathrm{e}}^{-\mathrm{j}30}{\mathrm{V}}_{\mathrm{an}}\right){\left(\frac{\begin{array}{c}\\ {\mathrm{I}}_{\mathrm{a}}\end{array}}{{\displaystyle \frac{\mathrm{\eta }}{\sqrt{3}}}{\mathrm{e}}^{\mathrm{j}30}}\right)}^{*} \\ S\text{'}=V_{an}{I}_a^{*} \\ S\text{'}=S \end{gathered}$$

Therefore, the power gain is 1.

Determine the impedance refered to as the primary side

(c)

The load impedance is given by the following:

$Z_L=\frac{V_{an}}{I_a}$

Using equations (3) and (4),

$$\begin{gathered} Z_L=\frac{V_{an}}{I_a} \\ =\frac{{\displaystyle \frac{V_{a\text{'}n\text{'}}}{{\displaystyle \frac{\eta }{\sqrt{3}}}{e}^{-j30}}}}{{\displaystyle \frac{\eta }{\sqrt{3}}}{e}^{-j30}{I}_a\text{'}} \\ =\frac{{\displaystyle \frac{1}{C_3}}}{C_3\text{'}}\frac{V_{a\text{'}n\text{'}}}{I_a} \\ =\frac{1}{C_3^2}{Z}_l\text{'} \end{gathered}$$

Therefore, the impedance gain is $\frac{1}{C_3^2}$.