Question

A single-phase transformer has 2000 turns on the primary winding and 500 turns on the secondary. Winding resistances are ${\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathit{\Omega }$, and ${\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{125}$; leakage reactance’s are${\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{8}$, and ${\mathit{X}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathit{\Omega }$. The resistance load on the secondary is $\mathbf{12}\mathbf{}\mathit{\Omega }$.

(a) If the applied voltage at the terminals of the primary is $\mathbf{1000}\mathbf{}\mathit{V}$, determine ${\mathit{v}}_{\mathbf{2}}$at the load terminals of the transformer, neglecting magnetizing current.

(b) If the voltage regulation is defined as the difference between the voltage magnitude at the load terminals of the transformer at full load and at no load in percent of full-load voltage with input voltage held constant, compute the percent voltage regulation.

Short answer

Answer

1. The load voltage is $244.086\angle -4.{66}^{°}V$.

The percentage regulation is $2.42\%$.

Step-by-step solution

Determine the formulas of winding resistance, leakage reactance, load and load resistance referred to primary, original load voltage, voltage regulation and turns ratio.

Write the formula of turns ratio.

$\mathit{a}\mathbf{=}\frac{{\mathbf{N}}_{\mathbf{1}}}{{\mathbf{N}}_{\mathbf{2}}}$ ……. (1)

Write the formula of secondary winding resistance referred to primary.

${\mathit{X}}_{\mathbf{2}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}{\mathit{X}}_{\mathbf{2}}$ ……. (2)

Write the formula of secondary leakage reactance referred to primary.

${\mathit{R}}_{\mathbf{L}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}{\mathit{R}}_{\mathbf{L}}$ ……. (3)

Write the formula of load resistance referred to primary.

${\mathit{R}}_{\mathbf{L}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}{\mathit{R}}_{\mathbf{L}}$ ……. (4)

Write the formula of equivalent resistance

${\mathit{R}}_{\mathbf{e}\mathbf{q}}\mathbf{=}{\mathit{R}}_{\mathbf{L}}^{\mathbf{\text{'}}}\mathbf{+}{\mathit{R}}_{\mathbf{2}}^{\mathbf{\text{'}}}\mathbf{+}{\mathit{R}}_{\mathbf{1}}$ ……. (5)

Write the formula of equivalent reactance

${\mathit{X}}_{\mathbf{e}\mathbf{q}}\mathbf{=}{\mathit{X}}_{\mathbf{2}}^{\mathbf{\text{'}}}\mathbf{+}{\mathit{X}}_{\mathbf{1}}$ ……. (6)

Write the formula of load voltage referred to primary

${\mathit{V}}_{\mathbf{L}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{V}}_{\mathbf{s}}\left(\frac{R_L^{\text{'}}}{R_{eq}+X_{eq}}\right)$ ……. (7)

Write the formula of original load voltage

${\mathit{V}}_{\mathbf{L}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{L}}^{\mathbf{\text{'}}}}{\mathbf{a}}$ ……. (8)

Write the formula of percentage voltage regulation

${\mathit{V}}_{\mathbf{R}}\mathbf{\%}\mathbf{=}\left(\frac{V_s-V_L^{\text{'}}}{V_L^{\text{'}}}\right)\mathbf{100}$ ……. (9)

Determine the load voltage.

(a)

Determine turns ratio.

Substitute 2000 for $N_1$and 500 for $R_2$in equation (1).

$$\begin{gathered} a=\frac{N_1}{N_2} \\ =\frac{2000}{500} \\ =4 \end{gathered}$$

Determine secondary winding resistance and reactance referred to primary winding.

Substitute $0.125\Omega$for $R_2$and 4 for a in equation (2)

$$\begin{gathered} R_2^{\text{'}}={\left(4\right)}^2\times 0.125 \\ =2\Omega \end{gathered}$$

Substitute $0.5\Omega$for $X_2$and 4 for a in equation (3)

$$\begin{gathered} X_2^{\text{'}}={\left(4\right)}^2\times 0.5 \\ =8\Omega \end{gathered}$$

Substitute $12\Omega$for $R_L$and 4 for a in equation (4)

$$\begin{gathered} R_L^{\text{'}}={\left(4\right)}^2\times 12 \\ =192\Omega \end{gathered}$$

Substitute $192\Omega$for $R_L^{\text{'}}$, $2\Omega$for $R_2^{\text{'}}$and $2\Omega$for $R_1$in equation (5)

$$\begin{gathered} R_{eq}=192+2+2 \\ =196\Omega \end{gathered}$$

Substitute role="math" localid="1655820498180" $8\Omega$for $X_2^{\text{'}}$and $8\Omega$for $X_1^{}$in equation (6)

$$\begin{gathered} X_{eq}=8+8 \\ =16\Omega \end{gathered}$$

Determine the load voltage.

Substitute $16\Omega$for $X_{eq}$, $196\Omega$for $R_{eq}$, $192\Omega$for $R_2^{\text{'}}$and 100 V for $V_s$in equation (7)

$$\begin{gathered} V_2^{\text{'}}=1000\times \frac{192}{196+j16} \\ =976.344\angle -4.{66}^{°}V \end{gathered}$$

Substitute $76.344\angle -4.{66}^{°}V$for and 4 for a in equation (8)

$$\begin{gathered} V_L=\frac{V_L^{\text{'}}}{a} \\ =\frac{976.344\angle -4.{66}^{°}}{4} \\ =244.086\angle -4.{66}^{°}V \end{gathered}$$

Thus, the load voltage is $244.086\angle -4.{66}^{°}V$.

Determine the voltage regulation.

(b)

Determine the voltage regulation.

Substitute $976.344V$for $V_2^{\text{'}}$and $1000V$for $V_s$in equation (9)

$$\begin{gathered} V_R\%=\frac{1000-976.344}{976.344}\times 100 \\ =2.42\% \end{gathered}$$

Thus, the percentage load regulation is $2.42\%$.