Question

An ideal transformer with N₁ = 1000 and N₂= 250 is connected with an impedance Z₂₂ across winding 2. If V₁ = 1000 $\mathbf{\angle }{\mathbf{0}}^{\mathbf{o}}$and I₁ = 5$\angle -{30}^0$, determine localid="1656740653429" ${\mathit{V}}_{\mathbf{2}}\mathbf{}\mathbf{,}\mathbf{}{\mathit{I}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{}{\mathit{Z}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{,}$and impedance localid="1656740657616" ${\mathit{Z}}_{\mathbf{2}}\mathbf{\text{'}}$, which is the value of localid="1656740661426" ${\mathit{Z}}_{\mathbf{2}}$referred to the primary side of the transformer.

Short answer

Answer

The secondary winding voltage is $250\angle 0^{°}V$, the secondary current is $20\angle {30}^{°}A$, the secondary impedance is $12.5\angle {30}^{°}\Omega$and the impedance referred to primary is $200\angle {30}^{°}\Omega$

Step-by-step solution

Determine the formulas of transformer ratio and load impedance.

Write the formula of transformer ratio in terms of induced voltage.

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

Write the formula of transformer ratio in terms of induced voltage.

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

Write the formula of impedance connected across secondary winding.

${\mathbf{Z}}_{\mathbf{2}}=\frac{{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{I}}_{\mathbf{2}}}$ ……. (3)

Write the formula of secondary impedance referred to the primary

${\mathit{Z}}_{\mathbf{2}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{Z}}_{\mathbf{2}}{\mathbf{\left(}\frac{{\mathbf{N}}_{\mathbf{1}}}{{\mathbf{N}}_{\mathbf{2}}}\mathbf{\right)}}^{\mathbf{2}}$ …… (4)

Determine the secondary winding voltage, secondary current, impedance and secondary impedance referred to the primary.

Determine the secondary winding voltage

Substitute $1000\angle 0^{°}V$for V₁, 1000 for $N_1$and 250 for $N_2$ in equation (1).

$\begin{array}{rcl}\frac{V_2}{1000\angle 0^{°}V}& =& \frac{250}{1000}\\ V_2& =& 250\angle 0^{°}V\end{array}$

Determine the secondary current

Substitute $5\angle -{30}^{°}A$for $I_1$, 1000 for $N_1$and 250 for$N_2$ in equation (2).

$\begin{array}{rcl}\frac{5\angle -{30}^{°}A}{I_2}& =& \frac{250}{1000}\\ I_2& =& 20\angle -{30}^{°}A\end{array}$

Determine the impedance connected across secondary winding

Substitute $20\angle -{30}^{°}A$for $I_2$and $250\angle 0^{°}V$for$V_2$in equation (3).

$$\begin{gathered} Z_2=\frac{250\angle 0^{°}V}{20\angle -{30}^{°}A} \\ =12.5\angle {30}^{°} \end{gathered}$$

Determine the impedance referred to the primary

Substitute $12.5\angle {30}^{°}\Omega$for $Z_2$, 1000 for $N_1$and 250 for $N_2$ in equation (4).

$$\begin{gathered} Z_2\text{'}=12.5\angle {30}^{°}\Omega \times {\left(\frac{1000}{250}\right)}^2 \\ =200\angle {30}^{°}\Omega \end{gathered}$$

Therefore, the secondary winding voltage is$150\angle 0^{°}V$, the secondary current is $20\angle -{30}^{°}A$, the secondary impedance is $12.5\angle {30}^{°}\Omega$and the impedance referred to primary is $200\angle {30}^{°}\Omega$