Question

A balanced Y-connected voltage source with ${\mathbf{E}}_{\mathbf{a}\mathbf{g}}\mathbf{=}\mathbf{277}\mathbf{\angle }{\mathbf{0}}^{\mathbf{0}}\mathbf{}\mathbf{V}$is applied to a balanced-Y load in parallel with a balanced-$\mathbf{∆}$ load where ${\mathbf{Z}}_{\mathbf{Y}}\mathbf{=}\mathbf{20}\mathbf{+}\mathbf{j}\mathbf{10}\mathbf{}\mathbf{\Omega }$and ${\mathbf{Z}}_{\mathbf{∆}}\mathbf{=}\mathbf{30}\mathbf{-}\mathbf{j}\mathbf{15}\mathbf{}\mathbf{\Omega }$. The Y load is solidly grounded. Using base values of ${\mathbf{S}}_{\mathbf{base}\mathbf{1}\mathbf{\varphi }}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{kVA}$and${\mathbf{V}}_{\mathbf{base}\mathbf{,}\mathbf{LN}}\mathbf{=}\mathbf{277}\mathbf{}\mathbf{V}$, calculate the source current${\mathbf{I}}_{\mathbf{a}}$ in per-unit and in amperes.

Short answer

The value of source current per unit is $0.9937\angle 9.{463}^0\mathrm{pu}$ and in amperes is

Step-by-step solution

Determine the formulas of base impedance, per-unit impedance, per-unit current and source current.

Write the formula of base impedance.

${\mathrm{Z}}_{\mathrm{base}}=\frac{{{\mathrm{V}}_{\mathrm{base}}}^2}{{\mathrm{S}}_{\mathrm{base}}}$ ……. (1)

Write the formula of per-unit impedance.

${\mathrm{Z}}_{\mathrm{per}-\mathrm{unit}}=\frac{{\mathrm{Z}}_{\mathrm{original}}}{{\mathrm{Z}}_{\mathrm{base}}}$ ……. (2)

Write the formula of per-unit current.

${\mathrm{I}}_{\mathrm{a}\left(\mathrm{pu}\right)}=\frac{{\mathrm{V}}_{\mathrm{per}-\mathrm{unit}}}{{\mathrm{Z}}_{\mathrm{y}\left(\mathrm{pu}\right)}\parallel {\mathrm{Z}}_{∆\left(\mathrm{pu}\right)}}$ …… (3)

Write the formula of base current.

${\mathrm{I}}_{\mathrm{a}\left(\mathrm{base}\right)}=\frac{{\mathrm{S}}_{\mathrm{base}}}{{\mathrm{V}}_{\mathrm{base}}}$ …… (4)

Write the formula of source current.

${\mathrm{I}}_{\mathrm{a}}={\mathrm{I}}_{\mathrm{per}-\mathrm{unit}}\times {\mathrm{I}}_{\mathrm{base}}$ …… (5)

Draw the single line diagram.

Determine the base impedance.

Substitute $277\mathrm{V}$for ${\mathrm{V}}_{\mathrm{base}}$and $10\mathrm{kVA}$for ${\mathrm{S}}_{\mathrm{base}}$in equation (1).

$\begin{array}{rcl}{\mathrm{Z}}_{\mathrm{base}}& =& \frac{{277}^2}{10\times {10}^3}\\ & =& 7.6729\mathrm{\Omega }\end{array}$

The Y-load impedance in per-unit is calculated using equation (2)

role="math" localid="1655297932067" $\begin{array}{rcl}{\mathrm{Z}}_{\mathrm{y}\left(\mathrm{pu}\right)}& =& \frac{\left(20+\mathrm{j}10\right)\mathrm{\Omega }}{7.6729\mathrm{\Omega }}\\ & =& 2.607+\mathrm{j}1.303\\ & =& 2.914\angle 26.{57}^0\end{array}$

The delta-load impedance in per-unit is calculated using equation (2).

$\begin{array}{rcl}{\mathrm{Z}}_{∆\left(\mathrm{pu}\right)}& =& \frac{\left(30-\mathrm{j}15\right)\mathrm{\Omega }}{7.6729\mathrm{\Omega }}\\ & =& 1.303-\mathrm{j}0.6516\\ & =& 1.437\angle -26.{57}^0\end{array}$

The per-unit source voltage is ${\mathrm{V}}_{\mathrm{pu}}=1\angle 0^0$.

The single-line diagram is drawn below:

Detremine the source current in per-unit and in amperes.

Determine the per-unit source current.

Substitute $1\angle 0^0$for ${\mathrm{V}}_{\mathrm{pu}}$, $2.914\angle 26.{57}^0$ for ${\mathrm{Z}}_{\mathrm{Y}\left(\mathrm{pu}\right)}$, and $1.437\angle -26.{57}^0$for ${\mathrm{Z}}_{∆\left(\mathrm{pu}\right)}$ in equation (3).

$\begin{array}{rcl}{\mathrm{I}}_{\mathrm{a}\left(\mathrm{pu}\right)}& =& \frac{1\angle 0^0}{2.914\angle 26.{57}^0\parallel 1.437\angle -26.{57}^0}\\ & =& \frac{1\angle 0^0}{1.071\angle -9.{463}^0}\\ & =& 0.9337\angle 9.{463}^0\mathrm{pu}\end{array}$

Determine the base source current

Substitute $277\mathrm{V}$for ${\mathrm{V}}_{\mathrm{base}}$ and $10\mathrm{kVA}$ for ${\mathrm{S}}_{\mathrm{base}}$ in equation (4).

$\begin{array}{rcl}{\mathrm{I}}_{\mathrm{a}\left(\mathrm{base}\right)}& =& \frac{10\times {10}^3}{277}\\ & =& 36.14\mathrm{pu}\end{array}$

Determine the original source current

Substitute$36.14\mathrm{pu}$for ${\mathrm{I}}_{\mathrm{a}\left(\mathrm{base}\right)}$ and role="math" localid="1655299458640" $0.9337\angle 9.{463}^0\mathrm{pu}$ for ${\mathrm{I}}_{\mathrm{a}\left(\mathrm{pu}\right)}$ in equation (5).

$$\begin{gathered} {\mathrm{I}}_{\mathrm{a}}=36.14\mathrm{pu}\times 0.9337\angle 9.{463}^0\mathrm{pu} \\ =33.71\angle 9.{463}^0\mathrm{A} \end{gathered}$$

The value of source current per unit is $0.9937\angle 9.{463}^0\mathrm{pu}$and in amperes is ${\mathrm{I}}_{\mathrm{a}}=33.71\angle 9.{463}^0\mathrm{A}$.