Question

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

Short answer

The value of source current per unit is $0.9337\angle 9.463°\text{\hspace{0.17em}pu}$ and in amperes is $I_a=33.71\angle 9.463°\text{\hspace{0.17em}A}$.

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.

${\mathrm{I}}_{\mathrm{a}\left(\mathrm{pu}\right)}=\frac{{\mathrm{V}}_{\mathrm{per}-\mathrm{unit}}}{{\mathrm{Z}}_{\mathrm{y}\left(\mathrm{pu}\right)}\left|\right|{\mathrm{Z}}_{\mathrm{\Delta }\left(\mathrm{pu}\right)}}$ …… (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. 

The base kVA power is$S_{base}=10\text{\hspace{0.17em}kVA}$ .

The base voltage is $V_{base}=277\text{\hspace{0.17em}V}$.

Determine the base impedance.

Substitute$277\text{\hspace{0.17em}V}$for $V_{base}$ and $10\text{\hspace{0.17em}kVA}$ for $S_{base}$ in equation (1).

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

$\begin{array}{c}{Z}_{Y\left(pu\right)}=\frac{(20+j10)\Omega }{7.6729\Omega }\\ =\text{2.607+j1.303}\\ =2.914\angle 26.57°\end{array}$

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

$\begin{array}{c}{Z}_{\Delta \left(pu\right)}=\frac{(30-j15)\Omega }{7.6729\Omega }\\ =1.303-j0.6516\\ =1.437\angle -26.57°\end{array}$

The per-unit source voltage is $V_{pu}=1\angle 0°$.

The single-line diagram is drawn below:

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

Determine the per-unit source current.

Substitute $1\angle 0°$ for $V_{pu}$, $2.914\angle 26.57°$for $Z_{Y\left(pu\right)}$, and $1.437\angle -26.57°$ for $Z_{\Delta \left(pu\right)}$in equation (3).

Determine the base source current

Substitute$277V$ for $V_{base}$ and$10kVA$ for$S_{base}$ in equation (4).

Determine the original source current

Substitute $36.14\text{\hspace{0.17em}pu}$ for $I_{a\left(base\right)}$and $0.9337\angle 9.463°\text{\hspace{0.17em}pu}$for $I_{a\left(pu\right)}$ in equation (5).

The value of source current per unit is $0.9337\angle 9.463°\text{\hspace{0.17em}pu}$ and in amperes is$I_a=33.71\angle 9.463°\text{\hspace{0.17em}A}$.