Question

A balanced three-phase 240-V source supplies a balanced three-phase load. If the line current ${\mathrm{l}}_{\mathrm{A}}$is measured to be 15 A and is in phase with the line-to-line voltage,localid="1655122963197" ${\mathbf{V}}_{\mathbf{BC}}$, find the per-phase load impedance if the load is (a) Y-connected, (b)localid="1655122973134" $\mathbf{∆}$-connected.

Short answer

(a) The per phase load impedance for Y- connected is$j9.23\Omega .$

(b) The per phase load impedance for - connected is$8.66\angle -60°A$.

Step-by-step solution

Calculate the per phase load impedance if the load is Y-connected.

(a)

The voltage between line A and neutral is given by

${\mathit{E}}_{\mathbf{A}\mathbf{N}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{L}\mathbf{L}}}{\sqrt{\mathbf{3}}}\mathbf{\angle }\mathbf{0}\mathbf{°}$

Here, $E_{LL}$is the line-to-line voltage.

Substitute the values and solve as

$$\begin{gathered} E_{AN}=\frac{240}{\sqrt{3}}\angle 0^0 \\ \begin{array}{ll}{E}_{AN}=138.57\angle 0^0& \end{array} \end{gathered}$$

Since voltage $E_{AN}$is the reference voltage and voltage $V_{AB,}{V}_{BC,}{V}_{CA}$have the phase shift of$\angle 30°,\angle -90°,\&\angle 150°,$respectively and current $l_A$is in the phase of $V_{BC}$. So, current $l_A$would be$15\angle -{90}^o,A$.

The impedance of the phase A is given by

$Z_A=\frac{E_{AN}}{I_A}$

Substitute the values and solve as

$\begin{array}{l}{Z}_A=\frac{138.56\angle 0^0,}{15\angle -{90}^0,}\\ Z_A=9.23\angle {90}^0,\end{array}$

Hence, the per phase load impedance for Y connected is$j9.23\Omega$.

Step 2: Calculate the per phase load impedance if the load is △- connected.

(b)

The current between line A and B is given by

$I_{AB}=\frac{I_A}{\sqrt{3}}\angle {30}^o$

Here, $l_A$is the line current.

Substitute the values and solve as

$\begin{array}{l}{l}_{AB}=\frac{15}{\sqrt{3}}\angle {\left(-90+30\right)}^o\\ l_{AB}=8.66\angle -{60}^{o}A\end{array}$

The load impedance AB is given by,

$Z_{AB}=\frac{E_{AB}}{I_A}$

Substitute the values and solve as,

$\begin{array}{l}{Z}_{AB}=\frac{240}{8.66\angle -{60}^{o}A}\\ Z_A=j27.71\Omega \end{array}$

Hence, the per phase load impedance for $Y$-connected is$j9.23\Omega$.