Question

A three-phase line, which has an impedance of $\mathbf{(}\mathbf{2}\mathbf{+}\mathbf{j}\mathbf{4}\mathbf{)}\mathbf{}\mathbf{\Omega }$per phase, feeds two balanced three-phase loads that are connected in parallel. One of the loads is Y-connected with an impedance of $\mathbf{(}\mathbf{30}\mathbf{+}\mathbf{j}\mathbf{40}\mathbf{)}\mathbf{}\mathbf{\Omega }$per phase, and the other is $\mathbf{∆}$-connected with an impedance of $\mathbf{(}\mathbf{60}\mathbf{-}\mathbf{j}\mathbf{45}\mathbf{)}\mathbf{}\mathbf{\Omega }$per phase. The line is energized at the sending end from a $\mathbf{60}\mathbf{}\mathbf{Hz}$, three-phase, balanced voltage source of $\mathbf{120}\sqrt{\mathbf{3}}\mathbf{V}$(rms, line-to-line). Determine (a) the current, real power, and reactive power delivered by the sending-end source; (b) the line-to-line voltage at the load; (c) the current per-phase in each load; and (d) the total three-phase real and reactive powers absorbed by each load and by the line. Check that the total three-phase complex power delivered by the source equals the total three-phase power absorbed by the line and loads.

Short answer

(a) The source current is $5\angle 0^0\mathrm{A}$, the real and reactive power delivered by source is $1800\mathrm{W}$ and$0\mathrm{VAR}$ .

(b) The load voltage is $193.64\mathrm{V}$.

(c) The source current in star load is $2.236\angle -63.{43}^0\mathrm{A}$and the current in delts load is $4.472\angle 26.{56}^0\mathrm{A}$.

(d) The total real power is $1800\mathrm{W}$ and the total reactive power is $0\mathrm{VAR}$.

Step-by-step solution

Determine the formulas of line current

Write the formula of phase voltage

${\mathbf{V}}_{\mathbf{ph}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{line}}}{\sqrt{\mathbf{3}}}$ ……. (1)

Write the formula of per-phase impedance in star connection

${\mathbf{Z}}_{\mathbf{y}}\mathbf{=}\frac{{\mathbf{Z}}_{\mathbf{∆}}}{\mathbf{3}}$ ……. (2)

Write the formula for source current

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

Write the formula for apparent power

$\mathbf{S}\mathbf{=}\mathbf{3}{\mathbf{V}}_{\mathbf{ph}}\mathbf{I}\mathbf{⭒}$ ……. (4)

Write the formula of current divison

${\mathbf{I}}_{\mathbf{y}}\mathbf{=}\mathbf{I}\mathbf{}\left(\frac{Z}{Z+Z_y}\right)$ ……. (5)

Determine the source current, real and reactive power.

(a)

Determine the phase voltage.

Substitute $120\sqrt{3}$ for ${\mathrm{V}}_{\mathrm{line}}$in equation (1).

$$\begin{gathered} {\mathrm{V}}_{\mathrm{ph}}=\frac{120\sqrt{3}}{\sqrt{3}} \\ =120\mathrm{V} \end{gathered}$$

Determine the per-phase impedance of delta load.

Substitute $(60-\mathrm{j}45)\mathrm{\Omega }$for ${\mathrm{Z}}_{∆}$equation (2).

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{y}}=\frac{{\mathrm{Z}}_{∆}}{3} \\ =\frac{(60-\mathrm{j}45)}{3} \\ =(20-\mathrm{j}15)\mathrm{\Omega } \end{gathered}$$

The equivalent resistance of circuit is

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{eq}}={\mathrm{Z}}_{\mathrm{L}}+{\mathrm{Z}}_{\mathrm{y}}\parallel \mathrm{Z} \\ =(2+\mathrm{j}4)+\left[(20-\mathrm{j}15)\parallel (30+\mathrm{j}40)\mathrm{\Omega }\right] \\ =24\mathrm{\Omega } \end{gathered}$$

Determine the source current

Substitute $24\mathrm{\Omega }$ for ${\mathrm{Z}}_{\mathrm{eq}}$and $120\mathrm{V}$for ${\mathrm{V}}_{\mathrm{ph}}$in equation (3).

$$\begin{gathered} \mathrm{I}=\frac{120\angle 0^0}{24\angle 0^0} \\ =5\angle 0^0\mathrm{A} \end{gathered}$$

The apparent power delivered by the source using equation (4) is,

$$\begin{gathered} \mathrm{S}=3{\mathrm{V}}_{\mathrm{ph}}{\mathrm{I}}^{★} \\ =3\times 120\times 5\angle 0^0 \\ =1800\angle 0^0\mathrm{VA} \end{gathered}$$

Therefore, the active power delivered by the source is $\mathrm{P}=1800\mathrm{W}$ and reactive power is$\mathrm{Q}=0\mathrm{VAR}$ .

Determine the line-to-line voltage at load.

(b)

The equivalent load resistance is,

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{load}}={\mathrm{Z}}_{\mathrm{y}}\parallel \mathrm{Z} \\ =\left[(20-\mathrm{j}15)\parallel (30+\mathrm{j}40)\mathrm{\Omega }\right] \\ =22.36\angle -10.3^0\mathrm{\Omega } \end{gathered}$$

Determine the phase-voltage at load.

$$\begin{gathered} \mathrm{V}={\mathrm{IZ}}_{\mathrm{load}} \\ =5\angle 0^0\times 22.36\angle -10.3^0 \\ =111.8\angle -10.3^0\mathrm{V} \end{gathered}$$

The line-to-line load voltage using equation (1) is

$$\begin{gathered} {\mathrm{V}}_{\mathrm{load}}=\sqrt{3}\times 111.8\angle -10.3^0 \\ =193.64\angle -10.3^0\mathrm{V} \end{gathered}$$

Determine the phase current in each load.

(c)

Determine the phase current in first load.

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{V}}{\mathrm{Z}} \\ =\frac{111.8\angle -10.3^0}{(30+\mathrm{j}40)} \\ =2.236\angle -63.{43}^0\mathrm{A} \end{gathered}$$

Determine the phase current in second load.

$$\begin{gathered} {\mathrm{I}}_{\mathrm{y}}=\frac{\mathrm{V}}{{\mathrm{Z}}_{\mathrm{y}}} \\ =\frac{111.8\angle -10.3^0}{(20+\mathrm{j}15)} \\ =4.472\angle -26.{56}^0\mathrm{A} \end{gathered}$$

Determine the real and reactive power absorbed by the load and line.

(d)

Determine the power absorbed the first load.

$$\begin{gathered} \mathrm{S}=3{\mathrm{VI}}^{★} \\ =3\times 111.8\angle -10.3^0\times 2.236\angle 63.{43}^0 \\ =749.95\angle 53.{13}^0\mathrm{VA} \\ =449.97\mathrm{W}+\mathrm{j}599.95\mathrm{VAR} \end{gathered}$$

Determine the power absorbed the f second load.

$$\begin{gathered} {\mathrm{S}}_{\mathrm{y}}=3{\mathrm{VI}}_{\mathrm{y}} \\ =3\times 111.8\angle -10.3^0\times 4.472\angle -26.{56}^0 \\ =1499.9\angle 36.{86}^0\mathrm{VA} \\ =1200\mathrm{W}+\mathrm{j}899.732\mathrm{VAR} \end{gathered}$$

Determine the power absorbed the line

role="math" localid="1655180712557" $$\begin{gathered} {\mathrm{S}}_{\mathrm{L}}=3{\mathrm{Z}}_{\mathrm{L}}{\left({{\mathrm{I}}_{\mathrm{ph}}}^{★}\right)}^2 \\ =3\times (2+\mathrm{j}4)\times {(5\angle 0^0)}^2 \\ =335.41\angle 63.{43}^0\mathrm{VA} \\ =150.025\mathrm{W}+\mathrm{j}299.98\mathrm{VAR} \end{gathered}$$

The total real and reactive power absorbed by load and line is,

$$\begin{gathered} {\mathrm{S}}_{\mathrm{T}}=\mathrm{S}+{\mathrm{S}}_{\mathrm{Y}}+{\mathrm{S}}_{\mathrm{L}} \\ =449.97\mathrm{W}+\mathrm{j}599.95\mathrm{VAR}+1200\mathrm{W}-\mathrm{j}899.732\mathrm{VAR}+150.25\mathrm{W}+\mathrm{j}299.98\mathrm{VAR} \\ =1799.995\mathrm{W}+\mathrm{j}0.198\mathrm{VAR} \end{gathered}$$

Therefore, the total power absorbed by the load and line is same as the power delivered by the source