Question

Consider the balanced three-phase system shown in Figure 2.34. Determine${\mathbf{v}}_{\mathbf{1}}\mathbf{}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{i}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$. Assume positive phase sequence.

Short answer

Therefore, the voltage ${\mathrm{v}}_1\left(\mathrm{t}\right)\mathrm{is}149.5\cos \left(\mathrm{\omega t}-3.{03}^{°}\right)$and the current $i_2\left(t\right)$is$129.43\cos \left(\mathrm{\omega t}+116.{97}^{°}\right)\mathrm{A}$.

Step-by-step solution

Determine the formula to calculate the voltagev1 (t)andi1 (t)

The impedance value in the star connection is given by,

${\mathbf{Z}}_{\mathbf{Y}}\mathbf{=}\frac{\mathbf{Z}}{\mathbf{3}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here,${\mathbf{Z}}_{\mathbf{Y}}$is the impedance of star connection and Z is the impedance of the delta connection.

The phase current in load 2 can be calculated as

${\mathbf{l}}_{\mathbf{A}\mathbf{\text{'}}\mathbf{B}\mathbf{\text{'}}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{A}\mathbf{\text{'}}\mathbf{B}\mathbf{\text{'}}}}{{\mathbf{Z}}_{\mathbf{A}\mathbf{\text{'}}\mathbf{B}\mathbf{\text{'}}}}$

Step 2: Calculate the values of the voltage andv1(t) and i2(t).

By using the equation (1) calculate the impedance in star connection.

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{y}}=\frac{-\mathrm{j}2}{3} \\ {\mathrm{Z}}_{\mathrm{y}}=-\mathrm{j}\frac{2}{3}\mathrm{\Omega } \end{gathered}$$

The per phase diagram can be drawn as

The impedance$\mathrm{j}1.0\mathrm{\Omega }\&-\mathrm{j}\frac{2}{3}\mathrm{\Omega }$are connected in the parallel.

$$\begin{gathered} {\mathrm{Z}}_1=\frac{\mathrm{j}1\times \left(\begin{array}{c}-\mathrm{j}{\displaystyle \frac{2}{3}}\end{array}\right)}{\mathrm{j}1-\mathrm{j}{\displaystyle \frac{2}{3}}} \\ {\mathrm{Z}}_1=\frac{{\displaystyle \frac{2}{3}}}{\mathrm{j}{\displaystyle \frac{1}{3}}} \\ {\mathrm{Z}}_1=\mathrm{j}2\mathrm{\Omega } \end{gathered}$$

The redraw the diagram as,

By using the voltage divider rule calculate the voltage${\mathit{V}}_{\mathbf{1}}$

localid="1655118048297" $$\begin{gathered} {\mathrm{V}}_1=\frac{-\mathrm{j}2}{-\mathrm{j}2+\mathrm{j}1+0.1+\mathrm{j}0.01}\times 100\angle 0^{°} \\ {\mathrm{V}}_1=\frac{-\mathrm{j}2}{0.1-\mathrm{j}1.89}\times 100\angle 0^{°} \\ {\mathrm{V}}_1=105.67\angle -3.{03}^{°}\mathrm{V} \end{gathered}$$

The phase voltage localid="1655118057388" $v_1\left(t\right)$can be calculated as,

localid="1655118067223" $$\begin{gathered} {\mathrm{v}}_1\left(\mathrm{t}\right)=\sqrt{2}\times 105.67\cos \left(\mathrm{\omega t}-3.{03}^{°}\right) \\ {\mathrm{v}}_1\left(\mathrm{t}\right)=149.5\cos \left(\mathrm{\omega t}-3.{03}^{°}\right) \end{gathered}$$

Hence the voltage islocalid="1655118108187" ${\mathrm{v}}_1\left(\mathrm{t}\right)\mathrm{is}149.5\cos \left(\mathrm{\omega t}-3.{03}^{°}\right)$.

The line voltage of load 1 can be calculated as

localid="1655118119496" $$\begin{gathered} {\mathrm{V}}_{\mathrm{A}\text{'}\mathrm{B}\text{'}}=\sqrt{3}{\mathrm{e}}^{\mathrm{j}30}{\mathrm{V}}_{\mathrm{A}\text{'}\mathrm{N}\text{'}} \\ {\mathrm{V}}_{\mathrm{A}\text{'}\mathrm{B}\text{'}}=\sqrt{3}\angle {30}^{°}(105.67\angle -3.{03}^{°}) \\ {\mathrm{V}}_{\mathrm{A}\text{'}\mathrm{B}\text{'}}=183.026\angle 26.{97}^{°}\mathrm{V} \end{gathered}$$

The phase current in load 2 can be calculated as,

localid="1655118134245" $$\begin{gathered} {\mathrm{l}}_{\mathrm{A}\text{'}\mathrm{B}\text{'}}=\frac{{\mathrm{V}}_{\mathrm{A}\text{'}\mathrm{B}\text{'}}}{{\mathrm{V}}_{\mathrm{A}\text{'}\mathrm{B}\text{'}}} \\ {\mathrm{V}}_{\mathrm{A}\text{'}\mathrm{B}\text{'}}=\frac{183.026\angle 26.{97}^{°}}{-\mathrm{j}2} \\ {\mathrm{V}}_{\mathrm{A}\text{'}\mathrm{B}\text{'}}=91.52\angle 116.{97}^{°}\mathrm{A} \end{gathered}$$

Calculate the current in the time domain as,

localid="1655118146442" $$\begin{gathered} {\mathrm{i}}_2\left(\mathrm{t}\right)=\sqrt{2}\times 91.52\cos \left(\mathrm{\omega t}+116.{97}^{°}\right) \\ {\mathrm{i}}_2\left(\mathrm{t}\right)=129.43\cos \left(\mathrm{\omega t}+116.{97}^{°}\right)\mathrm{A} \end{gathered}$$

Hence the currentlocalid="1655118156749" ${\mathrm{i}}_2\left(\mathrm{t}\right)\mathrm{is}129.43\cos \left(\mathrm{\omega t}+116.{97}^{°}\right)\mathrm{A}$.