Question

Calculate the source currents in Example 8.6 without using symmetrical components. Compare your solution method with that of Example 8.6. Which method is easier?

Short answer

The value of source current without using symmetrical component $I_A,I_B$and $I_C$are$25.10\angle -46.78°\text{\hspace{0.33em}A}$, $26.57\angle 73.76°\text{\hspace{0.33em}A}$and $25.66\angle -163.66°\text{\hspace{0.33em}A}$respectively.

Step-by-step solution

Write the given data from the question.

The line to ground voltages.

$\begin{array}{l}{V}_{ag}=277\angle 0°V\\ V_{bg}=260\angle -120°\text{\hspace{0.33em}V}\\ {\text{V}}_{cg}=295\angle 115°\text{\hspace{0.33em}V}\end{array}$

The per phase line impedance is$Z_L=1\angle 85°\Omega$ .

The per phase load impedance is $Z_Y=10\angle 40°\Omega$.

Calculate the source current without using symmetrical component.

Apply the Kirchhoff’s voltage law in upper loop.

$\begin{array}{l}{I}_C\left(Z_L+Z_Y\right)+\left(I_C+I_B\right)\left(Z_L+Z_Y\right)=V_{cg}-V_{ag}\\ I_C\left(Z_L+Z_Y\right)+I_C{Z}_L+I_C{Z}_Y+I_B{Z}_L+I_B{Z}_Y=V_{cg}-V_{ag}\\ I_C\left(Z_L+Z_Y+Z_L+Z_Y\right)+I_B\left(Z_L+Z_Y\right)=V_{cg}-V_{ag}\\ 2I_C\left(Z_L+Z_Y\right)+I_B\left(Z_L+Z_Y\right)=V_{cg}-V_{ag}\end{array}$ …... (1)

Apply the Kirchhoff’s voltage law in lower loop.

$\begin{array}{l}{I}_B\left(Z_L+Z_Y\right)+\left(I_C+I_B\right)\left(Z_L+Z_Y\right)=V_{ag}-V_{bg}\\ I_B\left(Z_L+Z_Y+Z_L+Z_Y\right)+I_C\left(Z_L+Z_Y\right)=V_{ag}-V_{bg}\\ 2I_B\left(Z_L+Z_Y\right)+I_C\left(Z_L+Z_Y\right)=V_{ag}-V_{bg}\end{array}$ …… (2)

Write the equation (1) and (2) into matrix form.

$\left[\begin{array}{cc}2\left(Z_L+Z_Y\right)& \left(Z_L+Z_Y\right)\\ \left(Z_L+Z_Y\right)& 2\left(Z_L+Z_Y\right)\end{array}\right]\left[\begin{array}{c}{I}_C\\ I_B\end{array}\right]=\left[\begin{array}{c}{V}_{cg}-V_{ag}\\ V_{bg}-V_{ag}\end{array}\right]$

Substitute$1\angle 85°\Omega$for $Z_L$, $10\angle 40°\Omega$for $Z_Y$, $277\angle 0°\text{\hspace{0.33em}V}$for $V_{ag}$, $260\angle -120°\text{\hspace{0.33em}V}$for $V_{bg}$and$295\angle 115°\text{\hspace{0.33em}V}$into above equation.’

$\begin{array}{l}\left[\begin{array}{cc}2\left(1\angle 85°+10\angle 40°\right)& \left(1\angle 85°+10\angle 40°\right)\\ \left(1\angle 85°+10\angle 40°\right)& 2\left(1\angle 85°+10\angle 40°\right)\end{array}\right]\left[\begin{array}{c}{I}_C\\ I_B\end{array}\right]=\left[\begin{array}{c}295\angle 115°-277\angle 0°\\ 260\angle -120°-277\angle 0°\end{array}\right]\\ \left[\begin{array}{cc}21.46\angle 43.78°& 10.73\angle 43.78°\\ 10.73\angle 43.78°& 21.46\angle 43.78°\end{array}\right]\left[\begin{array}{c}{I}_C\\ I_B\end{array}\right]=\left[\begin{array}{c}482.52\angle 146.35°\\ 465.13\angle -151.05°\end{array}\right]\end{array}$

Solve the matrix further as,

$\begin{array}{l}\left[\begin{array}{c}{I}_C\\ I_B\end{array}\right]={\left[\begin{array}{cc}21.46\angle 43.78°& 10.73\angle 43.78°\\ 10.73\angle 43.78°& 21.46\angle 43.78°\end{array}\right]}^{-1}\left[\begin{array}{c}482.52\angle 146.35°\\ 465.13\angle -151.05°\end{array}\right]\\ \left[\begin{array}{c}{I}_C\\ I_B\end{array}\right]=\left[\begin{array}{cc}0.062\angle -43.78°& -0.031\angle -43.78°\\ -0.031\angle -43.78°& 0.062\angle -43.78°\end{array}\right]\left[\begin{array}{c}482.52\angle 146.35°\\ 465.13\angle -151.05°\end{array}\right]\\ \left[\begin{array}{c}{I}_C\\ I_B\end{array}\right]=\left[\begin{array}{c}26.57\angle 73.76°\\ 25.66\angle -163.66°\end{array}\right]\end{array}$

The of the currents $I_B$is $26.57\angle 73.76°\text{\hspace{0.33em}A}$and$I_C$is$25.66\angle -163.66°\text{\hspace{0.33em}A}$.

According to the Kirchhoff’s law, the sum of all current at node N should be equal to zero.

$I_A+I_B+I_C=0$

$I_A=-\left(I_C+I_B\right)$

Substitute$26.57\angle 73.76°\text{\hspace{0.33em}A}$for $I_B$and$25.66\angle -163.66°\text{\hspace{0.33em}A}$for $I_C$into above equation.

$\begin{array}{l}{I}_A=-\left(26.57\angle 73.76°+25.66\angle -163.66°\right)\\ I_A=-25.10\angle 133.22°\\ I_A=25.10\angle -46.78°\text{\hspace{0.33em}A}\end{array}$

Hence, the value of source current without using symmetrical component $I_A,I_B$and$I_C$ are$I_A=25.10\angle -46.78°\text{\hspace{0.33em}A}$, $26.57\angle 73.76°\text{\hspace{0.33em}A}$and$25.66\angle -163.66°\text{\hspace{0.33em}A}$ respectively.