Equation 1$$ \frac{V_{1}}{5} + \frac{V_{1}-V_{2}}{10+j6} - 10\angle45^\circ = 0 $$ Equation 2 $$ -4V_{1} + \frac{V_{2}-V_{1}}{10+j6} + \frac{V_{3}}{-j2} + \frac{V_{3}}{8+j7} = 0 $$ Equation 3 $$ V_{2} - V_{3} = 20\angle30^\circ $$
This is what I did:
Equation 1:
$$V_{1}(\frac{1}{5})+\frac{V_{1}-V_{2}}{10+j6}* \frac{10-j6}{10-j6} -7.1+7.1j$$
Becomes:
$$V_{1}(\frac{1}{5})+\frac{10V_{1}-6V1j-10V_{2}+6V2j}{136} = 7.1-7.1j$$
$Real$ $$V_{1}(\frac{1}{5}+\frac{10}{136})+V_{2}(\frac{-10}{136})=7.1$$
$Imaginary$ $$V_{1}(\frac{-6}{136})+V_{2}(\frac{6}{136})=-7.1$$
Solve for $V_{1}$ and $V_{2}$:
$$ V_{1}=-23.67 $$ $$ V_{2}=-184.6 $$
And I pretty much did the same thing for Equation 2 after making Equation 3: $V_{3}=V_{2}-17.32-10j$
And now I have two values for $V_{1}$ and two values for $V_{2}$. But first I would like to know if the way I used to solve this is correct? If so what should I do with the two values for $V_{1}$ and $V_{2}$? Add them up and then calculate $V_{3}$ from equation 3? Please put me in the right way.
$\endgroup$ 41 Answer
$\begingroup$The variables $V_{1},V_{2},V_{3}$ are not real numbers, but complex numbers, as can be seen by solving the given system of complex equations. It can be solved algebraically using complex numbers, as commented above by Travis. Since $\cos 30{{}^\circ}=\frac{\sqrt{3}}{2},\sin 30{{}^\circ}=\frac{1}{2}$, we get from equation 3 \begin{equation*} V_{3}=V_{2}-10\sqrt{3}-j10. \end{equation*}
Then equation 2 becomes \begin{equation*} \left( -4-\frac{1}{10+j6}\right) V_{1}+\left( \frac{1}{10+j6}+\frac{1}{-2j}+ \frac{1}{8+7j}\right) V_{2}+\left( \frac{1}{-2j}+\frac{1}{8+7j}\right) \left( -10\sqrt{3}-10j\right) =0 \end{equation*}
or (rewriting the coefficients in the algebraic form) \begin{equation*} \left( -\frac{277}{68}+\frac{3}{68}j\right) V_{1}+\left( \frac{1109}{7684}+j \frac{3027}{7684}\right) V_{2}-\frac{80}{113}\sqrt{3}+\frac{495}{113} +j\left( -\frac{80}{113}-\frac{495}{113}\sqrt{3}\right) =0. \end{equation*}
Since $\cos 45{{}^\circ}=\sin 45{{}^\circ}=\frac{\sqrt{2}}{2}$, the equation 1 becomes \begin{equation*} \left( \frac{1}{5}+\frac{1}{10+j6}\right) V_{1}-\frac{1}{10+j6}V_{2}-5\sqrt{2 }-j5\sqrt{2}=0 \end{equation*}
or, simplifying the coefficients \begin{equation*} \left( \frac{93}{340}-j\frac{3}{68}\right) V_{1}-\left( \frac{5}{68}-\frac{3}{68}% j\right) V_{2}-5\sqrt{2}-j5\sqrt{2}=0. \end{equation*}
Solving the equivalent system of equations 1 and 2 \begin{equation*} \left\{ \begin{array}{c} \left( \frac{93}{340}-j\frac{3}{68}\right) V_{1}-\left( \frac{5}{68}-j\frac{3}{68} \right) V_{2}-5\sqrt{2}-j5\sqrt{2}=0 \\ \left( -\frac{277}{68}+\frac{3}{68}j\right) V_{1}+\left( \frac{1109}{7684}+j \frac{3027}{7684}\right) V_{2}-\frac{80}{113}\sqrt{3}+\frac{495}{113} +j\left( -\frac{80}{113}-\frac{495}{113}\sqrt{3}\right) =0 \end{array} \right. \end{equation*}
we obtain the solutions \begin{eqnarray*} V_{1} &=&\frac{7270}{42\,661}\sqrt{3}+\frac{327\,550}{42\,661}\sqrt{2}+\frac{ 42\,960}{42\,661}+i\left( \frac{7270}{42\,661}-\frac{90\,050}{42\,661}\sqrt{2 }-\frac{42\,960}{42\,661}\sqrt{3}\right) \\ &\approx &12.16-j4.558\,9 \end{eqnarray*}
and \begin{eqnarray*} V_{2} &=&\frac{73\,362}{42\,661}\sqrt{3}+\frac{237\,490}{42\,661}\sqrt{2}+ \frac{120\,156}{42\,661}+j\left( \frac{73\,362}{42\,661}-\frac{3289\,970}{ 42\,661}\sqrt{2}-\frac{120\,156}{42\,661}\sqrt{3}\right) \\ &\approx &13.668-j112.22 \end{eqnarray*}
From equation 3 we obtain $V_{3}$ \begin{eqnarray*} V_{3} &=&\frac{73\,362}{42\,661}\sqrt{3}+\frac{237\,490}{42\,661}\sqrt{2}+ \frac{120\,156}{42\,661}-10\sqrt{3}+j\left( \frac{73\,362}{42\,661}-\frac{ 3289\,970}{42\,661}\sqrt{2}-\frac{120\,156}{42\,661}\sqrt{3}-10\right) \\ &\approx &-3.\,652\,7-j122.22 \end{eqnarray*}
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