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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.

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    $\begingroup$ In equation (3) you can just solve for $V_3$ in terms of $V_2$ and substitute. $\endgroup$ – Travis Aug 28 '14 at 10:10
  • $\begingroup$ Thanks for spending your time. But I have from equation 1: $V_{1} = -23.67$ and $V_{2}= -184.6$ and from equation 2: $V_{1}=2.97$ and $V_{2}=20$. Now should I add these up to get the actual $V_{1}$ and $V_{2}$ ? $\endgroup$ – Osama Qarem Aug 28 '14 at 11:37
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    $\begingroup$ If you split up your equations into real and imaginary parts, you must do the same for the $V_i$. To do this, write $V_i = W_i + j X_i$, $i = 1, 2, 3$, distribute, and then break into real and imaginary parts. You don't have to do this though, you can simply perform arithmetic using complex numbers to solve the problem, just like you would for equations involving real coefficients and variables. $\endgroup$ – Travis Aug 28 '14 at 11:43
  • $\begingroup$ I'm sorry.. But I can't understand what you're trying to say, English isn't my first lang. so it might be me. Are you saying that the way I've solved for $V_{1}$ and $V_{2}$ in equation 1 is incorrect? $\endgroup$ – Osama Qarem Aug 28 '14 at 12:02
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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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