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HARD Solution - Kirchhoff's Mesh Equations

EASY Solution - Thevenin's Theorem **[ USED with Electron Current ]**

$V_{T}\ =\ Thevenin's\ \ Voltage$

Complex Circuit Solution

$V_{T}=V_{S}{\Biggl [}{\biggl (}{\frac {R_{4}}{R_{3}+R_{4}}}{\biggr )}-{\biggl (}{\frac {R_{2}}{R_{1}+R_{2}}}{\biggr )}{\Biggr ]}$

$V_{R}\ =\ Thevenin's\ \ Resistance$

$R_{T}={\frac {\ {\Bigl (}R_{1}+R_{4}{\Bigr )}{\Bigl (}R_{2}+R_{3}{\Bigr )}\ }{R_{1}+R_{2}+R_{3}+R_{4}}}$

.

$I_{5}={\frac {V_{T}}{\ R_{5}+R_{T}\ }}\qquad \qquad \qquad \qquad \ \ E_{5}=I_{5}R_{5}$ $I_{2}={\frac {\ R_{2}{\bigl (}\ V_{S}-I_{5}R_{2}{\bigr )}\ }{R_{1}+R_{2}}}+I_{5}R_{2}\qquad \ E_{2}=I_{2}R_{2}$

$I_{1}=I_{2}-I_{5}\qquad \qquad \qquad \qquad \qquad \ E_{1}=I_{1}R_{1}$

$I_{4}={\frac {\ E_{2}+E_{5}\ }{R_{4}}}\qquad \qquad \qquad \qquad \ \ \ \ E_{4}=I_{4}R_{4}$

$I_{3}=I_{4}+I_{5}\qquad \qquad \qquad \qquad \qquad \ \ E_{3}=I_{3}R_{3}$

$V_{1}=I_{2}R_{2}\qquad \qquad \qquad \qquad \qquad \ \ \ \ \ \ V_{2}=I_{4}R_{4}$

$This\ \ MUST\ \ Be\ \ True$

$E_{R5}=I_{5}R_{5}={\Bigl [}V_{2}-V_{1}{\Bigr ]}$

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