User:Tet-Math2/sandbox

The printable version is no longer supported and may have rendering errors. Please update your browser bookmarks and please use the default browser print function instead.

HARD Solution - Kirchhoff's Mesh Equations

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

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

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

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

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

.

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

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

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

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

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

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

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

HARD Solution - Kirchhoff's Mesh Equations

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

V T = T h e v e n i n ′ s V o l t a g e {\displaystyle V_{T}\ =\ Thevenin's\ \ Voltage}

V T = V S [ ( R 4 R 3 + R 4 ) − ( R 2 R 1 + R 2 ) ] {\displaystyle 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 = T h e v e n i n ′ s R e s i s t a n c e {\displaystyle V_{R}\ =\ Thevenin's\ \ Resistance}

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

I 1 = I 2 − I 5 E 1 = I 1 R 1 {\displaystyle I_{1}=I_{2}-I_{5}\qquad \qquad \qquad \qquad \qquad \ E_{1}=I_{1}R_{1}}

I 4 = E 2 + E 5 R 4 E 4 = I 4 R 4 {\displaystyle 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 E 3 = I 3 R 3 {\displaystyle I_{3}=I_{4}+I_{5}\qquad \qquad \qquad \qquad \qquad \ \ E_{3}=I_{3}R_{3}}

V 1 = I 2 R 2 V 2 = I 4 R 4 {\displaystyle V_{1}=I_{2}R_{2}\qquad \qquad \qquad \qquad \qquad \ \ \ \ \ \ V_{2}=I_{4}R_{4}}

T h i s M U S T B e T r u e {\displaystyle This\ \ MUST\ \ Be\ \ True}

E R 5 = I 5 R 5 = [ V 2 − V 1 ] {\displaystyle E_{R5}=I_{5}R_{5}={\Bigl [}V_{2}-V_{1}{\Bigr ]}}

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

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

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

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

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

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

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

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

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

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

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