Hexagonal Ohm

Using the principles of Thevenin equivalent resistance a circuit or part of a circuit, that contains only resistors, can be replaced by a single resistance value. By breaking the circuit down into smaller parts and then using the calculated Thevenin equivalent resistance in subsequent calculations the total equivalent resistance can be determined.

|| denotes parallel

- - denotes series. Calculating the equivalent resistance R,EF considering only the resistors R4, R10, R11.

R,EF = (R11 - - R10) || R4

$\frac { 1 }{ { R }_{ EF } } =\frac { 1 }{ { R }_{ 11 }+{ R }_{ 10 } } +\frac { 1 }{ { R }_{ 4 } } =\frac { 1 }{ 40 } +\frac { 1 }{ 20 } =\frac { 3 }{ 40 }$

$\ { R }_{ EF }=\frac {40 }{ 3 } \Omega$

Calculating the equivalent resistance R,CD considering only the resistances R3, R5, R9, R12, R,EF

R,CD = (R12 - - R9 || (R3 - - REF - - R5).

$\frac { 1 }{ { R }_{ CD } } =\frac { 1 }{ { R }_{ 12 }+{ R }_{ 9 } } +\frac { 1 }{ { R }_{ 3 }+{ R }_{ EF }+{ R }_{ 5 } } =\frac { 1 }{ 20+20 } +\frac { 1 }{ 20+\frac { 40 }{ 3 } +20 } =\frac { 1 }{ 40 } +\frac { 3 }{ 160 } =\frac { 7 }{ 160 }$

${ R }_{ CD }=\frac { 160}{7 } \Omega$

Calculating the equivalent resistance, R,AB considering the resistances R1, R2, R6, R7, R8, R,CD.

R,AB = R1 || (R7 - - R8) || (R2 - - R,CD - - R6) $\frac { 1 }{ { R }_{ AB } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 7 }+{ R }_{ 8 } } +\frac { 1 }{ { R }_{ 2 }+{ R }_{ CD }+{ R }_{ 6 } } =\frac { 1 }{ 20 } +\frac { 1 }{ 40 } +\frac { 1 }{ 20+\frac { 160 }{ 7 } +20 } =\frac { 40 }{ 440 }$

${ R }_{ AB }=11 \Omega$