Equivalent Resistance (H)

It is difficult to present solution without a diagram. I will try to use some notations. I hope they are useful.

$\begin {matrix} x+y & \text{x in series with y} \\ x||y & \text{x parallel with y} \\ _y^x\Delta z^+_+ & \text {x, y and z in a delta-connection} \\ x Y _{z+}^{y+} & \text{x,y, z in a wye-connection} \end {matrix}$

$\triangle$ -Y transformation is used in the solution.

Now from node 1 to node 2:

$R_{12}=[^3_4 \Delta 10 ^{+6} _{+8} ]\space ||\space [^3_4 \Delta 3 ^6 _8 ]+^3_4 \Delta 3 ||2||1$

$\quad \quad = [\frac{3\times 4}{3+10+4} Y ^{\frac{30}{17}+6} _{\frac{40}{17} + 8} ]\space ||\space [\frac{6}{5} Y ^{\frac{9}{10}+6} _{\frac{6}{5} + 8} ]+^3_4 \Delta 3 ||2||1$

$\displaystyle \quad \quad = [\frac{12}{17}+(\frac{30}{17}+6)||(\frac{40}{17}+8)]\space ||\space [\frac{6}{5}+(\frac{9}{10}+6)||(\frac{6}{5}+8)]$ $\displaystyle \quad \quad \quad + \frac{6}{5}+(\frac{9}{10}+2)||(\frac{6}{5}+1)$

$\displaystyle \quad \quad = \left[ \frac {12}{17} + \frac {\frac {30+102}{17}\times \frac {40+136}{17}}{\frac {30+102}{17} + \frac {40+136}{17}}\right] || \left[ \frac{6}{5}+\frac{\frac {69}{17}\times \frac {92}{17}}{\frac {69}{17}+ \frac {92}{17}} \right]+\frac{6}{5}+\frac{\frac {29}{10}\times \frac {11}{10}}{\frac {29}{10}+ \frac {11}{10}}$

$\displaystyle \quad \quad = \left[ \frac {12}{17} + \frac {528}{119} \right] || \left[ \frac{6}{5}+\frac {138}{35} \right]+\frac{6}{5}+\frac{319}{255}$

$\quad \quad = \dfrac {36}{7} || \dfrac {36}{7} + \dfrac {125}{51} = \dfrac {18}{7} + \dfrac {125}{51} = \dfrac {1793}{357}\Omega$

From node 2 to node 3:

$R_{23} = (6+3)||0||3||6 = 0\Omega \quad \Rightarrow R_{123} = \dfrac {1793}{357}\Omega$

From node 4 to node 3:

$R_{43} =[^5_{10} \Delta 15 ^{+5+} _{+5+} 15 Y ^{10}_{5} ] \space ||3||10||8$

$\quad \quad = [\frac{5}{3} Y ^{\frac{5}{2} + 5 + 5} _{5+5+\frac{5}{2}} Y\frac {5}{3}] \space ||3||10||8$

$\displaystyle \quad \quad = \left[\frac {5}{3} + \left( \frac {5}{2}+5+5 \right) || \left( 5+5+\frac {5}{2} \right) + \frac{5}{3} \right]|| 3||10||8$

$\displaystyle \quad \quad = \left[\frac {5}{3} +\frac {25}{4} + \frac{5}{3} \right]|| 3||10||8 = \frac{115}{12}||3||10||8 \Omega$

Therefore,

$R_{13} = R_{123}\space || \space R_{43} = \frac {1793}{357}||\frac{115}{12} || 3 || 10 || 8 = \dfrac {1}{\frac {357}{1793}+\frac {12}{115} + \frac {1}{3} + \frac {1}{10} + \frac{1}{8}}$

$\quad \quad \quad \quad \quad \quad \quad \quad = \boxed{1.16}$

$\frac{1793}{357}$ // $\frac{2760}{1829}$

$\displaystyle \frac{\frac{1793}{357} \frac{2760}{1829}}{\frac{1793}{357} + \frac{2760}{1829}}$ = (1.160377112947940039163208250395+) $\Omega$ {Just to provide a precise exact figure.}

$\frac{4948680}{4264717}$ = $\frac{2^3 \times 3 \times 5 \times 11 \times 23 \times 163}{1021 \times 4177}$ for explicit prime factors.

Answer: $\boxed {1.16}$ $\Omega$

1. Split the circuit up into multiple parts (keep track of where they will go)
   1. Resolve each 'module' using the Y- $\Delta$ Transform and simplifying. This should be easy as they will be separated into parallel and series parts.
   2. Put it back together and hope you have not made a mistake.

You will need lots of working!

It is helpful to write out the formulae for the Y- $\Delta$ transform!