Find the résistance between point A and point D of the network as shown below :

Explanation

(a) For the net resistance between A and D, equivalent circuit can be drawn as

Here `R_2 and R_6` are in series and let their net resistance be `R_(26)`
`R_(26) = R_2 + R_6`
`R_4 = 3 + 6 = 9Omega`
Also `R_(26) and R_(5)` are in parallel and let their net resistance be `R_(45)`
`1/R_P=1/R_(26)+1/R_(45)`
`1/R_P =1/9+1/14=(14+9)/(9xx14)=23/126`
`R_P = 126/23 = 5.48 Omega`
Now `R_1 , R_P and R_8` are in parallel and let their net resistance be R.
`:. R = R_(1)+R_P + R_8`
`=2+5.48 +9 = 16.48 Omega`
(b) For the net resistance between B and D, equivalent circuit can be drawn as :
Here `R_2 ,R_4`and `R_5` are in series and let their net resistance be `R_(245)`
`R_(245) = R_2 + R_4 + R_5`
`= 3 + 6 + 8`
`R_(245)= 17Omega`
Now `R_(245) and R_6` are in parallel and let their resistance be `R_P`.

`:. 1/R_p =1/R_(245) +1/R_6 =1/17 +1/6`
`1/R_P =(6+17)/(17xx6)=23/102`
`R_P = 102/23=4.43Omega`
Now `R_3, R_P and R_8` are in series and let together their net resistance be R.
`R= R_3 + R_4 + R_8`
= 4 + 4.43 +9
`implies R= 17.43 Omega`