Calculate the equivalent resistance between the points P and of the network shown in the figure given below :

Explanation

`R_5 and R_6` are in series and their net resistance `R_(56)` is given by
`R_(56) = R_5 + R_6 = 4 + 2 = 6Omega`
`R_(56) and R_4` are in parallel and their net resistance `R_(456)` is given by
`1/(R_(456))+1/R_4+1/R_(56)=1/3+1/6`
`=(2+1)/6 =3/6 =1/2`
`1/(R_(456)) =(2+1)/6 =3/6 =1/2`
`R_(456)` is in series with `R_3` and their net resistance `R_(3456)` is given by
`R_(3456) = R_3+ R_(456) = 4 + 2 = 6Omega`
Now `R_(3456)` in in parallel with `R_2` and let their net resistance be `R_(23456)`.
`1/(R_(2356))=1/R_2 +1/(R_(3456))=1/3 +1/6`
`implies R_(23456)=2Omega`
Now `R_(23456)` is in series with `R_1` and their net resistance R is given by
`R= R_(1) + R_(23456)`
`R = 4+2 = 6Omega`