net resistance..???

there is only one 1Ω resistor ...rest are 2Ω...

as it is a balanced w bridge hence total resistance will be 8/6=1.33 (approx)

o did this in this manner....
at the bottom...(right hand side)...the 2Ω and 2Ω resistors are in parallel...so there the net R is 1...
next since that's a complete wheat stone bridge...bcoz..
R₁R₂=R₃R₄ i.e. 12=12...
so, there is zero voltage at the centre..so we are left with 2Ω and2Ω resistors in series which are in parallel with a series connection of 1Ω and 1Ω resistors...
therefore,
2Ω +2Ω=4Ω(series connection at the top)
1Ω+1Ω=2Ω(series connection at the bottom)
4Ω and 2Ω are in parallel..so net resistance will be 43Ω..

i did this like this...plzz..correct if there is any mistake...

any 1 to check it...

i think u r correct