Question

Can anyone solve this question..???

Answer 1

Using similarity of load of resistances , point O is mid point , We can break circuit from point O as load of both sides(left and right) of point O is equal .

After breaking circuit ,

R2 is in series with R3

R' = R2 + R3 = r + r = 2r

Again,

R' is in parallel with R1

1/R'' = 1/R' + 1/R1 = 1/2r + 1/r

R'' = 2r/3

Now, R'' , R4 and R7 are in series.

R''' = 2r/3 + r + r = 8r/3

R5 and R6 are in series,

R'''' = r + r = 2r

R'''' and R''' are in parallel

Hence, For Equivalent Resistance

1/R = 1/R''' + 1\R'''

1/R = 3/8r + 1/2r

1/R = 7/8r

R = 8r/7 Ω