Find `R_(net)` between A and B.

Options

A. `60Omega`
B. `40Omega`
C. `70Omega`
D. `20Omega`

Answer

B

Explanation

,
In the circuit, the branch EGHF have three resistance of `10Omega`, `10Omega` and `10Omega`, respectively which are connected in series combination. So, their equivalent resistance is given by
`R_(1) = 10+20+10 = 40Omega`
This `R_(1)` resistance is parallel with `40Omega` resistance which is connected in the branch EF. So, their equivalent resistance
`R_(2) = (40 xx 40)/(40+40) = 20Omega`
Now, circuit becomes

Now, t in the branch CEFD `10Omega, 20Omega` and `10Omega` resistance are connect in series combination their equivalent resistance is given by
`R_(3) = 10Omega + 20Omega+40Omega=40Omega`
This `R_(3)` is parallel with `40Omega` resistance which is present in branch CD. Their equivalent resistance.
`R_(4) = (40 xx 40)/(40+40) = 20Omega`
Now, circuit becomes ltrbgt
The net resistance between A and B, `R_(net) = 10 +20 +10 = 40Omega`