Question

 

Answer 1

The resistors of values 20 Ω, 10 Ω and 20 Ω are connected in series. Therefore, their net resistance is:
R = 20 Ω + 10 Ω + 20 Ω
R = 50 Ω 

This combination of resistors are connected in parallel with that of value 30 Ω. Therefore, the resistance R' is:
1/R' = (1/50) + (1/30)
1/R' = (3 + 5)/150
R' = 150/8
R' = 18.75 Ω

The resistance, R' is connected in series with the two resistors of values 10 Ω each. Hence, the total resistance between A and B is:
18.75 + 10 + 10 = 38.75 Ω