Question

in the given circuit the equivatent ressistance between the points and b in a)9 b) 11.6 c)14.5 d)21.2​

Answer 1

Answer:

hey here is your answer

The correct option (B) 11.6 Explanation: 3Ω in series with 2Ω = 3 + 2 = 5Ω This 5Ω is parallel to 6Ω ∴ R' = {(5 × 6)/(5 + 6)} = (30/11)Ω This R' will be in series with 7Ω hence R" = 7 + (30/11) = (107/11)Ω circuit can be redrawn as Here RAB = 9Ω series with parallel combination of 5Ω, RΩ, (107 / 11)Ω ∴ RAB = 9 [(1/5) + (1/12) + {1/(107/11)}]–1 = 9 [(1/5) + (1/12) + (11/107)]–1 = 9 + {1/(0.38)} = 9 + 2.58 = 11.58Ω = 11.6Ω

hope it help you

Answer 2

Sorry I couldn't find any circuit here...