In the given figure the capacitors `C_(1), C_(3), C_(4),C_(5)` have a capaciance `4 muF` each if the capaitor `C_(2)` has a capacitance `10 muF`, then effective capacitance between `A` and `B` will be

Options

A. `2 muF`
B. `4 muF`
C. `6 muF`
D. `8 muF`

Answer

B

Explanation

The given network can be replaced through that shown below.
As `(C_(4))/(C_(3)) = (C_(1))/(C_(5))`
This is a balanced wheatstone bridge arrangement in which no charge flows through `C_(2)`, hence `C_(2)` is not considered.
Capacitors `C_(1)` and `C_(5)` are in series
`(1)/(C') = (1)/(C_(1)) + (1)/(C_(5))`
`rArr C' = (C_(1)C_(5))/(C_(1) + C_(5)) = (4 xx 4)/(4 + 4) = 2muF`
Also `C_(3)` and `C_(4)` are in series, hence
`C'' = (C_(3)C_(4))/(C_(3) + C_(4)) =(4 xx 4)/(4 + 4) = 2muF`
Now, `C'` and `C''` are in parallel, hence net capacitance between `A` and `B` is
`C_(AB) = 2+2 = 4muF`