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 mu F `
B. `6mu F `
C. `4 muF `
D. ` 8 mu F `

Answer

C

Explanation

Capacitance of capacitors ` C_(1),C_(3),C_(4) ,C_(5)=4 mu F ` each and capacitance of capacitor ` C_(2)=10 mu F ` . If a battery is applied across A and B, the point b and c will be at the same potential since `(C_(1) =C_(4)=C_(3)=C_(5)=4 muF `). Therefore no charge flows through ` C_(2)` . We have the capacitor ` C_(1) and C_(5)` in series . Therefore their equivalent capacitance ,
` C'=(C_(1) xx C_(5))/(C_(1)+C_(5))=( 4 xx 4)/( 4+4)= 2 mu F ` .
Similarly , ` C_(4) and C_(3)` are in series . Therefore their equivalent capacitance , `C''=(C_(3) xx C_(4))/(C_(3) +C_(4))=( 4 xx 4)/(4+4)=2 mu F `
Now C' and C" are in parallel. Therefore effective capacitance between A and B .
`=C' + C'' =2+2=4 muF`.