How do you arrange four equal capacitor of `4muF` to get effective capacitance `3 muF` ?

Options

A. three in series, one in parallel
B. two in parallel, two in series
C. three in parallel, one in series
D. all four in series

Answer

C

Explanation

If three condensers are in parallel then
`C_(P)=4+4+4+=12uF,`
Now `C_(P) and C_(4)` are in series then
`C_(s)=(C_(P)*C_(4))/(C_(P)+C_(4))=(12xx4)/(12+4)=48/10=3muF`