Find the equivalent capacitance between P and Q for the configuration shown below in the figure (a).

Explanation

The capacitors 1 `mu` F and `3muF` are connected in parallel and `6muF` and `2muF` are also separately connected in parallel . So these parallel combinations reduced to equivalent single capacitances in their respective positions as shown in the figure (b) .
`C_(eq)=1muF+3muF=4muF`
`Ceq=6muF+2muF=8muF`
From the figure (b) we infer that the tow `4muF` capacitors are connected in series and the two `8muF` capatiors are connected in series . By using formula for the series we can reduce to their equivalent capacitances as shown in figure (c) .
`(1)/(C)=(1)/(4)+(1)/(4)=(1)/(2) implies C_(eq) =2muF "and " (1)/(C_(eq))=(1)/(8)+(1)/(8)=(1)/(4)implies C_(eq)=4muF`
From the figure (c) we infer that `2muF` and `4muF` are connected in parallel . So the equivalent capacitance is given in the figure (d) .
`(C_(eq)=2muF+4muF=6muF`
Thus the combination of capacitances in figure (a) can be replaced by a single capacitance `6muF.`