In fig, the values of capacitances are as follow
`C_(1) = C_(2) = C_(3) = C_(4) = 4mu F, C_(5) = 5 mu F` Calculate the equivalent capacitance between the points P and Q. If a battery of 10 volt is connected between these points, what will be the charge on each capacitor ?

Explanation

The given circuit can be redrawn as shown in Fig.
Clearly the capacitors form wheatstone bridge arrangement.
As `C_(1) = C_(2) = C_(3) = C_(4) :. (C_(1))/(C_(2)) = (C_(3))/(C_(4))`

Hence the bridge is balanced. The points R and S are at the same potential. No charge can accumulate on `C_(5)`, which thus becomes ineffective.
Between P and Q, two series combinations `(C_(1), C_(2)) and (C_(3), C_(4))` are connected in parallel. If C' and C" are the equivalent capacitances of these combinations, then from
`(1)/(C') = (1)/(C_(1))+ (1)/(C_(2)) = (1)/(4) + (1)/(4) = (2)/(4) , C' = 2 muF`
Similarly,`C" = 2mu F`
Hence equivalent capacitance between P and Q is `C = C' + C" = 2 + 2 = 4 muF`
As pot diff. between P and Q is V = 0 volt and equivalent capacitance of `C_(1) and C_(2)` in series is `2 muF`.
`:.` Charge on each of capacitors `C_(1) and C_(2)` is `= 2xx10 = 20 muC`.
Similarly, charge on each of capacitors `C_(3) and C_(4)` is also `20 muC`. charge on `C_(5) = 0`