Find the equialent capacitance of the combination of capacitors between the points `A` and `B` as shown in Fig. Also calculate the total charge that flows in the circuit, when a `100 V` battery is connected between the points `A` and `B`.

Answer

`20 mu F ; 2xx10^(3) C`

Explanation

In Fig, `C_(5), C_(6)` are in parallel.
`:. C_(56) = C_(5) + C_(6) = 10+10 = 20 mu F`,
Also `C_(2), C_(3), C_(4)` are in series
`:. (1)/(C_(234)) = (1)/(C_(2)) + (1)/(C_(3) + (1)/(C_(4))) = (3)/(60) = (1)/(20)`
`C_(234) = 20 mu F`
Now, `C_(234)` and `C_(56)` are in parallel.
`:.` Their combined capacity is
`C' = C_(234) + C_(56) = 20+20 = 40 mu F`
Now, `C_(1)` and `C'` are in series.
Therefore, capacity of the combination is
`C = (C_(1) C')/(C_(1) + C') = (40xx40)/(40 + 40) = 20 mu F`
Total charge that flows in the circuit, when a battery of `100 V` is conneced between `A` snf `B` is `q = CV = (20xx10^(-6))xx100 = 2xx10^(-3) C`