Several capacitors are connected as shown in. If the charge on the `5 muF` capacitor is `120 muC`, the potential between points A and D` is
.

Answer

`5.33 V`

Explanation

Charge across `5muF` capacitor is `10muC`,
Therefore, potential difference across `5 muF`. Cpacitor is
`q/C=(10xx10xx10^(-6))/(5xx10^(-6))-=2 V`
As `3,4`, and `5muF` capacitors are in parallel, so potential difference across each of the them equals `2V`.
Charge on `3 muF` capacitor is `CV=3xx10^(-6)xx2=6 muC`
Charge on `4 muF` capacitor is `4xx10^(-6)xx2=8 muC`
Total charge flowing in upper branch of circuit is
`10 muC+6muC+ 8muC=24 muC`
Potential difference across `C_(2)` is `24 muC/F=6 V`.
Total potential differecnce across AB is `6+2=8 V`. Equivalent capacitance of lower branch of circuit is
`(6xx3)/(6+3)=2 muF`
So, charge flowing is `2xx10^(-6)xx8=16 muC`
herefore, potential difference between `A` and `C`is
`(16muC)/(3 muF)=(16)/3=5.33 V`.