Two capacitors of capacitance of `6 muF and 12 muF` are connected in series with a battery. The voltage across the `6 muF` capacitor is 2V. Compute the total battery voltage.

Explanation

Given `C_(1)=6muF,C_(2)=12muF,V_(1)=2V`
Charge on capacitor `C_(1)` is `q_(1)=C_(1)V_(1)=(6muF)xx2(V)=12muC`
In series arrangement charge on each capacitor remains the same, so
charge on `C_(2)` is also `q_(2)=q_(1)=12muC`.
`therefore` Potential difference across `C_(2)` is `V_(2)=(q_(2))/(C_(2))=(120muC)/(12muF)=1V`.
Total battery voltage `V=V_(1)+V_(2)=2+1=3V`