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

`C_(1) = 6 muF`
`C_(2) = 12 muF`
Net capacitance in series combination is given by
`1/C= 1/C_(1) + 1/C_(2)`
`1/C =1/6 + 1/12 =(2+1)/12 = 3/12 = 1/4`
`rArr C = 4 muF`
`V_(1) = 2V` = Potential difference across `6 muF` capacitor i.e.,`C_(1)`
Let q is charge on capacitor `C_(1)`.

`therefore C_(1) =q/V_(1)`
`rArr q=C_(1)V_(1)`
`rArr q = 6 xx 2 = 12 muF`
We know `C = q/V`
Total battery voltage `=V = q/C =12/4 = 3V`