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 2 V.Compute the total battery voltage.

Explanation

In series arrangement of capacitors `V_1/V_2 = C_2/C_1`
`rArr V_2 = (C_1 V_1)/(C_2) = (6muF xx 2V)/(12muF) = 1V`
`:.` Total voltage across the capacitors combination
` = V_1 + V_2 = 2 + 1 =3V`
`:.` Battery voltage = 3 V