Three capacitors having capacitances `1 muF,2muF " and "3 muF` are joined in series. A potential difference of 1100 V is applied to the combination. Find the charge and potential difference across each capacitor.

Explanation

If C be the equivalent capacitance of the combination, then
`(1)/(C )=(1)/(1)+(1)/(2)+(1)/(3)=(11)/(6) " or, "C= (6)/(11)muF = (6)/(11)xx10^(-6)F`
Total charge of the combination,
`Q= CV = (6)/(11)xx10^(-6)xx1100 = 6xx10^(-4)C`
Since the capacitors are connected in series, charge on each capacitor is equal to the total charge, i.e., `6xx10^(-4)C`.
Potential difference across the plates of the first capacitor,
`V_1 = (Q)/(C_1)= (6xx10^(-4))/(2xx10^(-6))= 600V`
Similarly, for the other two capacitors, respectively,
`V_2 = (Q)/(C_2)= (6xx10^(-4))/(2xx10^(-6)) = 300V`
and `V_3= (Q)/(C_3)= (6xx10^(-4))/(3xx10^(-6))= 200V`.