Two capacitors of capacitances `20muF " and " 60muF` are connceted in series. If potential difference between the two ends of the combination is 40 V, calculate the terminal potential differences of each capacitor?

Explanation

If C be the equivalent capacitance of the combination, then
`C=(20xx60)/(20+60)= 15 muF = 15 xx10^(-6)F`
Total charge of the combination,
`Q= CV= 15xx10^(-6)xx40= 6xx 10^(-4)C`
Since the two capacitors are connected in series, charge on each capacitor is equal to the total charge of the combination, i.e., `6xx10^(-4)C`.
Potential difference between the two plates of the capacitor having capacitance `C_1` is,
`V_1= (Q)/(C_1)= (6xx10^(-4))/(20xx10^(-6)) = 30V`
Again, potential difference between the two plates of the capacitor having capacitance `C_2`is,
`V_2 =(Q)/(C_2)= (6xx10^(-4))/(60xx10^(-6)) = 10V`.