Two capacitors `3muF` and `6muF` are connected in series across a potential difference of `120V`. Then the potential difference across `3muF` capacitor is:

Options

A. 50 V
B. 60 V
C. 70 V
D. 80 V

Answer

D

Explanation

`V_(1)=Q/C_(1)=(C_(s)V)/C_(1)=((C_(1)C_(2))/(C_(1)+C_(2)))V/C_(1)`
`=(3xx10^(-6)xx6xx10^(-6)xx120)/(9xx10^(-6)xx3xx10^(-6))=80V`