Two capacitors of `10 pF` and `20 pF` are connected to 200 V and 100 V sources, respectively. If they are connected by the wire, then what is the common potential of the capacitors?

Options

A. 133.3V
B. 150V
C. 300V
D. 400V

Answer

A

Explanation

Given `C_(1)= 10 pF = 10 xx 10^(-12)F`
`C_(2)= 20pF = 20 xx 10^(-12)F`
`V_(1)= 200V, V_(2)=100V`
Using `rArr q_(1)= V_(1)C_(1)`
`q_(2)= V_(2)C_(2)`
Common potential of capacitors
`V=(q_(1) + q_(2))/(C_(1) + C_(2)) = (V_(1)C_(1) + V_(2)C_(2))/(C_(1) + C_(2))`
`=(200 xx 10 xx10^(-12)+ 100 xx 20 xx 10^(-12))/(10 xx 10^(-12) +20 xx 10^(-12))= (4000)/(30)= 133.3V`