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

Answer

A

Explanation

Given, `C_(1)=10pF=10xx10^(-12)F,`
`C_(2)=20pF=20xx10^(-12)F,`
`V_(1)=200V and V_(2)=100V`
where, `C_(1)=` Capacitance of Ist capacitor
`C_(2)=` Capacitance of Iind capacitor
`V_(1)=` Voltage across Ist capacitor
`V_(2)=` Voltage across IInd capacitor
We know that, `V_(1)=(q_(1))/(C_(1)) and V_(2)=(q_(2))/(C_(2))`
`rArr" "q_(1)=V_(1)C_(1) and q_(2)=V_(2)C_(2)`
So, 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))`
`=(200xx10xx10^(-12)+100xx20xx10^(-12))/(10xx10^(-12)+20xx10^(-12))`
`=(200xx10+100xx20)/(10+20)`
`=(2000+2000)/(30)`
`=(4000)/(30)=133.3V`