The ratio of energy stored in capacitors `C_(1)` and `C_(2)`

Options

A. `9/8`
B. `8/9`
C. `5/6`
D. `6/5`

Answer

A

Explanation

The resistances each `R` are redundant, no current flows through them.

p.d. across = `C_(2) : V_(2) = 4E`
`p.d. "across" C_(1) : V_(A) + 2E - V_(1) = V_(B) implies V_(A) -V_(B)+2E = V_(1)`
`4E+2E=V_(1) implies V_(1) = 6E`
`(U_1)/(U_2) = (1/2 C_1V_(1)^(2))/(1/2 C_2V_(2)^(2)) = ((6)^(2))/(2(4)^(2)) = 9/8`
Total charge on capacitors `Q_(i) = C_(1)V_(1) + C_(2)V_(2) = 14 CE`.