(a) Find potential difference between `A` and `B` and `C`.

(b) Find potential difference across capacitor `C`.


(i) Find ratio of energy stored in capacitors, `C_(1)` and `(C_2)`
(ii) If battery of emf of `2E` is short -circuited, find charge flown in battery of emf `4E`.

Explanation

(a)
`V_(1) = 2/(6+2)xx100=25V = V_A-V_B`
`V_2 = 100-V_1=75V= V_(B)-V_(C)`

`i=(2-1)/(20+10) = 1/30 A`
From lowest branch `AXB`
`V_A - V_B = 2-ixx20=2-1/30xx20=4/3A`
Branch `A Y B`
`V_(A) = V' -1 = V_B` ltbr. `V' = (V_A-V_B) -1 = 4/3-1 = 1/3 V`
(c )
`V_(A) -V_(B) = 4E`
Middle branch `A X B`
`V_(A) +2E-V_(1)=V_(B)`
`V_(1)=V_(A)-V_(B)+2E = 4E+2E = 6E`
p.d. across `C_(2)`:
`V_(2) = V_(A) - V_(B)+2E = 4E + 2E =6E`
(i) `(U_1)/(U_2)= (1/2C_(1)V_(1)^(2))/(1/2 C_(2)V_(2)) = (C(6E)^(2))/(2C(4E)^(2))= 9/8`
Total charge on capacitors
`Q = C_(1)V_(1) + C_(2)V_(2) = Cxx6E+2Cxx4E`
(ii )The battery of emf `2E` is short-circuited i.e. replace it by a wire

Total charge on capcitors
`Q' = Cxx4E+2Cxx4E=12CE`
Charge flown in the battery of emf `4E` ltbr. `Q - Q' = 2CE`.