(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`.

Options

A. `CE`
B. `2CE`
C. `3CE`
D. `4 CE`

Answer

B

Explanation

When battery of emf `2E` is short - circuited i.e. repalce it by a wire.

p.d across each capacitor, `V_(1) , V_(2) = 4E`
Total charge on capacitors `Q_(f) = C_(1)V_(1) + C_(2)V_(2) = 12 CE`
Amount of charge flown in cell of emf `4E`
`Q_(i)- Q_(f) = 2CE`.