What amount of heat will be generated in the circuit shown in Fig. after the swich `Sw` is shifted from position 2 ?

Options

A. `(epsilon_(0)C C_(0))/((2C+C_(0)))`
B. `(epsilon^(2)C C_(0))/((C+C_(0)))`
C. `(epsilon^(2)C C_(0))/((C+2C_(0)))`
D. `(2epsilon^(2)C C_(0))/((C+2C_(0)))`

Answer

A

Explanation

The charge on the condensers in position `1` are shown. Here `(q)/(C)=(q_(0))/(C_(0))=(q+q_(0))/(C+C_(0))`
and `(q+q_(0))((1)/(C+C_(0))+(1)/(C))=epsilon`
or `(q+q_(0))=(C(C+C_(0))epsilon)/(C_(0)+2C)`
Hence, `q=(C^(2) epsilon)/(C_(0)+2C) and q_(0)=(C C_(0) epsilon)/(C_(0) +2C)`.
(a)
(b)
After the switch thrown to position `2`, the charges changes as shown in (Fig-b). A charge `q_(0)` has flown in the right loop through the two condensers and a charge `q_(0)` through the cell, Because of the symmetery of the problem there is no change in the energy stored in the condensers. Thus
`H` (Heat product) = Energy delivered by the cell
`= Delta Q epsilon =q_(0) epsilon=(C C_(0) epsilon^(2))/(C_(0)+2C)`.