In when the switch is swapped from `1` to `2`, find the heat produced in the circuit.
.

Answer

`(E^(2)C C_(0))/((2C+C_(0)))`.

Explanation

For (a), `C_(eq)=((C+C_(0))C)/(2C+C_(0))` and `Q=C_(eq)E`
`:. Q_(1)=(C_(0)Q)/(C+C_(0))` and `Q_(2)=(CQ)/(C+C_(0))`
.
Charge flowing through battery on reconnection, as in (b)
`q=Q-Q_(2)=Q-(CQ)/(C+C_(0))=(C_(0)Q)/(C+C_(0))=(C_(0)E)/(C+C_(0))C_(eq)`
Total energy on capacitors before and after reconnection is the same . Hence, whole work done by battery will go in the form of heat, so, heat generted is
`W_(b)=qE=(C_(0)E^(2))/(C+C_(0))xx((C+C_(0))C)/((2C+C_(0)))=(CC_(0)E^(2))/(2C+C_(0))`
Alternatively:
Heat `=sum((DeltaQ)^(2))/(2C)=2[((Q-Q_(2))^(2))/(2C)]+((2Q_(1))^(2))/(2C_(0))`
(Charge floeing through the switch is `2Q_(1)`.
`(E^(2)C C_(0))/((2C+C_(0)))`.