A capacitor of capacitance `C = 5.00 muF` is connected to a source of constant `emf xi = 200 V` (Fig). Then the swich `Sw` was thrown over from contact 1 to conatact 2. Find the amount of heat generated in a resistance `R_(1) = 500 Omega` if `R_(2) = 300 Omega`

Explanation

When switch 1 is closed maximum charge accumulaated on the capacitor,
`q_(max) = C xi` ......(1)
and when switch 2 is closed, at any arbitary instant of time,
`(R_(1) + R_(2)) (-(dq)/(dt)) = q//C`,
because capacitor is discharging.
or, `int_(q_(max))^(q) (1)/(q) = - (1)/((R_(1) + R_(2)) C) int_(0)^(t) dt`
Intergrating, we get
In `(q)/(q_(max)) e = (-t)/((R_(1) + R_(2))C)` or, `q = q_(max) e = (-t)/((R_(1) + R_(2))C)` ....(2)
Differentiating with respect to time,
`i (f) = (dq)/(dt) = q_(max) e^((-t)/((R_(1) + R_(2))C)) (-(1)/((R_(1) + R_(2)) C))`
or, `i(t) = (C xi)/((R_(1) + R_(2))C) e (-t)/((R_(1) + R_(2))C)`
Negative sign is ignored, as we are not interested in the direction of the current.
thus, `i(t) = (xi)/((R_(1) + R_(2))) e (-1)/((R_(1) + R_(2)) C)` ....(3)
When the swich `(Sw)` is at the position 1, the charge (maximum) accumalted on the capacitor is,
`q = C xi`
When teh `Sw` is thown to position 2, the capacitor starts discharging and as a result the electric energy stored turns into heat energy tho' the resistors `R_(1)` and `R_(2)` (during a very long interval of time). Thus from the energy conservation, the total heat liberated tho' the resistors.
`H = U_(i) = (q^(2))/(2C) = (1)/(2) C xi^(2)`
During the process of discharging of the capacitor, the current tho' the resitors `R_(1)` and `R_(2)` is the same at all teh moments of time, thus
`H_(1) prop R_(1)` and `H_(2) prop R_(2)`
So, `H_(1) = (H R_(1))/((R_(1) + R_(2)))` (as `H = H_(1) + H_(2))`
Hence, `H_(1) = (1)/(2) (CR_(1))/(R_(1) + R_(2)) xi^(2)`