A charged capacitor `C_1` is discharged through a resistance R by putting switch S in position 1 of the circuit as shown in fig.5.201. when the discharge current reduces to `i_omega`, the switch is sudenly shifted to position 2. Calculate the amount of heat liberated in resistor R starting form this instant. Also calculate current I through the circuit as a function of time.

Explanation

Let the charge on capacitor `C_1 be q_0` when the switch was
shifted from position 1 to position 2. just before shifting of
switch, the circuit was as shown in fig 5.202

`(q_0)/(C_1) - I_0R =0 or q_0 = I_0 RC_1`
When the switch is shifted from position 1 to 2,
the capacitor `C_1` continues to be discharged while `C_2` starts
charging. Let at time t, after shiffting of switch to position 2,
charge on capacitor `C_2` be q and let current through the circuit
be I.
Therefore, charge remaining on `C_1` is equal to `(q_0 - q)` as
shown in fig 5.202 (b). Applying Kirchhoff's voltage law on
the circuit shown in fig. 5.202 (b) we get
`(q)/(C_2) + IR - ((q_0 - q))/(C_1) = 0`
or ` IR = (q_0 - q)/(C_1) - (q)/(C_2) =((q_0 C_2 - qC_2) -qC_1)/(C_1C_2)`
But current , `I= dq/dt` (rate of increase of chargen `C_2`
or `R = (dp)/(dt) = (q_0C_2 -q(C_1+C_2))/(C_1C_2)`
or `(dp)/( q_0C_2 - q (C_1+C_2)) = (dt)/(RC_1xxC_2)`
But at t = 0, q= 0,
`int_0^q (dq)/(q_0C_2 - q(C_1+C_2)) = int_0^1 (dt)/(RC_1C_2)`
From the above equation, we get
`q = ((q_0C_2)/(C_1+C_2)) [1-e^(-((C_1+C_2)/(RC_1C_2))t)]`
Substituting `q_0 = I_0 RC_1,` we get
`q = I_0(RC_1C_2)/(C_1+C_2) [1-e^(-((C_1+C_2)/(RC_1C_2))t)]`
But current
`I = (dq)/(dt) = I_0e^(-(C_1+C_2)/(RC_1C_2)t)`
in a steady state, the common potential difference across
capacitors is given by
`V = (q_0 +0)/(C_1+C_2) = (I_0RC_1)/(C_1+C_2) ("using "V = (C_1V_1+C_2V_2)/(C_1+C_2))`
Initially energy stored in `C_1` was
`U_1 = (q_0^2)/(2C_1) =(1)/(2) I_0^2R^2C_1`
In steady state, energy stored in two capacitors is
`U_2 = (1)/(2)C_1V^2 +(1)/(2) C_2V^2 = (1)/(2) (C_1 +C_2) (I_0^2R^2C_1^2)/((C_1+C_2)^2) = (I_0^2R^2C_1^2)/(2(C_1+C_2))`
Heat generated across resistor R is equal to the loss of
energy stored ni capacitors during redistribution of charge, i.e.,
`U_1 = U_2 =(I_0^2R^2C_1C_2)/(2(C_1+C_2))`