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_0`, the switch is suddenly 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.

Answer

A, B, C

Explanation

Capacitor `C_1` will discharge according to the equation
`q=q_0e^((-t)/tau_C)` …………..i
Here, `tau_C=C_1R`
and discharging current.
`i=(-dq)/(dt)=q_0/tau_C.e^(-t/tau_C)=(q_0e^((-t)/tau_C)/(C_1R))`……….ii
At the given instant `i=i_0`
Therefore from eqn ii
`q_0e^(-t/tau_C)=i_0C_1R` at this instant.
or charge `C_1` at this instant will be
`q=(i_0C_1R)` [From eqn ii]
Now this charge q will later on distribute in `C_1` and `C_2` .
`:. U_i=q^2/(2C_1+C_2)=U_f`
`:.` Heat genetrated in resistance.
`H=U_i=U_f`
Susbtituging values of `q:U_i` and `U_f` we get
`H=((I_0R)^2C_1C_2)/((2(C_1+C_2))`