In the LCR circuit shown in Fig., the ac driving voltage is `upsilon = upsilon_(m) sin omega t`.
(i) Write down the equation of motion for q (t).
(ii) At `t = t_(0)`, the source stops and R is short circuited.
Now write down how much energy is stored in each of L and C.
(iii) Describe subsequent motion of charges.

Explanation

(i) In the circuit shown in Fig
When driving voltage `upsilon = upsilon_(m) sin omega t` is applied, the valid equaiton for the circuit with usual notation is
`L (d^(2) q)/(dt^(2)) + R (dq)/(dt) + (q)/(C ) = upsilon_(m) sin omega t`
This is the equation of motion for q (t).
Let `q = q_(m) sin (omega t + phi) = - q_(m) cos (omega t + phi)`
`:. i = i_(m) sin (omega t + phi) = q_(m) omega sin (omega t + phi)`
`i_(m) = (upsilon_(m))/(Z) = (upsilon_(m))/(sqrt(R^(2) + (X_(C) - X_(L))^(2)))` and `phi = (tan^(-1)) ((X_(C ) - X_(L)))/(R )`
(ii) When source stops at `t = t_(0)` and R is short circuited, erergy is stored in L and C
`U_(L) = (1)/(2) L i^(2) = (1)/(2) L [(upsilon_(m))/(sqrt(R^(2) + (X_(C ) - X_(L))^(2)))]^(2) sin^(2) (omega t_(0) + phi)`
and `U_(C ) = (1)/(2) (q^(2))/(C ) = (1)/(2C) [ q_(m)^(2) cos^(2) (omega t_(0) + phi)] = (1)/(2 C) (i_(m)^(2))/(omega^(2)) cos^(2) (omega t_(0) + phi)`
`U_(C ) = (1)/(2C) [(upsilon_(m))/(sqrt(R^(2) + (X_(C ) - X_(L))^(2)))]^(2) (1)/(omega^(2)) cos^(2) (omega t_(0) + phi)`
(iii) When R is short circuited, the circuit becomes an LC oscillation. The capacitor will go no discharging till all erergy goes to L, and back to C again. This keeps on repeating indefinitely.