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.

Answer

A

Explanation

Consider the R-L-C circuit shown in the adjacent diagram.

Given,
Let current at any instant be I
Applying KVL in the given circuit
`iR + L(di)/(dt) + q/C -V_(m) sinomegat=0`……….(i)
Now, we can write `i=(dq)/(dt) rArr (di)/(dt) = (d^(2)q)/(dt)^(2)`
From Eq. (i) `(dq)/(dt)R + L(d^(2)q)/(dt)^(2) + q/C=V_(m)sinomegat`
`rArr L(d^(2)q)/(dt)^(2)+ R(dq)/(dt)+ q/C=V_(m) sin omegat`
This is the required equation of variation (motion) of charge.
b) Let `q=q_(m)sin(omegat +phi)=-q_(m)cos(omegat + phi)`
`i=i_(m)sin(omegat+phi) = q_(m)omegasin(omegat+phi)`
`i_(m) = V_(m)/Z= V_(m)/(sqrt(R^(2)+(X_(C)-X_(L))^(2)`
`phi = tan^(-1)(X_(c)-X_(L))/R`
Where R is short circuited at `t=t_(0)`, energy is stored in L and C.
`U_(L) = 1/2Li^(2)=1/2LV_(m)/(sqrt(R^(2)+(X_(C)-X_(L)))^(2)] sin^(2)(omegat_(0)+phi)`
and `U_(C) = 1/2 xx q^(2)/C= 1/(2C)[q^(2)mcos^(2)(omegat_(0)+ phi)]`
`=1/(2C)[V_(m)/sqrt(R^(2)+(X_(C)-X_(L))^(2))^(2)]`
`=1/(2C) xx (i_(m)/omega)^(2) cos^(2)(omegat_(0) + phi)`
`=(i^(2)m)/(2Comega^(2))cos^(2)(omegat_(0)+phi)` `(therefore i_(m) = q_(m)omega`]
`=1/(2C) [V_(m)/(sqrt(R^(2)+(X_(C)-X_(L))^(2))^(2)]] (cos^(2)(omegat_(0)+phi)`
c) When R is a short circuited, the circuit becomes an L-C oscillaor. The capacitor will go on discharging and all energy will go to L and back and forth. Hence, there is oscillation of energy from electrostatic to magnetic and magnetic to electrostatic.