For an LCR circuit driven at frequency `omega`, the equation reads `L (di)/(dt) + Ri + (q)/(C ) = upsilon_(i) = upsilon_(m) sin omega t`
(i) Multiply the equation by I and simplify where possible. (ii) Interpret each term physically.
(iii) Cast the equation in the form of a conservation of energy statement.
(iv) Intergrate the equation over one cycle to find that the phase difference between `upsilon` and i must be acute.

Answer

A

Explanation

Consider the L-C-R circuit. Applying KVL for the loop, we can write.

`rArr L(di)/(dt) + q/C + iR= V_(m)sinomegat`……….(i)
Multiplying both sides by `i`, we get
`Li(di)/(dt) + q/Ci + i^(2)R=(V_(m)i)sin omegat=Vi`............(ii)
where `Li(di)/(dt)=d/(dt)(1/2Li^(2))` = rate of change of energy stored in an inductor.
`Ri^(2)`= joule heating loss
`q/Ci= d/(dt)(q^(2)/(2C))`= rate of change of energy stored in the capacitor.
Vi= rate at which driving force pours in energy. It goes into (i) ohmic loss and ii) increase of stored energy.
Hence, Eq. (i) is in the form of conservation of energy statement. Integrating both sides of Eq. (ii) with respect to time over one full cycle `(0to T)` we may write.
`int_(0)^(t) d/(dt) ( 1/2Li^(2)+q^(2)/(2C))` dt + `int_(0)^(T)Ri^(2)dt= int_(0)^(T)Vidt`
`rArr 0+(+ve)=int_(0)^(T)Vidt`
`rArr int_(0)^(T) dt gt 0` if phase difference between V and i is a constant and acute angle.