Question

For an $LCR$ circuit driven at frequency $\omega$, the equation read $L\frac{di}{dt}+Ri+\frac{q}{c}=v_i=v_m\sin \omega t$

1) Multiply the equation by $i$ and simplify where possible.
2) Interpret each term physically.
3) Cast the equation in the form of a conservation of energy statement.
4) Integrate the equation over one cycle to find that the phase difference between $v$ and $i$ must be acute.

Answer 1

Given,
$L\frac{di}{dt}+Ri+\frac{q}{c}=v_i$

Multiplying by $i,$

$Li\frac{di}{dt}+R{i}^2+\frac{qi}{c}=vi$ ---(i)

$Li\frac{di}{dt}=\frac{d}{dt}\left(\frac{1}{2}L{i}^2\right)$

$Li\frac{di}{dt}=$ rate of change of energy stored in an inductor

$R{i}^2=$ joule heating loss

$\frac{q}{c}i=\frac{d}{dt}\left(\frac{q^2}{2c}\right)$

$\frac{q}{c}i$ = rate of change of energy stored in the capacitor

$vi=$ rate of at which driving force pours in energy

It goes into ohmic loss and increase of stored energy
Equation (i) is in the form of conservation of energy statement.

${\int }_0^{T}dt\frac{d}{dt}(\frac{L}{2}{i}^2+\frac{q^2}{c})+{\int }_0^{T}R{i}^{2}dt={\int }_0^{T}vidt0+(+ve)={\int }_0^{T}vidt$

${\int }_0^{T}vidt>0$ if phase difference is a constant and an acute angle.