For an $LCR$ circuit driven at frequency $\omega $,the equation reads $L\frac{di}{dt} + Ri + \frac{q}{C} = V_i = V_m \sin \omega t$.
$(a)$ Multiply the equation by $i$ and simplify where possible.
$(b)$ Interpret each term physically.
$(c)$ Cast the equation in the form of a conservation of energy statement.
$(d)$ Integrate the equation over one cycle to find that the phase difference between $V$ and $i$ must be acute.

(N/A) The given equation for an $LCR$ circuit is:
$L \frac{di}{dt} + Ri + \frac{q}{C} = V_m \sin \omega t$
$(a)$ Multiplying by $i$ on both sides:
$Li \frac{di}{dt} + i^2 R + \frac{q}{C} i = V_m i \sin \omega t$
Since $i = \frac{dq}{dt}$,we can write:
$\frac{d}{dt} \left( \frac{1}{2} L i^2 \right) + i^2 R + \frac{d}{dt} \left( \frac{q^2}{2C} \right) = Vi$
$(b)$ Physical interpretation:
$Li \frac{di}{dt} = \frac{d}{dt} (\frac{1}{2} L i^2)$ is the rate of change of magnetic energy stored in the inductor.
$i^2 R$ is the rate of Joule heating (power dissipation) in the resistor.
$\frac{q}{C} i = \frac{d}{dt} (\frac{q^2}{2C})$ is the rate of change of electric energy stored in the capacitor.
$Vi$ is the instantaneous power supplied by the source.
$(c)$ The equation $\frac{d}{dt} (\frac{1}{2} L i^2 + \frac{q^2}{2C}) + i^2 R = Vi$ represents the conservation of energy,where the power supplied by the source equals the sum of the rate of change of stored energy and the power dissipated as heat.
$(d)$ Integrating over one cycle $T = \frac{2\pi}{\omega}$:
$\int_0^T \frac{d}{dt} (\frac{1}{2} L i^2 + \frac{q^2}{2C}) dt + \int_0^T i^2 R dt = \int_0^T Vi dt$
Since the energy stored in $L$ and $C$ is periodic,the integral of the derivative over one cycle is zero.
$0 + I_{rms}^2 R T = \int_0^T Vi dt$
Since $I_{rms}^2 R T > 0$,the average power $\int_0^T Vi dt$ must be positive. This implies $\cos \phi > 0$,meaning the phase difference $\phi$ must be acute.