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.

Explanation

To solve the problem step by step, we will address each part of the question systematically.

### Step-by-Step Solution:

**(i) Multiply the equation by I and simplify where possible.**

Given the equation:
\[ L \frac{di}{dt} + Ri + \frac{q}{C} = \upsilon_i = \upsilon_m \sin(\omega t) \]

Multiply both sides by \( I \):
\[ I \left( L \frac{di}{dt} + Ri + \frac{q}{C} \right) = I \upsilon_m \sin(\omega t) \]

This expands to:
\[ L I \frac{di}{dt} + R I^2 + \frac{qI}{C} = I \upsilon_m \sin(\omega t) \]

**(ii) Interpret each term physically.**

1. **\( L I \frac{di}{dt} \)**: This term represents the rate of change of energy stored in the inductor. It signifies how the current change affects the magnetic field and energy in the inductor.

2. **\( R I^2 \)**: This term represents the power loss due to resistance (Joule heating). It indicates how much energy is dissipated as heat in the resistor.

3. **\( \frac{qI}{C} \)**: This term represents the rate of change of energy stored in the capacitor. It relates to how the charge on the capacitor affects the electric field and energy stored in it.

4. **\( I \upsilon_m \sin(\omega t) \)**: This term represents the power supplied by the source. It shows how the input voltage drives the current in the circuit.

**(iii) Cast the equation in the form of a conservation of energy statement.**

Rearranging the equation gives:
\[ L I \frac{di}{dt} + R I^2 + \frac{qI}{C} - I \upsilon_m \sin(\omega t) = 0 \]

This can be interpreted as a conservation of energy statement:
- The energy increase in the inductor (first term) plus the energy stored in the capacitor (third term) must equal the energy lost in the resistor (second term) and the energy supplied by the source (fourth term).

**(iv) Integrate the equation over one cycle to find that the phase difference between \( \upsilon \) and \( I \) must be acute.**

Integrate the equation over one cycle \( T \):
\[ \int_0^T \left( L I \frac{di}{dt} + R I^2 + \frac{qI}{C} - I \upsilon_m \sin(\omega t) \right) dt = 0 \]

This leads to:
\[ \int_0^T L I \frac{di}{dt} dt + \int_0^T R I^2 dt + \int_0^T \frac{qI}{C} dt = \int_0^T I \upsilon_m \sin(\omega t) dt \]

The left-hand side represents energy stored and dissipated, while the right-hand side represents energy supplied.

Since \( R I^2 \) is always positive and the integral of \( I \upsilon_m \sin(\omega t) \) over one cycle is positive, it implies that the energy supplied is greater than the energy dissipated, which can only happen if the phase difference \( \phi \) between \( \upsilon \) and \( I \) is acute (i.e., \( 0 < \phi < 90^\circ \)).