Q.For an LCR circuit driven at frequency $\omega$, the equation reads $L\dfrac{di}{dt} + Ri + \dfrac{q}{C} = v_i = v_m \sin\omega t$.

Concept: Power Dissipation in Resistors — the $Ri^2$ term represents instantaneous power loss as heat, and its cycle average must be supplied by the source.

(i) Multiply by $i$:

Since $i = dq/dt$, the third term becomes $\frac{q}{C} \frac{dq}{dt}$.

(ii) Physical interpretation:

• $L i \frac{di}{dt} = \frac{d}{dt}\left(\frac12 L i^2\right)$ — rate of change of magnetic energy stored in the inductor.
• $R i^2$ — instantaneous power dissipated as heat in the resistor.
• $\frac{q}{C} \frac{dq}{dt} = \frac{d}{dt}\left(\frac{q^2}{2C}\right)$ — rate of change of electric energy stored in the capacitor.
• $v_m i \sin\omega t$ — instantaneous power supplied by the source.

(iii) Conservation of energy statement:

The rate of increase of stored energy plus the dissipation rate equals the input power. …

Concept and Intuition

The LCR circuit is a beautiful example of energy conversion and conservation. When we drive it with an AC source, energy flows back and forth between the inductor's magnetic field and the capacitor's electric field, while the resistor steadily converts electrical energy into heat. The key insight is that power — the rate of energy transfer — is always $P = vi$ for any circuit element. By multiplying the circuit equation by $i$, we transform a voltage-balance equation into a power-balance equation, which reveals exactly where energy goes at every instant.

Step-by-Step Solution

1. Multiply the equation by $i$

The given equation is:

Multiplying every term by $i$:

2. Simplify the first and third terms

The first term can be rewritten using the chain rule. Notice that:

For the third term, recall that current is the rate of change of charge: $i = \frac{dq}{dt}$. So:

3. Write the simplified power equation

Substituting these back:

Or equivalently:

4. Physical interpretation of each term

5. Cast as a conservation of energy statement

Rearranging:

This reads: The power supplied by the source equals the power dissipated in the resistor plus the rate at which energy is stored in the inductor and capacitor. It is a statement of conservation of energy — no energy is created or destroyed, only converted between forms.

6. Integrate over one complete cycle

Integrate both sides from $t = 0$ to $t = T$, where $T = \frac{2\pi}{\omega}$ is the time period:

The last integral is the net change in stored energy over a full cycle. Since the circuit returns to the same state after one complete cycle (steady-state AC), the stored energy at $t=0$ and $t=T$ is identical. Therefore:

So we are left with:

$\int_0^T v_m i \sin\omega t \, dt = \int_0^T Ri^2 \, dt$ …

Method: Converting a Circuit's Differential Equation into an Energy-Balance Statement

Use this whenever you're asked to interpret a circuit equation physically, or to relate a circuit's phase angle to whether it dissipates or stores energy.

Steps

Step 1: Start from the loop's voltage equation

Write Kirchhoff's voltage law for the circuit as given, e.g.

This is a statement about voltages, not yet about energy or power.

Step 2: Multiply every term by the current $i$

Since power is $p = vi$ for any two-terminal element, multiplying the whole equation by $i$ converts each voltage-drop term into an instantaneous power term. This is the key trick: it turns a circuit (voltage) equation into an energy (power) equation.

Step 3: Recognise each term as the time-derivative of a stored energy

Use $Li\frac{di}{dt} = \frac{d}{dt}\left(\frac12 Li^2\right)$ and, since $i = \frac{dq}{dt}$, $\frac{q}{C}i = \frac{d}{dt}\left(\frac{q^2}{2C}\right)$. The resistive term $Ri^2$ does not simplify this way — it stays as an irreversible loss, because $R>0$ always makes it non-negative.

Step 4: Integrate over one full cycle to eliminate the stored-energy terms …