Obtain an expression for average power of AC over a cycle.

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Q. Obtain an expression for average power of AC over a cycle.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2026Subjective· 3mImportance★★★★★

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Concept understanding — Average Power Absorption

Average Power Absorption – From Intuition to Precision

Think of pushing a child on a swing. You don't push constantly — you push only when the swing is moving away from you, and you time your push to add energy each time. Some pushes land perfectly, others might be slightly off. Over several minutes, what matters is not the force at any single instant, but the net energy you transferred averaged over time.

That's the core idea behind average power absorption: how much energy, on average, is being delivered per unit time to a device or system, even when the instantaneous power fluctuates wildly.

The Intuitive Picture

Consider a light bulb connected to household AC supply. The voltage oscillates 50 times per second (in India). At the peak of the voltage cycle, the bulb glows brightest; when voltage crosses zero, the bulb goes dark for an instant. But you don't see flickering — your eyes average out the rapid changes. What you perceive as "brightness" corresponds to the average power the bulb absorbs.

Similarly, when you charge a phone battery, the power drawn isn't constant — it's high when the battery is low, then tapers off. The "charging speed" you care about is the average power over the charging session.

The Precise Definition

P_{\text{avg}} = \frac{1}{T} \int_{0}^{T} p(t) \, dt

Where:

$p(t)$ is the instantaneous power at time $t$ (in watts)
$T$ is the time period over which we average (in seconds)

For a resistor with a sinusoidal voltage $v(t) = V_m \sin(\omega t)$ and current $i(t) = I_m \sin(\omega t)$ (since they're in phase), the instantaneous power is:

p(t) = v(t) \cdot i(t) = V_m I_m \sin^2(\omega t)

This is always positive (since $\sin^2$ is never negative) but it oscillates between $0$ and $V_m I_m$ . The average over one complete cycle gives:

P_{\text{avg}} = \frac{V_m I_m}{2}

Why This Matters for Exams

The most common mistake students make is confusing peak power with average power. A 100 W bulb doesn't draw 100 W at every instant — it draws about 200 W at the voltage peak and 0 W at the zero crossing. The 100 W rating is the average power it's designed to dissipate safely.

Watch out

Never use $P = VI$ directly with AC peak values unless you divide by 2 (for sinusoidal waveforms). The correct formula for average power in a resistor is $P_{\text{avg}} = \frac{V_m I_m}{2} = V_{\text{rms}} I_{\text{rms}}$ , where $V_{\text{rms}} = V_m / \sqrt{2}$ .

The General Case (Phase Differences)

When voltage and current are not in phase — as in circuits with inductors or capacitors — the instantaneous power can become negative during parts of the cycle (energy flows back to the source). The average power then becomes:

$P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos \phi$ …

Why this formula?

Average Power Absorption: Why the Formula Holds

Let's build this from first principles — understanding why average power is what it is, not just memorising the formula.

1. Instantaneous Power — The Starting Point

For any circuit element, instantaneous power is always:

p(t) = v(t) \cdot i(t)

This is the fundamental definition: power at an instant is voltage times current at that same instant.

2. Why We Need an Average

In AC circuits, both $v(t)$ and $i(t)$ vary sinusoidally with time. So $p(t)$ also varies — often at twice the frequency of the original signals.

Instantaneous power oscillates between zero and a peak value.
What matters for real energy consumption is the average over a complete cycle.

Hence, we define:

P_{\text{avg}} = \frac{1}{T} \int_0^T p(t) \, dt

where $T$ is the time period of the AC waveform.

3. The Key Derivation (Step-by-Step)

Step 1: Write the sinusoidal forms

Let:

$v(t) = V_m \cos(\omega t + \theta_v)$
$i(t) = I_m \cos(\omega t + \theta_i)$

Here $\theta_v$ and $\theta_i$ are phase angles. The phase difference is:

\phi = \theta_v - \theta_i

Step 2: Instantaneous power

p(t) = V_m I_m \cos(\omega t + \theta_v) \cos(\omega t + \theta_i)

Use the trigonometric identity:

\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]

So:

p(t) = \frac{V_m I_m}{2} \left[ \cos(\theta_v - \theta_i) + \cos(2\omega t + \theta_v + \theta_i) \right]

Step 3: Average over one cycle

The average of $\cos(2\omega t + \text{constant})$ over a full cycle is zero — because it's a sinusoid symmetric about zero.

Only the constant term survives:

P_{\text{avg}} = \frac{V_m I_m}{2} \cos(\phi)

4. The Standard Form Using RMS Values

Recall:

$V_{\text{rms}} = \frac{V_m}{\sqrt{2}}$
$I_{\text{rms}} = \frac{I_m}{\sqrt{2}}$

Therefore:

\frac{V_m I_m}{2} = V_{\text{rms}} I_{\text{rms}}

So the final formula is:

\boxed{P_{\text{avg}} = V_{\text{rms}} \, I_{\text{rms}} \, \cos \phi}

5. What $\cos \phi$ Really Means

$\phi$ is the phase difference between voltage and current.
$\cos \phi$ is called the power factor.
Why it appears: Only the component of current in phase with voltage contributes to average power. The quadrature (90° out-of-phase) component averages to zero.

| $\phi$ | $\cos \phi$ | Interpretation | …

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Averaging the instantaneous power $vi$ over a full AC cycle (with current lagging/leading voltage by phase φ) gives $P_{avg}=V_{rms}I_{rms}\cos\phi$ , where $\cos\phi$ is the power factor.

Derivation

Let the applied voltage and resulting current in an AC circuit be

$v = V_m\sin\omega t, \qquad i = I_m\sin(\omega t-\phi)$

where $\phi$ is the phase difference between them.

Instantaneous power:

$p = vi = V_mI_m\sin\omega t\,\sin(\omega t-\phi)$

Using the product-to-sum identity $\sin A\sin B = \dfrac12[\cos(A-B)-\cos(A+B)]$ :

$p = \dfrac{V_mI_m}{2}\Big[\cos\phi - \cos(2\omega t-\phi)\Big]$

Averaging over a full cycle. The term $\cos(2\omega t-\phi)$ oscillates at twice the frequency and averages to zero over a complete cycle, leaving

$P_{avg} = \dfrac{V_mI_m}{2}\cos\phi$

In terms of rms values. Since $V_{rms}=V_m/\sqrt2$ and $I_{rms}=I_m/\sqrt2$ : …

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Q.Obtain an expression for average power of AC over a cycle.

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