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.

Tamil Nadu DgeTamil 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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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 21 on this concept.

CBSE 2025Set ANNUAL1 markMCQ
$Q. The average power dissipated in a pure inductor is$
(i) VI^2
(ii) zero
(iii) (1/2)VI
(iv) VI^2/4
›Reveal solutionSolution

A pure inductor has phase angle 90 degrees, so average power = V I cos(90) = 0.

Average power in an AC circuit is $P_{avg} = V_{rms} I_{rms}\cos\phi$ . In a purely inductive circuit the current lags the applied voltage by exactly $90^\circ$ , so $\cos\phi = \cos 90^\circ = 0$ . Hence $P_{avg} = 0$ : over a full cycle e …

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CBSE 2024Set A1 markMCQ
$Q. If the phase difference between alternating current and e.m.f. is φ, then the value of power factor is (A) cos φ (B) cos^2 φ (C) sin φ (D) tan φ$
›Reveal solutionSolution

Power factor = cos φ.

The average power dissipated in an AC circuit is

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

where φ is the phase difference between the applied emf and the current. The factor $\cos\phi$ that determines what fraction of the apparent power $V_{rms}I_{rms}$ is actually consumed is called the power factor.

…

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CBSE 2024Set A1 markMCQ
$Q. In AC circuit, power is lost in only (A) resistance (B) inductance (C) capacitance (D) all of these$
›Reveal solutionSolution
Only resistance dissipates (real) power in AC; ideal L and C do not.

In an AC circuit the average power is $P = V_{rms}I_{rms}\cos\phi$ .

In a resistor, current and voltage are in phase (φ = 0), so cos φ = 1 and power $I_{rms}^2R$ is dissipated as heat. …
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CBSE 2024Set A1 mark
$Q. Write the value of power factor for a pure resistive circuit.$
›Reveal solutionSolution

In a pure resistor, V and I are in phase, so power factor cos φ = cos 0° = 1.

The average power dissipated in an AC circuit is $P_{avg} = V_{rms}I_{rms}\cos\phi$ , where $\cos\phi$ is called the power factor and φ is the phase difference between the current and the voltage.

In a purely resistive circuit, the current is always in phase with the applied voltage (φ = 0°), because a resistor offers no reactance. Therefore:

$\cos\phi = \cos 0^\circ = 1$ …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. For an alternating current i = I0 sinωt passing through a resistor R, how much is the average power loss due to Joule heating ?$
(a) I0²R
(b) (1/2) I0²R
(c) 4 I0²R
(d) 2 I0²R
›Reveal solutionSolution

Average power in a resistor over one AC cycle is Irms²R, and Irms = I0/√2 for a sinusoidal current.

For i = I0 sin ωt through a resistor R, the instantaneous power is p = i²R = I0²R sin²ωt.

Averaging sin²ωt over a full cycle gives 1/2, so:

Pavg = I0²R × (1/2) = (1/2) I0²R …

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CBSE 2024Set ANNUAL1 mark
$Q. What is wattless current?$
›Reveal solutionSolution

The quadrature (reactive) current component, $I_{rms}\sin\phi$ , that transfers no net energy.

In an AC circuit with a phase difference $\phi$ between voltage and current, the average power is $P = V_{rms}I_{rms}\cos\phi$ . The current can be resolved into two components: one in phase with the voltage, $I_{rms}\cos\phi$ (which does work / consumes power), and one $90°$ out of phase with the voltage, $I_{rms}\sin\phi$ (the quadrature component). This second component does no net work over a complete cycle — it is called the wattless current (or idle current). It occurs purely in ideal (resistance-free) inductive or capacitive circuits, w …

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CBSE 2023Set MODEL1 markMCQ
$Q. In an LCR circuit, the power factor becomes 0 (zero) when:$
(a) $R = 0$
(b) $\omega L = \omega C$
(c) $\omega L = \dfrac{1}{\omega C}$
(d) $\left(\omega L - \dfrac{1}{\omega C}\right) = R$
›Reveal solutionSolution

Power factor is zero only for a purely reactive (resistance-less) AC circuit.

In an LCR circuit, the power factor is $\cos\phi = \dfrac{R}{Z}$ , where $Z=\sqrt{R^2+(\omega L-\tfrac{1}{\omega C})^2}$ . The power factor becomes zero (i.e. $\phi = 90^\circ$ , purely reactive circuit, average power consumed is zero) only when $R=0$ . …

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CBSE 2023Set B1 mark
$Q. Write True or False: The average power supplied to an inductor over one complete cycle is zero.$
›Reveal solutionSolution

Because current and voltage in a pure inductor are 90° out of phase, the average power delivered to it over one full cycle works out to zero.

For a pure inductor in an AC circuit, current lags voltage by $\pi/2$ . The instantaneous power is $p = vi = v_0 i_0 \sin(\omega t)\sin(\omega t - \pi/2) = -v_0 i_0 \sin(\omega t)\cos(\omega t)$ , which when averaged over a complete cycle integrates to zero because $\langle \sin\omega t\cos\omega t\rangle = 0$ over a full period. Physically, the inductor stores energy in its magnetic field during one quarter cycle …

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CBSE 2022Set I1 markMCQ
$Q. The expression for real average power in an alternating current circuit is (A) P_av = E_rms cos φ (B) P_av = I_rms cos φ (C) P_av = E_rms I_rms sin φ (D) P_av = E_rms I_rms cos φ$
›Reveal solutionSolution

Average power in AC = E_rms × I_rms × cos φ, where cos φ is the power factor.

In an AC circuit the instantaneous voltage and current are generally out of phase by an angle φ. Averaging the instantaneous power e·i over a full cycle gives the real (active) power:

$P_{av} = E_{rms}\,I_{rms}\cos\varphi$

…

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CBSE 2022Set I1 markMCQ
$Q. Power factor is equal to (A) R/Z (B) Z/R (C) R.Z (D) W.L$
›Reveal solutionSolution

Power factor cos φ = R/Z.

In a series AC circuit the impedance triangle has resistance R along the base, net reactance (X_L − X_C) as the perpendicular, and impedance Z as the hypotenuse. The phase angle φ between voltage and current satisfies:

$\cos\varphi = \frac{R}{Z}$

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. In purely inductive or capacitive circuit, power factor (cos φ) is ____.$
(a) 0
(b) 1
(c) -1
›Reveal solutionSolution

In a purely inductive or purely capacitive AC circuit, current and voltage are 90° out of phase, so the power factor is zero.

…

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CBSE 2021Set A1 markMCQ
$Q. Which of the following relations is correct for power factor? (A) Power factor = True average power \times Apparent average power (B) Power factor = Apparent average power / True average power (C) Power factor = True average power / Apparent average power (D) Power factor = ½ [ True average power \times Apparent average power ]$
›Reveal solutionSolution

Power factor = true (real) power ÷ apparent power = cos φ.

In an ac circuit the apparent power is V_rms·I_rms, while the true (real) power actually consumed is V_rms·I_rms·cos φ, where φ is the phase angle between voltage and current. The power factor is the ratio of these:

$\text{power factor} = \frac{\text{True power}}{\text{Apparent power}} = \cos\varphi$

…

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

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