A device 'X' is connected to an ac source. Over one complete cycle, three curves are plotted on common axes of voltage, current and power against time $t$, labelled A, B and C. Curve A oscillates at twice the frequency of the other two and is symmetric about the time axis, with equal areas above and below it (so its average over the cycle is zero). Curves B and C are sinusoids of the same period as each other but shifted by a quarter cycle (a phase difference of $90^\circ$), so that wherever one of them is maximum the other is zero. (a) Which curve shows the power consumed over a full cycle? (b) What is the average power consumption over a cycle? (c) Identify the device 'X'.

A device 'X' is connected to an ac source. Over one complete cycle, three curves are plotted on common axes of

NCERT Exemplar · Q15

t

(a) Which curve shows the power consumed over a full cycle?

(b) What is the average power consumption over a cycle?

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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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Curves B and C are the voltage and current, a quarter cycle ( $90^\circ$ ) apart. Their product, the instantaneous power, is curve A: it swings at double frequency and is symmetric about the time axis, so its average is zero. A device that gives a $90^\circ$ voltage-current phase difference is a pure reactance — an ideal inductor or capacitor.

Concept: instantaneous vs average power

Let the voltage be $v=v_m\sin\omega t$ and the current $i=i_m\sin(\omega t\pm\tfrac{\pi}{2})=\pm i_m\cos\omega t$ (a $90^\circ$ phase difference). The instantaneous power is

p=vi=\pm v_m i_m\sin\omega t\cos\omega t=\pm\frac{v_m i_m}{2}\sin 2\omega t.

This has twice the frequency of $v$ and $i$ and is symmetric about zero.

(a) Which curve is power?

The curve that oscillates at double frequency and is symmetric about the time axis (equal positive and negative loops) is the power curve — that is curve A. Curves B and C, being single-frequency sinusoids $90^\circ$ apart, are the applied voltage and the current.

(b) Average power over a cycle

The average of $\sin 2\omega t$ over a full cycle is zero, so

\langle p\rangle=\frac{v_m i_m}{2}\,\langle\sin 2\omega t\rangle=0.

Equivalently, $\langle p\rangle=V_{rms}I_{rms}\cos\phi$ with $\phi=90^\circ$ , and $\cos 90^\circ=0$ . The equal positive and negative loops of curve A cancel.

(c) Identify device 'X' …

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

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$V=100\sin(100t)$
›Reveal solutionSolution
This tests the average-power formula for AC circuits, $P_{avg}=V_{rms}I_{rms}\cos\phi$ , using the peak values and phase angle read off the given $V(t)$ and $I(t)$ expressions. Answer: 2.5 W.

Concept and Intuition

In a pure AC circuit, instantaneous power oscillates, but the useful (average) power dissipated depends on how much the current is in phase with the voltage. If current lags/leads voltage by $\phi$ , only the in-phase component $I_0\cos\phi$ contributes to the time-averaged power — this is captured by the power factor $\cos\phi$ . The RMS values are used because they represent the equivalent DC values that would dissipate the same average power.

Step-by-Step Solution

Given $V=100\sin(100t)$ V $\Rightarrow V_0=100$ V, phase $=0$ .

Given $I=100\sin\left(100t+\dfrac{\pi}{3}\right)$ mA $\Rightarrow I_0=100$ mA $=0.1$ A, and current leads voltage by $\phi=\pi/3=60^\circ$ .

RMS values: $V_{rms}=\dfrac{V_0}{\sqrt2}=\dfrac{100}{\sqrt2}$ V, $I_{rms}=\dfrac{I_0}{\sqrt2}=\dfrac{0.1}{\sqrt2}$ A. …
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AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
$50 \sin(50t)$
›Reveal solutionSolution
Average AC power uses $P=V_{rms}I_{rms}\cos\phi$ ; with a $\pi/4$ phase lag between current and voltage, the power dissipated comes out to about 0.884 W.

Concept and Intuition

In an AC circuit with a phase difference $\phi$ between voltage and current (as happens with any reactive element), only the in-phase component of current does real work — hence the $\cos\phi$ (power factor) term. Using peak values, $P_{avg} = \dfrac{V_0 I_0}{2}\cos\phi$ , equivalent to $V_{rms}I_{rms}\cos\phi$ .

Step-by-Step Solution

From $V=50\sin(50t)$ V, $V_0 = 50$ V; from $I=50\sin(50t+\pi/4)$ mA, $I_0 = 50\ \mathrm{mA} = 0.05$ A, and phase difference $\phi=\pi/4$ .

Average power: $P = \dfrac{V_0 I_0}{2}\cos\phi = \dfrac{50\times0.05}{2}\cos(45^\circ)$ .

$\dfrac{50\times0.05}{2} = \dfrac{2.5}{2} = 1.25$ . …
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AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
$\Omega$
›Reveal solutionSolution
For a series LR circuit, impedance is the hypotenuse of the $R$ - $X_L$ right triangle; here $Z=100\ \Omega$ and power factor $=R/Z=0.6$ .

Concept and Intuition

In a series AC circuit with resistance $R$ and inductive reactance $X_L$ , the voltage across $R$ and across $L$ are 90° out of phase, so they combine as vectors (phasors) at right angles. The impedance is the phasor sum:
$Z=\sqrt{R^2+X_L^2}$
The power factor is the cosine of the phase angle between current and applied voltage, given by $\cos\phi = R/Z$ (since the resistor is the only element that actually dissipates real power).

Step-by-Step Solution

$Z = \sqrt{R^2+X_L^2} = \sqrt{60^2+80^2} = \sqrt{3600+6400}=\sqrt{10000}=100\ \Omega$ . …
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AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$\dfrac{1}{3}$
›Reveal solutionSolution
With $X_L=2R$ and $X_C=3R$ , the net reactance magnitude equals R itself, making the impedance triangle a 45°-45°-90° triangle and giving power factor $1/\sqrt2$ .

Concept and Intuition

In a series LCR circuit, the power factor is $\cos\phi=\dfrac{R}{Z}$ , where $Z=\sqrt{R^2+(X_L-X_C)^2}$ is the total impedance. The resistor is the only element that dissipates real power; the reactive elements only shift the phase, and the power factor tells us what fraction of the apparent power (VI) is actually real (dissipated) power.

Step-by-Step Solution

Net reactance: $X=X_L-X_C=2R-3R=-R$ ; its magnitude is $R$ (the circuit is net capacitive, but only the magnitude matters for the power factor).

Impedance: $Z=\sqrt{R^2+X^2}=\sqrt{R^2+R^2}=\sqrt{2R^2}=R\sqrt2$ . …
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AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
$\sqrt{2}$
›Reveal solutionSolution
This tests the relation between power factor and the R–X triangle in a series LR ac circuit; the ratio $R:X_L$ works out to $1:\sqrt3$ .

Concept and Intuition

In a series R–L ac circuit, the resistance $R$ and inductive reactance $X_L$ combine as perpendicular sides of a right triangle whose hypotenuse is the impedance $Z = \sqrt{R^2+X_L^2}$ . The phase angle $\phi$ by which current lags the applied voltage satisfies $\cos\phi = R/Z$ and $\tan\phi = X_L/R$ . The power factor $\cos\phi$ tells us directly what fraction of $Z$ is resistive, which pins down the angle and hence the ratio of the two triangle legs.

Step-by-Step Solution

Given power factor $\cos\phi = R/Z = 0.5$ , so $\phi = 60^\circ$ .

In the impedance triangle, $\tan\phi = \dfrac{X_L}{R}$ . …
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AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
$Q. A resistor of resistance R and an inductor of inductive reactance R are connected in series to an ac source. A capacitor of capacitive reactance 2R is then connected in series with L and R. The ratio of the power factors of LR and LCR circuits is (A) 1:1 (B) 1:2 (C) 1:3 (D) 2:3$
›Reveal solutionSolution
Both the LR circuit and the resulting LCR circuit end up with the same net reactance magnitude equal to $R$ , so their power factors are identical, giving a ratio of 1:1.

Concept and Intuition

Power factor is $\cos\phi = \dfrac{R}{Z}$ , where $Z=\sqrt{R^2+X^2}$ and $X$ is the net reactance (inductive minus capacitive). Adding a capacitor doesn't just add resistance to the circuit's opposition — it can partially cancel the inductive reactance, and here it happens to leave the net reactance magnitude unchanged.

Step-by-Step Solution

LR circuit: $X_L=R$ (given). Impedance $Z_1=\sqrt{R^2+X_L^2}=\sqrt{R^2+R^2}=R\sqrt2$ . Power factor: $\cos\phi_1 = \dfrac{R}{Z_1}=\dfrac{R}{R\sqrt2}=\dfrac{1}{\sqrt2}$ .

LCR circuit: capacitor with $X_C=2R$ added in series. Net reactance $=X_L-X_C = R-2R=-R$ (magnitude $R$ ). Impedance $Z_2=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{R^2+R^2}=R\sqrt2$ . …
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AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
$Q. An inductor is connected to an ac source of frequency 50 Hz. The frequency of the instantaneous power developed in the circuit is (A) 25 Hz (B) 50 Hz (C) 100 Hz (D) 200 Hz$
›Reveal solutionSolution
This tests how instantaneous power in an AC circuit relates to the source frequency. Answer: 100 Hz.

Concept and Intuition

For an inductor on an AC source, $v=V_0\sin(\omega t)$ and $i=I_0\sin(\omega t - \pi/2)$ (current lags voltage by 90°). The instantaneous power is $p=vi=V_0 I_0 \sin(\omega t)\sin(\omega t-\pi/2)$ . Using the product-to-sum trigonometric identity, this simplifies to a term containing $\cos(2\omega t)$ — i.e. the power oscillates at twice the angular frequency of the source. This is a general feature: the instantaneous power in any AC circuit (resistive, inductive, or capacitive) oscillates at twice the source frequency, since it's built from products of sinusoids of the same base frequency.

Step-by-Step Solution

Source frequency: $f = 50\ \text{Hz}$ , so $\omega = 2\pi f$ .

Voltage and current both vary as $\sin(\omega t + \text{phase})$ .

Instantaneous power $p(t) = v(t)i(t)$ is a product of two sinusoids of the same frequency $\omega$ . …
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Concept understanding — Average Power Absorption

Average Power Absorption – From Intuition to Precision

The Intuitive Picture

The Precise Definition

Why This Matters for Exams

The General Case (Phase Differences)

Why this formula?

Average Power Absorption: Why the Formula Holds

1. Instantaneous Power — The Starting Point

2. Why We Need an Average

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

Step 1: Write the sinusoidal forms

Step 2: Instantaneous power

Step 3: Average over one cycle

4. The Standard Form Using RMS Values

5. What $\cos \phi$ Really Means

Concept: instantaneous vs average power

(a) Which curve is power?

(b) Average power over a cycle

(c) Identify device 'X' …

Method: Identifying Voltage, Current and Power Curves from Their Relative Frequency and Phase

Steps

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept: instantaneous vs average power

(a) Which curve is power?

(b) Average power over a cycle

(c) Identify device 'X' …

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