An alternating current generator has an internal resistance $R_g$ and an internal reactance $X_g$. It is used to supply power to a passive load consisting of a resistance $R_g$ and a reactance $X_L$. For maximum power to be delivered from the generator to the load, the value of $X_L$ is equal to (a) zero. (b) Xg. (c) – Xg. (d) Rg.

An alternating current generator has an internal resistance Rg and an internal reactance Xg. It is used to sup

NCERT Exemplar · Q2

R_g

(a) zero.

(b) Xg.

(d) Rg.

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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

where $\phi$ is the phase angle between voltage and current. The $\cos \phi$ term is called the power factor, and it tells you what fraction of the apparent power ( $V_{\text{rms}} I_{\text{rms}}$ ) is actually absorbed as real power.

Important

Average power is the only power that does useful work or produces heat. Reactive power (associated with $\sin \phi$ ) oscillates between source and load without being consumed.

A Quick Check for Yourself

If a device draws 5 A (rms) from a 230 V (rms) supply with a power factor of 0.8, the average power absorbed is:

P_{\text{avg}} = 230 \times 5 \times 0.8 = 920 \text{ W}

Not $1150$ W — that would be the apparent power, which includes the energy that sloshes back and forth without being used.

Average power absorption is simply: over a full cycle, how much energy stays in the load, divided by the time. Everything else is just accounting for the waveform shape and phase.

Average power in an AC circuit, P_avg = V_rms I_rms cos φ, including the role of the power factor, is a core topic of the NCERT Class 12 Physics chapter on alternating current, tested in CBSE boards, JEE Main and NEET. Students searching "average power in ac circuit formula and power factor class 12 physics" will find this peak-versus-average-power distinction matches the NCERT-prescribed reasoning.

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
0°	1	Pure resistor — maximum power
90°	0	Pure inductor or capacitor — zero average power
Between	Between 0 and 1	Mixed R-L-C circuit

6. Special Cases (Exam-Ready)

For a pure resistor ( $\phi = 0$ ):

P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} = I_{\text{rms}}^2 R = \frac{V_{\text{rms}}^2}{R}

For a pure inductor or capacitor ( $\phi = \pm 90^\circ$ ):

P_{\text{avg}} = 0

Energy is stored and returned each cycle — no net absorption.

7. The Core Insight

Average power absorption is not about peak values — it's about the overlap between voltage and current waveforms.

The $\cos \phi$ factor quantifies this overlap. When voltage and current are perfectly aligned (in phase), all power is useful. When they are orthogonal, no net power is absorbed — only reactive power flows.

This is why power factor correction (adding capacitors to offset inductive loads) is so important in real power systems: it maximises $\cos \phi$ and thus minimises wasted current.

For maximum power transfer from an AC generator to a passive load, the load reactance must cancel the generator's internal reactance. The required value is $X_L = -X_g$ .

The key to this problem is understanding that in AC circuits, power is only dissipated in resistive components — reactances (inductors and capacitors) store and return energy, they don't consume it on average. So when we talk about "power delivered to the load," we mean the power dissipated in the load's resistive part $R_g$ .

But here's the twist: the load also has a reactive part $X_L$ , and the generator has its own internal reactance $X_g$ . These reactances don't consume power, but they do affect how much current flows through the circuit, and therefore how much power reaches the resistor.

The classic Maximum Power Transfer Theorem for DC circuits says: maximum power is delivered to the load when the load resistance equals the source resistance. For AC circuits, the theorem extends: maximum power transfer occurs when the load impedance is the complex conjugate of the source impedance. That is, if the source has impedance $Z_g = R_g + jX_g$ , then the load should have $Z_L = R_g - jX_g$ (assuming the load resistance is also $R_g$ as given).

Let's see why this works.

Set up the circuit. The generator has internal impedance $Z_g = R_g + jX_g$ . The load has impedance $Z_L = R_g + jX_L$ (note: the problem says the load resistance is also $R_g$ , and its reactance is $X_L$ ). They are connected in series, so the total impedance seen by the generator's internal voltage $V$ is:

Z_{\text{total}} = Z_g + Z_L = (R_g + jX_g) + (R_g + jX_L) = 2R_g + j(X_g + X_L).

Find the current. The current flowing through the circuit is:

I = \frac{V}{Z_{\text{total}}} = \frac{V}{2R_g + j(X_g + X_L)}.

The magnitude of this current is:

|I| = \frac{|V|}{\sqrt{(2R_g)^2 + (X_g + X_L)^2}}.

Power delivered to the load. Only the resistive part of the load dissipates power. The power delivered to the load is:

P = |I|^2 R_g = \frac{|V|^2 R_g}{(2R_g)^2 + (X_g + X_L)^2}.

Maximise the power. The numerator is constant (for a fixed $V$ and $R_g$ ). To maximise $P$ , we need to minimise the denominator. The denominator is:

D = 4R_g^2 + (X_g + X_L)^2.

The term $(X_g + X_L)^2$ is always non-negative. It is minimised when it equals zero, i.e., when:

X_g + X_L = 0 \quad \Rightarrow \quad X_L = -X_g.

At this point, the denominator is at its smallest value $4R_g^2$ , and the power is:

P_{\text{max}} = \frac{|V|^2 R_g}{4R_g^2} = \frac{|V|^2}{4R_g}.

Watch out

A common mistake is to think that maximum power occurs when the load resistance equals the source resistance and the load reactance equals the source reactance. That would give $X_L = X_g$ , which actually adds the reactances and reduces current. The correct condition is cancellation: $X_L = -X_g$ .

Tip

Think of it this way: the reactances don't consume power, but they do restrict current. If you let the load reactance oppose the generator's internal reactance, they cancel each other out, allowing maximum current to flow through the resistive parts. It's like having two opposing springs in a mechanical system — if they're equal and opposite, they neutralise each other and the system moves freely.

Check the result. With $X_L = -X_g$ , the total impedance becomes purely resistive: $Z_{\text{total}} = 2R_g$ . The circuit behaves like a DC circuit with two equal resistors in series, which is exactly the condition for maximum power transfer in the DC case. This confirms the result.

✓Final answer

X_L

Method: Applying the Maximum Power Transfer Theorem to an AC Source and Load

This method solves any "what load condition maximises delivered power" question for an AC generator driving a passive load — the AC generalisation of the DC maximum-power-transfer theorem.

Steps

Step 1: Write the total series impedance of source and load.

With the generator's own impedance $Z_g = R_g + jX_g$ and the load's impedance $Z_L = R_L + jX_L$ in series,

Z_{\text{total}} = Z_g + Z_L = (R_g + R_L) + j(X_g + X_L)

Step 2: Recall that only the resistive part of the load dissipates real power.

Reactances (inductors and capacitors) store and return energy over a cycle but do not consume it on average — so the quantity to maximise is the power delivered to $R_L$ specifically, not the total apparent power drawn from the source.

Step 3: Write the power delivered to the load's resistance in terms of current magnitude.

P = |I|^2 R_L = \frac{|V|^2 R_L}{(R_g+R_L)^2 + (X_g+X_L)^2}

Since the reactive part only ever appears as $(X_g+X_L)^2$ in the denominator and never helps the numerator, it can only reduce power — never increase it.

Step 4 (Applying to this problem): Minimise the denominator by cancelling reactances first, then optimise resistance.

The reactive term is minimised (zero) when $X_L = -X_g$ — the load reactance exactly cancels the source's own reactance, leaving a purely resistive total impedance. This is the AC version of the "conjugate match" condition $Z_L = Z_g^{*}$ . (If the load resistance itself were also free to vary, a second, separate step — setting $R_L = R_g$ — would apply, but many exam questions fix $R_L$ and ask only about the reactive part, as here.)

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 energy is stored in and returned from the magnetic field, with no net dissipation (this is the wattless component).

✓Final answer
$(ii) zero.$
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.

For a purely resistive circuit φ = 0, so cos φ = 1 (maximum power).

For a purely inductive or capacitive circuit φ = 90°, so cos φ = 0 (wattless current, no average power).

✓Final answer
$(A) cos φ.$
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.

In a pure inductor or pure capacitor, the current leads/lags the voltage by 90° (φ = 90°), so cos φ = 0. Energy is alternately stored and returned to the source each half-cycle, and the net average power is zero (a 'wattless' current).

Therefore, real power is lost only in the resistance.

✓Final answer
$(A) resistance.$
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:

$\cos\phi = \cos 0^\circ = 1$

This means all the supplied power is actually consumed (dissipated as heat) in a pure resistor — none of it is reactive power.

✓Final answer
$Power factor = 1 (unity) for a pure resistive circuit.$
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

This matches Pavg = Irms²R with Irms = I0/√2, since Irms² = I0²/2.

✓Final answer
$Average power = (1/2) I0²R — option (b).$
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, where the current is entirely $90°$ out of phase with the voltage, so no power is consumed at all.

✓Final answer
$I_{rms}\sin\phi$
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 $R=0$ .

(Note: the paper prints this option digit indistinctly as "0"; taken here as the power factor being zero, matching option (i).)

✓Final answer
$R = 0$
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 and returns exactly that energy to the circuit during the next quarter cycle — there is no net energy dissipation in an ideal (resistance-less) inductor.

✓Final answer
$True — average power over one complete AC cycle for a pure inductor is zero.$
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:

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

The factor cos φ is the power factor. The term E_rms I_rms is the apparent power, and E_rms I_rms sin φ is the wattless (reactive) power, which averages to zero over a cycle. Options (a) and (b) are dimensionally incomplete (a single quantity cannot be power), and (c) uses sin φ, which gives reactive not real power.

✓Final answer
$(D) P_av = E_rms I_rms cos φ.$
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:

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

Since cos φ is the power factor, its value is R/Z. It is dimensionless and lies between 0 and 1 (0 for a purely reactive circuit, 1 for a purely resistive circuit).

✓Final answer
$(A) R/Z.$
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.

Power factor is defined as $\cos\varphi$ , where φ is the phase difference between the applied voltage and the resulting current. In a purely inductive circuit, current lags voltage by 90° (φ = 90°). In a purely capacitive circuit, current leads voltage by 90° (φ = 90° again, opposite sign of lead/lag). In both cases $\cos 90° = 0$ . Physically this means the average power consumed over a cycle, $P_{avg} = V_{rms}I_{rms}\cos\varphi$ , is zero — no net energy is dissipated; energy is only stored and returned every half-cycle. This is why such circuits carry 'wattless current'.

✓Final answer
$\cos\varphi = 0$
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:

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

It lies between 0 and 1 and tells what fraction of the supplied power does useful work.

✓Final answer
$(C) Power factor = True average power / Apparent average power.$

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