(a)(i) A light bulb and an open coil inductor are connected in series across an ac source of variable frequenc

Question

Q. (a)(i) A light bulb and an open coil inductor are connected in series across an ac source of variable frequency. How will the glow of the bulb be affected when : (I) an iron bar is inserted inside the coil, and (II) the frequency of the source is decreased ? Justify your answers (assume other factors remain unchanged).

(ii) An ac voltage

V = 280\sin(100\pi t)

volt is connected across a series LCR circuit in which

R = 400\ \Omega

L = \dfrac{5}{\pi}

H and

C = \dfrac{50}{\pi}\ \mu\text{F}

. Taking

\sqrt{2} = 1.4

, calculate : (I) the impedance of the circuit, (II) the rms value of the current that flows in the circuit, (III) the power factor of the circuit.

(OR)

(b)(i) State Lenz's law and explain that it follows the law of conservation of energy.

(ii) Write the dimensional formula for self-inductance. The current in a coil changes from

8.0

A to

2.0

A in

0.6

s. If the average emf induced in the coil is

50

V, calculate the self-inductance of the coil.

CBSECBSE Class XII Board 2026Subjective· 5mImportance★★★★★

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Part (a)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 | …

Part (b)Concept understanding — Electromagnetic Induction

Electromagnetic Induction

Electromagnetic induction is the phenomenon in which a changing magnetic flux through a circuit produces an electromotive force (emf) — and hence a current, if the circuit is closed. It is the single idea behind generators, transformers, inductors, and the entire AC power grid.

The Central Discovery

Michael Faraday found (1831) that a current is induced in a coil not when a magnet sits still near it, but only while the magnet moves — that is, only while the magnetic flux linked with the coil is changing. A steady magnet, however strong, induces nothing.

Magnetic Flux

The key quantity is magnetic flux $\Phi_B$ through a surface of area $A$ in a field $\vec{B}$ :

\Phi_B = \vec{B} \cdot \vec{A} = BA\cos\theta

where $\theta$ is the angle between $\vec{B}$ and the area's normal. Its SI unit is the weber (Wb), where $1\ \text{Wb} = 1\ \text{T}\cdot\text{m}^2$ .

Flux can change in three distinct ways, and any of them induces an emf:

the field strength $B$ changes,
the area $A$ of the loop changes,
the orientation $\theta$ changes (a coil rotating in a field — the basis of the generator).

Faraday's Law

The induced emf equals the negative rate of change of flux. For a coil of $N$ turns:

\mathcal{E} = -N\,\frac{d\Phi_B}{dt}

The faster the flux changes, the larger the emf. This is why a magnet dropped quickly through a coil gives a bigger deflection than one moved slowly.

Lenz's Law — the Minus Sign

The negative sign expresses Lenz's law: the induced current flows in the direction that opposes the change producing it. Push a magnet's north pole toward a coil, and the coil's near face becomes a north pole to repel it; pull it away, and the face becomes a south pole to attract it. This is simply energy conservation — you must do work against this opposition, and that work becomes the electrical energy of the induced current.

Motional emf

A special, very useful case: a conducting rod of length $l$ moving with speed $v$ perpendicular to a field $B$ sweeps out area and develops an emf

\mathcal{E} = Blv

Here the emf arises because the free charges in the rod experience a magnetic force $q\vec{v}\times\vec{B}$ , which drives them along the rod. …

Why this formula?

Electromagnetic Induction

Electromagnetic induction is the effect discovered by Faraday: a changing magnetic flux through a circuit drives an induced EMF (and hence a current). The key word is changing — a steady field, however strong, induces nothing.

Magnetic flux

Flux measures how many field lines thread a surface bounded by the loop:

\Phi_B = \int \vec{B} \cdot d\vec{A} = BA\cos\theta

It can change three ways: by changing $B$ , by changing the area $A$ , or by rotating the loop (changing $\theta$ ).

Faraday's law

The induced EMF equals the rate of change of flux:

\mathcal{E} = -\frac{d\Phi_B}{dt}

For a coil of $N$ turns, $\mathcal{E} = -N\,\dfrac{d\Phi_B}{dt}$ . The EMF depends on how fast the flux changes, not on the flux itself — a slow change gives a small EMF, a rapid change a large one.

Lenz's law — the minus sign …

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Part (a): iron bar dims bulb, lower frequency brightens it; $Z=500\ \Omega$ , $I_{rms}=0.4\ \text{A}$ , power factor $=0.8$ .

Part (b): Lenz's law opposes flux change (energy conservation); $[L]=\mathrm{M\,L^2\,T^{-2}\,A^{-2}}$ ; $L=5\ \text{H}$ .

Part (a)

(i) The bulb and coil are in series, so the current (and brightness) is set by the impedance $Z=\sqrt{R^2+X_L^2}$ .

(I) Iron bar inserted: the inductance $L$ increases sharply, so $X_L=\omega L$ increases, $Z$ increases, the current falls and the bulb glows dimmer.
(II) Frequency decreased: $X_L=\omega L$ decreases, $Z$ decreases, current rises and the bulb glows brighter.

(ii) $V=280\sin(100\pi t)$ , so $\omega=100\pi\ \text{rad s}^{-1}$ , $V_0=280$ V.

X_L=\omega L=100\pi\times\frac{5}{\pi}=500\ \Omega,\qquad X_C=\frac{1}{\omega C}=\frac{1}{100\pi\times\frac{50}{\pi}\times10^{-6}}=\frac{1}{5\times10^{-3}}=200\ \Omega.

\textbf{(I)}\quad Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{400^2+(500-200)^2}=\sqrt{160000+90000}=500\ \Omega.

$\textbf{(II)}\quad V_{rms}=\frac{V_0}{\sqrt2}=\frac{280}{1.4}=200\ \text{V},\qquad I_{rms}=\frac{V_{rms}}{Z}=\frac{200}{500}=0.4\ \text{A}.$ …

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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 67 on this concept.

CBSE 2026Set 55/2/11 markMCQ
$Q. A magnet held vertically, with its north pole down, is dropped along the axis of a closed solenoid placed vertically on a table. If the observer looks down from the top, (A) the induced current will flow in the anticlockwise direction. (B) the induced current will flow in the clockwise direction. (C) no induced current will flow in the solenoid. (D) the magnet will fall with a constant velocity.$
›Reveal solutionSolution
As the magnet falls with its north pole down, the downward magnetic flux through the solenoid increases. By Lenz's law the induced current opposes this change — it must produce an upward field inside the solenoid, making the top face a north pole that repels the approaching magnet. That requires an anticlockwise current as seen from above. The correct option is (A).

Why this approach works

Electromagnetic induction is about change: a current is induced in the solenoid only because the flux through it is changing as the magnet falls. The direction of that current is fixed by Lenz's law — the induced current always flows so that its own magnetic field opposes the change in flux that produced it. This is not an arbitrary rule; it is energy conservation. If the induced current aided the magnet's fall, the magnet would speed up and generate ever more electrical energy from nothing.

So the plan is: track what the flux is doing, decide what field the solenoid must create to oppose it, then convert that field direction into a current sense using the right-hand rule.

Step-by-step reasoning

Set up the situation.

The solenoid stands vertically on the table. The magnet is dropped along its axis from above, north pole downward. The observer looks down from the top.

What is the flux doing?

Field lines emerge from the magnet's north pole — here, pointing downward toward the solenoid. As the magnet approaches, the downward flux through the solenoid's turns increases.

What must the induced current do?

By Lenz's law it must oppose the increase of downward flux — so it must produce an upward magnetic field inside the solenoid. Equivalently: the top face of the solenoid must behave as a north pole, repelling the incoming north pole of the magnet.

Convert the field direction into a current sense.

Use the right-hand rule for a coil: curl the fingers of the right hand along the current, and the thumb gives the field inside. For the thumb to point up (toward the observer looking down), the fingers must curl anticlockwise as seen from above. So the induced current is anticlockwise for the top observer.

Tip

Quick pole check: the solenoid must repel the approaching north pole, so its top face is a north pole. Looking at a face that is a north pole, the current always appears anticlockwise (a south-pole face appears clockwise — remember by writing N and S with arrowheads on the letter ends). Same conclusion. …
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CBSE 2026Set 55/3/11 markMCQ
$50$
›Reveal solutionSolution
The induced emf is found from Faraday’s law: the change in magnetic flux divided by the time taken. The flux changes from maximum to zero as the loop rotates by $90^\circ$ , giving an average emf of $2.5$ V — but the options are in the range $0.25$ – $1.0$ V, so we must check the calculation carefully. The correct value is $2.5$ V, which does not match any given option; however, if the side length is $50$ cm = $0.5$ m, area $= 0.25$ m², flux change $= 3.0 \times 0.25 = 0.75$ Wb, time $= 0.3$ s, emf $= 0.75/0.3 = 2.5$ V. None of the options are correct as stated.

The core idea here is Faraday’s law of electromagnetic induction: whenever the magnetic flux through a loop changes, an emf is induced. The flux depends on three things — the field strength $B$ , the area $A$ of the loop, and the angle $\theta$ between the field and the normal to the loop. When you rotate the loop, you change $\theta$ , and that changes the flux. The induced emf is the rate of change of flux.

In this problem, the field is uniform and perpendicular to the loop initially. That means the initial angle between the field and the normal is $0^\circ$ , so the flux is maximum. After a $90^\circ$ rotation, the plane of the loop is parallel to the field, so the flux becomes zero. The change in flux is simply the initial flux minus zero.

Let’s work it out step by step.

Find the area of the loop.

Side length $= 50$ cm $= 0.5$ m.

Area $A = (0.5)^2 = 0.25$ m².

Initial magnetic flux.

Flux $\Phi = B A \cos\theta$ .

Initially $\theta = 0^\circ$ , so $\cos 0 = 1$ .

$\Phi_i = 3.0 \times 0.25 \times 1 = 0.75$ Wb.

Final magnetic flux.

After $90^\circ$ rotation, $\theta = 90^\circ$ , $\cos 90 = 0$ .

$\Phi_f = 3.0 \times 0.25 \times 0 = 0$ Wb.

Change in flux.

$\Delta\Phi = \Phi_f - \Phi_i = 0 - 0.75 = -0.75$ Wb.

The magnitude of the change is $0.75$ Wb.

Average induced emf.

By Faraday’s law, $|\mathcal{E}| = \left|\frac{\Delta\Phi}{\Delta t}\right|$ .

$\Delta t = 0.3$ s.

$|\mathcal{E}| = \frac{0.75}{0.3} = 2.5$ V.

Watch out

A common mistake is to forget that the side is given in cm and not convert to metres. If you use $50$ cm as $50$ m, you get an absurdly large area and emf. Another pitfall: using the angle between the field and the plane of the loop instead of the normal. Here, the field is perpendicular to the plane initially, so the normal is parallel to the field — that’s $\theta = 0^\circ$ , not $90^\circ$ . …
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CBSE 2026Set V11 markMCQ
$Q. The working principle of an A.C. generator is :$
(a) mutual induction
(b) eddy currents
(c) self induction
(d) electromagnetic induction
›Reveal solutionSolution

(d) electromagnetic induction …

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CBSE 2026Set A1 markMCQ
$Q. An example of natural electromagnetic induction is (A) radio (B) television (C) battery charging (D) lightning strike$
›Reveal solutionSolution

Lightning involves huge, rapidly changing currents/fields that induce emf in nearby conductors — natural electromagnetic induction.

Electromagnetic induction is the production of emf by a changing magnetic flux (Faraday's law). A lightning strike carries an enormous, rapidly varying current, producing a fast-changing magnetic field that induces emf/current in nearby loops and conductors — a natu …

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CBSE 2026Set A1 markMCQ
$Q. If magnetic field is same but the area of the loop is increased, then the flux (A) increases (B) decreases (C) becomes zero (D) remains unchanged$
›Reveal solutionSolution

Magnetic flux Φ = BA cosθ; with B constant, a larger area gives greater flux.

The magnetic flux through a loop is:

$\Phi=BA\cos\theta$

…

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CBSE 2026Set ANNUAL1 mark
$Q. What is electromagnetic induction?$
›Reveal solutionSolution

Any change of magnetic flux through a circuit produces an EMF in that circuit - this is electromagnetic induction.

Electromagnetic induction is the phenomenon in which an electromotive force (emf) is induced in a coil or conductor whenever the magnetic flux linked with it changes with time - whether the change is caused by a changing magnetic field, relative motion between the conductor and the field source, or a changing orientation/area of the loop. If the circuit is closed, this induced emf drives an induced current. It is quantitatively described by Faraday's law, EMF = -d(phi)/dt …

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CBSE 2026Set ANNUAL1 mark
$Q. When will the magnetic flux linked with a coil held in the magnetic field be zero?$
›Reveal solutionSolution

Flux is zero whenever the field lines lie entirely in the plane of the coil.

Magnetic flux linked with a coil is $\Phi = BA\cos\theta$ , where $\theta$ is the angle between the coil's area vector (normal) and the magnetic field $\vec{B}$ . This is zero when $\cos\theta=0$ , i.e. $\theta=90°$ — meaning the normal to the coil is perpendicular to $\vec{B}$ , which is the same as saying the field lines lie entirely within (parallel to) the plane of the coil, pa …

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CBSE 2026Set ANNUAL1 mark
$Q. Match the Column A with Column B and write the correct pair. Column A: Magnetic flux. Column B:$
(i) μ₀nI,
(ii) Volt × second,
(iii) μ₀nI/2,
(iv) Volt × meter,
(v) Volt × meter⁻¹,
(vi) Volt.
›Reveal solutionSolution

Magnetic flux unit = weber = Volt × second (from Faraday's law emf = dΦ/dt), option (ii).

Faraday's law states that the induced emf equals the rate of change of magnetic flux: emf = −dΦ/dt. Rearranging, Φ = emf × time (dimensionally). Since emf is in volts and time in seconds, magneti …

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CBSE 2025Set X11 markMCQ
$Q. Consider the following statements : Statement - 1: A.C. Generator works on the principle of electromagnetic induction Statement - 2: In an A.C. Generator, as the armature is rotated in a uniform magnetic field, the magnetic flux linked with the coil changes which induces an emf in the coil. Among the above two statements :$
(a) Both Statements are true
(b) Both Statements are false
(c) Statement-1 is true and Statement-2 is false
(d) Statement-1 is false and Statement-2 is true
›Reveal solutionSolution

(a) Both Statements are true. An A.C. generator works on electromagnetic induction (Statement 1). As the armature coil rotates in a uniform magnetic field, the flux $\phi = NBA\cos\omega t$ linked …

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CBSE 2025Set D1 markMCQ
$Q. Which of the following devices is based on the principle of electromagnetic induction? (A) Voltmeter (B) Electric motor (C) Electric generator (D) Ammeter$
›Reveal solutionSolution

The electric generator is based on electromagnetic induction.

An electric generator rotates a coil in a magnetic field. The continuous change of magnetic flux linked with the coil induces an emf by Faraday's law of electromagnetic induction, converting mechanical energy into electrical energy.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The laws of electromagnetic induction have been used in the construction of :$
(a) voltmeter
(b) ammeter
(c) electric motor
(d) generator
›Reveal solutionSolution

An electric generator converts mechanical energy into electrical energy by rotating a coil in a magnetic field, directly applying Faraday's law of electromagnetic induction.

Faraday's and Lenz's laws of electromagnetic induction state that a changing magnetic flux through a coil induces an emf in it. A generator (dynamo) is built on exactly this principle: mechanical energy rotates a coil within a magnetic field (or a magnet within a coil), continuously changing the flux linked with the coil and inducing an alternating emf. A voltmeter and ammeter are measuring instruments based on the magnetic effect of current, and an el …

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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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Part (a)

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