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Topic 12 of 14

Periodic EMI and the AC generator

Spin a coil in a magnetic field and its flux rises and falls forever, giving an alternating emf. This file connects the turning coil to its sine-wave output and to the parts of a real generator.

NCERTAllen module pages 130–132Periodic EMI and AC generator

A coil turning in a field

Flux and emf of a coil rotating at constant speedLeft: the coil seen side-on turns steadily in a horizontal field. Right: as it turns, the flux follows a cosine curve and the emf a sine curve. When the flux is largest the emf is zero, and when the flux is zero the emf is largest.coil seen edge-on; green = its normalφeφ = NBA cos ωte = NBAω sin ωtone full turn = one full cycle
Periodic EMI. A steadily turning coil sees its flux rise and fall smoothly. The emf follows the slope of the flux, so it is a quarter-cycle ahead: zero when the flux peaks, largest when the flux passes through zero.
Picture it

Watch the shadow of a Ferris wheel seat on the ground. It slides smoothly from one side to the other and back, fastest in the middle and pausing at the ends. The flux through a turning coil behaves like that shadow's position, and the emf behaves like the shadow's speed: fastest (largest) exactly when the position is passing through the middle.

In exam language

A coil of N turns and area A turns at constant ω in a uniform field B, starting with its plane perpendicular to B (θ = ωt):

φ = NBA cos ωt = φ₀ cos ωt, φ₀ = NBA e = −dφ/dt = NBAω sin ωt = e₀ sin ωt, e₀ = NBAω = φ₀ω i = e/R = I₀ sin ωt, I₀ = NBAω/R
ωtPlane of coilFlux φEmf e
0Perpendicular to B+NBA (max)0
π/2Parallel to B0+NBAω (max)
πPerpendicular to B−NBA0
3π/2Parallel to B0−NBAω
2πPerpendicular to B+NBA0
Trap

Largest flux does not mean largest emf. They are 90° out of step: φ is largest when the plane faces the field, e is largest when the plane lies along the field.

Frequency and speed

Doubling the rotation speedDoubling omega doubles the peak emf and also doubles the frequency: twice as tall and twice as many cycles in the same time.etspeed ω: peak e₀, 1 cyclespeed 2ω: peak 2e₀, 2 cycles
Twice as fast. Turning the coil twice as fast makes the flux change twice as quickly (double the peak emf) and also repeat twice as often (double the frequency).

The emf completes one cycle for every turn of the coil, so its frequency equals the rotation frequency: f = ω/2π. At 3000 rpm the coil makes 50 turns per second, so the output is 50 Hz.

ChangePeak emf e₀ = NBAωFrequency
ω doubled× 2× 2
N doubled× 2unchanged
B or A doubled× 2unchanged
N doubled, ω halvedunchangedhalved
Trap

ω is the angular speed in rad/s. The frequency of the emf is f = ω/2π, the number of turns per second. Mixing the two changes the answer by 2π.

The AC generator

Parts of an AC generatorA rectangular coil, the armature, turns between the N and S poles of a magnet. Its ends connect to two slip rings that turn with it. Fixed carbon brushes press on the rings and carry the alternating current to a bulb, which brightens twice in each turn.NSarmature coilslip ringscarbon brushes (fixed)bulbmechanical → electricalbright twice per turn
AC generator. Mechanical work turns the armature, periodic EMI produces an alternating emf, and slip rings with brushes pass it to the outside circuit without twisting the wires.
PartWhat it does
Field magnetProvides the steady magnetic field B
ArmatureThe coil (usually on a soft iron core) in which the emf is induced as it turns
Slip ringsTwo brass rings fixed to the coil's ends; they turn with the coil
Carbon brushesStay still and press on the rings, carrying current to the load

Energy conversion: mechanical energy (turning the shaft against the magnetic braking of the induced current) becomes electrical energy.

Losses in a generator (Allen p. 132)

LossCauseReduced by
Copper lossI²R heating in the armature windingThicker, low-resistance wire
Flux leakageSome flux does not link the armatureGood design of poles and core
Iron loss: hysteresisRepeated magnetising of the coreSoft iron with a narrow hysteresis loop
Iron loss: eddy currentsCurrents induced in the solid coreLaminated cores (Topic 13)
Mechanical lossFriction in bearings and brushesLubrication
Correction to the Allen module

The list of armature functions on page 131 starts with "transport current across the field, resulting in shaft torque". That describes a motor. In a generator the armature is where the emf is induced; the torque on it is the magnetic braking that the turning force must overcome.

Beyond these pages: average emf

Over the half turn from the plane facing the field (φ = +NBA) to facing it the other way (φ = −NBA), the average emf is 2NBA ÷ (T/2) = 2e₀/π ≈ 0.64 e₀. Over a full turn the average is zero. This is not on these Allen pages but appears in NEET questions.

Formula sheet

FormulaMeaningWhen to useWatch out
φ = NBA cos ωtFlux through the rotating coilStarting with the plane ⊥ BMax NBA
e = NBAω sin ωtInduced emfAny instantMax when the plane ∥ B
e₀ = NBAω = φ₀ωPeak emfGenerator amplitudeω in rad/s
I₀ = NBAω/RPeak currentLoad resistance RInclude coil resistance if given
f = ω/2πOutput frequencyGiven rpm or rev/sEquals rotation frequency
eavg = 2e₀/πAverage over half a turnFrom plane ⊥ B to plane ⊥ BZero over a full turn
Phase difference 90°Between φ and eGraph and phase questionsφ max ↔ e = 0

Allen pages 130–132 checked

ItemCheck
φ = NBA cos ωt, e = NBAω sin ωt, i = (NBAω/R) sin ωtCorrect
90° phase difference; φ max when plane ⊥ B, e max when plane ∥ BCorrect
Table of φ and e at ωt = 0, π/2, π, 3π/2, 2πCorrect values (some labels loosely worded)
Frequency of emf = frequency of rotationCorrect
Generator parts: armature, slip rings, brushesCorrect
Armature function (1): shaft torqueDescribes a motor, not a generator
Losses: copper, flux leakage, iron (hysteresis, eddy), mechanicalCorrect

NEET practice: 29 questions

Watch units of rotation (rad/s, rev/s, rpm, Hz) and the difference between the peak, the average and the rms value. Every number was recalculated in Python.

Q1Numerical
A 100-turn coil of area 0.05 m² rotates at 50 rad/s in a uniform 0.2 T field, about an axis perpendicular to the field. The peak emf is
  1. (A)50 V
  2. (B)5 V
  3. (C)500 V
  4. (D)25 V
Show the solution
Given
N = 100, A = 0.05 m², ω = 50 rad/s, B = 0.2 T
Asked
e₀
Concept
Peak emf of a rotating coil.
Formula
e₀ = NBAω
Baby steps
  1. 100 × 0.2 × 0.05 × 50 = 50 V.
Answer
(A) 50 V
Why not the others
5 V and 500 V shift a power of ten. 25 V halves.
Shortcut
Multiply the four.
Where it went wrong
Arithmetic slip with 0.05.
Q2Numerical
A 50-turn coil of area 0.02 m² rotates at 600 rpm in a 0.5 T field, about an axis perpendicular to the field. The peak emf is about
  1. (A)5 V
  2. (B)300 V
  3. (C)31 V
  4. (D)63 V
Show the solution
Given
N = 50, A = 0.02 m², 600 rpm, B = 0.5 T
Asked
e₀
Concept
Convert rpm to rad/s.
Formula
e₀ = NBAω, ω = 2π × rpm/60
Baby steps
  1. 600 rpm = 10 rev/s → ω = 20π rad/s.
  2. e₀ = 50 × 0.5 × 0.02 × 20π = 10π ≈ 31 V.
Answer
(C) 31 V
Why not the others
300 V uses ω = 600. 5 V uses ω = 10. 63 V doubles.
Shortcut
rpm ÷ 60 × 2π.
Where it went wrong
Putting rpm in directly.
Q3Numerical
A 200-turn coil of area 0.1 m² rotates in a 0.05 T field to generate emf at 50 Hz. The peak emf is about
  1. (A)314 V
  2. (B)50 V
  3. (C)100 V
  4. (D)628 V
Show the solution
Given
N = 200, A = 0.1 m², B = 0.05 T, f = 50 Hz
Asked
e₀
Concept
ω = 2πf.
Formula
e₀ = NBA(2πf)
Baby steps
  1. NBA = 1 Wb.
  2. ω = 100π ≈ 314 rad/s.
  3. e₀ ≈ 314 V.
Answer
(A) 314 V
Why not the others
50 V uses f as ω. 100 V uses 2f. 628 V doubles.
Shortcut
NBA × 2πf.
Where it went wrong
Forgetting 2π.
Q4Numerical
A large coil of 500 turns and mean radius 10 m rotates at 2 rad/s about a horizontal diameter where the vertical component of the earth's field is 4 × 10⁻⁵ T. Its resistance is 12.56 Ω. The peak induced current is about
  1. (A)0.5 A
  2. (B)1 A
  3. (C)2 A
  4. (D)12.6 A
Show the solution
Given
N = 500, r = 10 m, ω = 2 rad/s, BV = 4 × 10⁻⁵ T, R = 12.56 Ω
Asked
I₀
Concept
Rotation about a horizontal axis makes the vertical field's flux change.
Formula
I₀ = NBAω/R
Baby steps
  1. A = π × 100 = 314 m².
  2. e₀ = 500 × 4 × 10⁻⁵ × 314 × 2 = 12.57 V.
  3. I₀ = 12.57/12.56 ≈ 1 A.
Answer
(B) 1 A
Why not the others
0.5 A forgets ω = 2. 2 A doubles. 12.6 A is the emf in volts.
Shortcut
e₀ ≈ 4π V; R = 4π Ω.
Where it went wrong
Using the horizontal component.
Q5Numerical
A generator coil has a peak emf of 40 V. Its plane starts perpendicular to the field. The instantaneous emf when it has turned through 30° is
  1. (A)40 V
  2. (B)34.6 V
  3. (C)20 V
  4. (D)zero
Show the solution
Given
e₀ = 40 V, ωt = 30°
Asked
e
Concept
e = e₀ sin ωt when starting from the plane ⊥ B.
Formula
e = e₀ sin ωt
Baby steps
  1. sin 30° = 0.5.
  2. e = 20 V.
Answer
(C) 20 V
Why not the others
34.6 V uses cos 30°. 40 V is the peak. Zero is the starting value.
Shortcut
Start ⊥ B → sine.
Where it went wrong
Using cos because the flux uses cos.
Q6Numerical
A coil rotating at 100π rad/s produces a peak emf of 100 V. The maximum flux linkage of the coil is about
  1. (A)3.14 Wb
  2. (B)0.32 Wb
  3. (C)1.0 Wb
  4. (D)0.016 Wb
Show the solution
Given
e₀ = 100 V, ω = 100π rad/s
Asked
φ₀ = NBA
Concept
e₀ = φ₀ω.
Formula
φ₀ = e₀/ω
Baby steps
  1. φ₀ = 100/(100π) = 1/π ≈ 0.32 Wb.
Answer
(B) 0.32 Wb
Why not the others
3.14 Wb multiplies by π. 1.0 Wb forgets π. 0.016 Wb divides by 2π again.
Shortcut
Peak emf ÷ ω.
Where it went wrong
Using f instead of ω.
Q7Concept
The emf of a coil rotating in a uniform field is at its maximum value when the plane of the coil is
  1. (A)at 45° to the field
  2. (B)perpendicular to the field
  3. (C)parallel to the field
  4. (D)at any angle, since the emf is constant
Show the solution
Given
Rotating coil
Asked
Orientation for e = e₀
Concept
Emf is largest when flux changes fastest, which is when flux passes through zero.
Formula
e = e₀ sin ωt
Baby steps
  1. Plane ∥ B → φ = 0.
  2. That is where φ changes fastest.
  3. e = e₀.
Answer
(C) parallel to the field
Why not the others
Perpendicular gives maximum flux but zero emf. 45° gives 0.71 e₀. The emf is not constant.
Shortcut
Zero flux ↔ maximum emf.
Where it went wrong
Confusing maximum flux with maximum emf.
Q8Numerical
A generator gives a peak emf of 60 V across a total resistance of 20 Ω. The peak current is
  1. (A)3 A
  2. (B)1200 A
  3. (C)0.33 A
  4. (D)40 A
Show the solution
Given
e₀ = 60 V, R = 20 Ω
Asked
I₀
Concept
Ohm's law at the peak.
Formula
I₀ = e₀/R
Baby steps
  1. 60/20 = 3 A.
Answer
(A) 3 A
Why not the others
1200 A multiplies. 0.33 A inverts. 40 A subtracts.
Shortcut
Peak over R.
Where it went wrong
Ratio reversal.
Q9Concept
The angular speed of a generator coil is doubled. The peak emf and the frequency of the output become
  1. (A)peak unchanged, frequency doubled
  2. (B)peak doubled, frequency unchanged
  3. (C)both doubled
  4. (D)both unchanged
Show the solution
Given
ω → 2ω
Asked
e₀ and f
Concept
e₀ = NBAω and f = ω/2π both contain ω.
Formula
—
Baby steps
  1. e₀ doubles.
  2. f doubles.
Answer
(C) both doubled
Why not the others
Each other option misses one of the two dependences.
Shortcut
Speed changes both height and rate.
Where it went wrong
Changing only the amplitude.
Q10Concept
A generator coil is rewound with twice the number of turns and turned at half the angular speed. The peak emf and frequency become
  1. (A)peak doubled, frequency halved
  2. (B)peak unchanged, frequency halved
  3. (C)peak halved, frequency unchanged
  4. (D)both unchanged
Show the solution
Given
N → 2N, ω → ω/2
Asked
e₀ and f
Concept
e₀ ∝ Nω, f ∝ ω.
Formula
e₀ = NBAω
Baby steps
  1. Nω unchanged → e₀ unchanged.
  2. ω halved → f halved.
Answer
(B) peak unchanged, frequency halved
Why not the others
The other options mishandle either Nω or f.
Shortcut
Track e₀ and f separately.
Where it went wrong
Assuming frequency depends on N.
Q11Numerical
A generator has a peak emf of 157 V. Its average emf over the half turn from the plane facing the field to the plane facing the field the other way is about
  1. (A)157 V
  2. (B)111 V
  3. (C)100 V
  4. (D)zero
Show the solution
Given
e₀ = 157 V, half turn between flux extremes
Asked
eavg
Concept
Average emf = total flux change ÷ time.
Formula
eavg = 2NBA/(T/2) = 2e₀/π
Baby steps
  1. 2e₀/π = 314/3.14 = 100 V.
Answer
(C) 100 V
Why not the others
111 V is e₀/√2 (the rms value). 157 V is the peak. Zero is the full-turn average.
Shortcut
2/π ≈ 0.64.
Where it went wrong
Using the rms value for the average.
Q12Numerical
A generator produces 50 Hz AC. The time from a moment of maximum flux through the coil to the next moment of maximum emf is
  1. (A)10 ms
  2. (B)5 ms
  3. (C)20 ms
  4. (D)2.5 ms
Show the solution
Given
f = 50 Hz
Asked
Time from φ max to e max
Concept
φ and e are a quarter cycle apart.
Formula
t = T/4
Baby steps
  1. T = 1/50 = 20 ms.
  2. T/4 = 5 ms.
Answer
(B) 5 ms
Why not the others
10 ms is half a period. 20 ms is a full period. 2.5 ms is an eighth.
Shortcut
90° = quarter period.
Where it went wrong
Using half a period.
Q13Numerical
A generator coil turns at 3000 rpm. The frequency of the alternating emf is
  1. (A)100 Hz
  2. (B)3000 Hz
  3. (C)314 Hz
  4. (D)50 Hz
Show the solution
Given
3000 rpm
Asked
f
Concept
Output frequency = rotation frequency.
Formula
f = rpm/60
Baby steps
  1. 3000/60 = 50 Hz.
Answer
(D) 50 Hz
Why not the others
3000 Hz uses rpm directly. 314 Hz is ω. 100 Hz doubles.
Shortcut
Revolutions per second.
Where it went wrong
Forgetting to divide by 60.
Q14Concept
At the instant the emf of a rotating coil is half its peak value, the flux through it (starting from the plane perpendicular to the field) is
  1. (A)equal to its maximum
  2. (B)half its maximum
  3. (C)zero
  4. (D)about 0.87 of its maximum
Show the solution
Given
e = e₀/2
Asked
φ/φ₀
Concept
e ∝ sin ωt, φ ∝ cos ωt.
Formula
sin²ωt + cos²ωt = 1
Baby steps
  1. sin ωt = ½ → ωt = 30°.
  2. cos 30° = 0.87.
Answer
(D) about 0.87 of its maximum
Why not the others
Half assumes φ and e move together. Zero and maximum are the quarter-cycle points.
Shortcut
Use the 30° angle.
Where it went wrong
Assuming φ and e are in phase.
Q15Concept
In an AC generator, the slip rings
  1. (A)let current pass from the rotating coil to the stationary circuit without twisting the wires
  2. (B)convert alternating current into direct current
  3. (C)provide the magnetic field
  4. (D)reduce eddy current losses
Show the solution
Given
Generator parts
Asked
Role of slip rings
Concept
They rotate with the coil and are touched by fixed brushes.
Formula
—
Baby steps
  1. The coil turns continuously.
  2. Fixed wires would twist.
  3. Rotating rings with sliding brushes solve this.
Answer
(A) let current pass from the rotating coil to the stationary circuit without twisting the wires
Why not the others
Converting AC to DC is a split-ring commutator's job. The field comes from the magnet. Lamination reduces eddy losses.
Shortcut
Rings: a rotating connection.
Where it went wrong
Confusing slip rings with a commutator.
Q16Concept
The brushes of an AC generator are usually made of
  1. (A)copper, and they rotate with the coil
  2. (B)carbon, and they rotate with the coil
  3. (C)iron, and they stay stationary
  4. (D)carbon, and they stay stationary
Show the solution
Given
Generator parts
Asked
Brushes
Concept
Brushes press on the rotating rings and connect to the external circuit.
Formula
—
Baby steps
  1. Carbon conducts and wears smoothly.
  2. The brushes are fixed; the rings rotate.
Answer
(D) carbon, and they stay stationary
Why not the others
Rotating brushes would defeat the purpose. Iron and copper are not the standard brush materials.
Shortcut
Rings rotate, brushes stay.
Where it went wrong
Mixing up which part turns.
Q17Concept
An AC generator works on the principle of
  1. (A)Joule heating
  2. (B)the magnetic effect of current
  3. (C)electromagnetic induction
  4. (D)the photoelectric effect
Show the solution
Given
Generator
Asked
Working principle
Concept
A rotating coil in a field has a periodically changing flux.
Formula
e = NBAω sin ωt
Baby steps
  1. Changing flux induces emf: periodic EMI.
Answer
(C) electromagnetic induction
Why not the others
Magnetic effect of current is the motor's principle. Joule heating and the photoelectric effect are unrelated.
Shortcut
Generator ↔ induction; motor ↔ force on current.
Where it went wrong
Swapping generator and motor.
Q18Concept
The phase difference between the flux through a rotating coil and the emf induced in it is
  1. (A)45°
  2. (B)0°
  3. (C)180°
  4. (D)90°
Show the solution
Given
φ = φ₀ cos ωt, e = e₀ sin ωt
Asked
Phase difference
Concept
Cosine and sine differ by a quarter cycle.
Formula
—
Baby steps
  1. cos ωt and sin ωt are 90° apart.
Answer
(D) 90°
Why not the others
0° would put the maxima together. 180° would make them opposite. 45° has no basis.
Shortcut
Quarter cycle.
Where it went wrong
Assuming they peak together.
Q19Concept
Iron losses in an AC generator consist of
  1. (A)I²R loss in the load
  2. (B)copper loss and flux leakage
  3. (C)friction and air resistance
  4. (D)hysteresis loss and eddy current loss in the core
Show the solution
Given
Generator losses
Asked
What iron loss includes
Concept
Iron loss happens in the soft iron core.
Formula
—
Baby steps
  1. Repeated magnetising → hysteresis loss.
  2. Induced currents in the core → eddy loss.
Answer
(D) hysteresis loss and eddy current loss in the core
Why not the others
Copper loss is in the winding. Friction is mechanical loss. The load's I²R is useful output.
Shortcut
Iron = hysteresis + eddy.
Where it went wrong
Listing copper loss as iron loss.
Q20Concept
The armature core of an AC generator is laminated in order to
  1. (A)increase the flux leakage
  2. (B)reduce eddy current losses
  3. (C)reduce the copper loss
  4. (D)increase the frequency
Show the solution
Given
Laminated core
Asked
Purpose
Concept
Thin insulated sheets break up large eddy current loops.
Formula
—
Baby steps
  1. Eddy currents need big closed paths in solid metal.
  2. Insulated laminations cut those paths.
Answer
(B) reduce eddy current losses
Why not the others
Flux leakage is not wanted. Copper loss is in the winding. Frequency depends on speed.
Shortcut
Lamination ↔ eddy currents.
Where it went wrong
Thinking lamination reduces winding resistance.
Q21Graph
The flux through a generator coil varies with time as shown. Which graph shows the induced emf (with sign)?
Graph of φ against tφt
  1. (A)Graph of e against tet
  2. (B)Graph of e against tet
  3. (C)Graph of e against tet
  4. (D)Graph of e against tet
Show the solution
Given
φ = φ₀ cos ωt
Asked
e(t)
Concept
e = −dφ/dt.
Formula
e = φ₀ω sin ωt
Baby steps
  1. d(cos ωt)/dt = −ω sin ωt.
  2. e = −dφ/dt = +φ₀ω sin ωt: starts at zero and rises.
Answer
(A) the graph in option A
Why not the others
The cosine copies φ. The −sin graph forgets the minus sign in e = −dφ/dt. The rectified curve never goes negative.
Shortcut
−d(cos)/dt = +sin.
Where it went wrong
Dropping one of the two minus signs.
Q22Graph
Which graph shows the peak emf of a generator coil against its angular speed?
  1. (A)Graph of e₀ against ωe₀ω
  2. (B)Graph of e₀ against ωe₀ω
  3. (C)Graph of e₀ against ωe₀ω
  4. (D)Graph of e₀ against ωe₀ω
Show the solution
Given
e₀ = NBAω
Asked
e₀ against ω
Concept
Linear.
Formula
e₀ ∝ ω
Baby steps
  1. Straight line through the origin.
Answer
(B) the graph in option B
Why not the others
The parabola squares ω. The flat line ignores ω. The root curve has no basis.
Shortcut
Direct proportion.
Where it went wrong
Squaring ω.
Q23Graph
The top curve is the emf of a generator turning at angular speed ω. Which lower curve shows the emf when it turns at 2ω, on the same time axis?
Graph of e against tet
  1. (A)Graph of e against tet
  2. (B)Graph of e against tet
  3. (C)Graph of e against tet
  4. (D)Graph of e against tet
Show the solution
Given
ω → 2ω
Asked
New e(t)
Concept
e₀ ∝ ω and f ∝ ω.
Formula
e = NBA(2ω) sin 2ωt
Baby steps
  1. Twice the height.
  2. Twice as many cycles.
Answer
(A) the graph in option A
Why not the others
Same height with two cycles changes only frequency. Double height with one cycle changes only amplitude. The last one halves the frequency.
Shortcut
Taller and faster.
Where it went wrong
Doubling only one of the two.
Q24Graph
Which graph shows the emf of a rotating coil against the angle θ turned from the position where its plane is perpendicular to the field, over one full turn?
  1. (A)Graph of e against θeθ
  2. (B)Graph of e against θeθ
  3. (C)Graph of e against θeθ
  4. (D)Graph of e against θeθ
Show the solution
Given
θ = ωt from plane ⊥ B
Asked
e against θ
Concept
e = e₀ sin θ.
Formula
e = e₀ sin θ
Baby steps
  1. 0 at θ = 0.
  2. +e₀ at 90°, 0 at 180°, −e₀ at 270°, 0 at 360°.
Answer
(D) the graph in option D
Why not the others
Cosine starts at the maximum (that is the flux). The triangle is not sinusoidal and never goes negative. A constant emf is a DC source.
Shortcut
Sine from the perpendicular start.
Where it went wrong
Drawing the flux graph.
Q25Assertion–reason
Assertion (A): The emf of a rotating coil is maximum when its plane is parallel to the magnetic field.
Reason (R): The flux through the coil is zero at that instant.
  1. (A)Both A and R are true, and R is the correct explanation of A.
  2. (B)Both A and R are true, but R is not the correct explanation of A.
  3. (C)A is true, but R is false.
  4. (D)A is false, but R is true.
Show the solution
Given
A: e max when plane ∥ B. R: φ = 0 then.
Asked
Truth and link
Concept
Emf depends on the rate of change of flux, not its value.
Formula
e = −dφ/dt
Baby steps
  1. A is true.
  2. R is true.
  3. But zero flux is not the reason; the reason is that φ changes fastest there. R does not explain A.
Answer
(B) Both A and R are true, but R is not the correct explanation of A.
Why not the others
(A) treats zero flux as the cause. (C) and (D) need a false statement.
Shortcut
Value versus rate.
Where it went wrong
Thinking zero flux causes maximum emf.
Q26Assertion–reason
Assertion (A): The frequency of the emf from an AC generator equals the rotation frequency of its coil.
Reason (R): The flux through the coil goes through one complete cycle in each rotation.
  1. (A)Both A and R are true, and R is the correct explanation of A.
  2. (B)Both A and R are true, but R is not the correct explanation of A.
  3. (C)A is true, but R is false.
  4. (D)A is false, but R is true.
Show the solution
Given
A: f(emf) = f(rotation). R: one flux cycle per turn.
Asked
Truth and link
Concept
e follows φ.
Formula
f = ω/2π
Baby steps
  1. A is true.
  2. R is true.
  3. One flux cycle per turn → one emf cycle per turn. R explains A.
Answer
(A) Both A and R are true, and R is the correct explanation of A.
Why not the others
(B) denies a direct link. (C) and (D) need a false statement.
Shortcut
Cycle for cycle.
Where it went wrong
Choosing (B) by habit.
Q27Assertion–reason
Assertion (A): The slip rings of an AC generator convert alternating current into direct current.
Reason (R): The slip rings rotate along with the armature coil.
  1. (A)Both A and R are true, and R is the correct explanation of A.
  2. (B)Both A and R are true, but R is not the correct explanation of A.
  3. (C)A is true, but R is false.
  4. (D)A is false, but R is true.
Show the solution
Given
A: slip rings rectify. R: rings rotate with coil.
Asked
Truth and link
Concept
Slip rings just connect; they do not rectify.
Formula
—
Baby steps
  1. R is true.
  2. A is false: the output stays AC.
  3. A false, R true.
Answer
(D) A is false, but R is true.
Why not the others
(A), (B) and (C) need A true.
Shortcut
Commutator rectifies; slip rings don't.
Where it went wrong
Confusing slip rings with a split-ring commutator.
Q28Assertion–reason
Assertion (A): Laminating the armature core of a generator reduces iron loss.
Reason (R): Lamination reduces the eddy currents induced in the core.
  1. (A)Both A and R are true, and R is the correct explanation of A.
  2. (B)Both A and R are true, but R is not the correct explanation of A.
  3. (C)A is true, but R is false.
  4. (D)A is false, but R is true.
Show the solution
Given
A: lamination reduces iron loss. R: it reduces eddy currents.
Asked
Truth and link
Concept
Eddy loss is part of iron loss.
Formula
—
Baby steps
  1. A is true.
  2. R is true.
  3. Lower eddy currents → lower iron loss: R explains A.
Answer
(A) Both A and R are true, and R is the correct explanation of A.
Why not the others
(B) denies the direct link. (C) and (D) need a false statement.
Shortcut
Eddy loss ⊂ iron loss.
Where it went wrong
Choosing (B) out of caution.
Q29Two statements
Statement I: The flux through a rotating coil and the emf induced in it differ in phase by 90°.
Statement II: The flux through the coil is maximum at the instant the emf is maximum.
  1. (A)Both Statement I and Statement II are true.
  2. (B)Both Statement I and Statement II are false.
  3. (C)Statement I is true, but Statement II is false.
  4. (D)Statement I is false, but Statement II is true.
Show the solution
Given
Phase relation
Asked
Which are true
Concept
φ max ↔ e = 0.
Formula
φ ∝ cos, e ∝ sin
Baby steps
  1. Statement I is true.
  2. Statement II is false.
Answer
(C) Statement I is true, but Statement II is false.
Why not the others
Both-true fails on II. Both-false fails on I. I-false-II-true reverses them.
Shortcut
Quarter-cycle apart.
Where it went wrong
Assuming they peak together.

Answer key

1 A
2 C
3 A
4 B
5 C
6 B
7 C
8 A
9 C
10 B
11 C
12 B
13 D
14 D
15 A
16 D
17 C
18 D
19 D
20 B
21 A
22 B
23 A
24 D
25 B
26 A
27 D
28 A
29 C

Spread across letters: A 8, B 7, C 7, D 7. No letter repeats more than twice in a row. Question mix: Numerical 10, Concept 10, Graph 4, Assertion–reason 4, Two statements 1. Balancing seed 3000.