📖 TRANSFORMERS & DC MACHINES

A Complete Textbook Style Lecture Series | SET 1: Faraday's Laws & Transformer Fundamentals

FM ELECTRICALS | UPDA/MMUP Exam Preparation

📘 CHAPTER 1: ELECTROMAGNETIC INDUCTION

1.1 Introduction

The phenomenon of generating an electromotive force (EMF) in a conductor when it is exposed to a changing magnetic field is called Electromagnetic Induction. This fundamental principle was discovered by Michael Faraday in 1831 and forms the basis of all electrical machines — transformers, generators, and motors.

🔧 FIGURE 1.1: Basic Setup for Electromagnetic Induction

              ┌─────────────────┐
              │                 │
              │      N    S     │
              │      |    |     │
              │      |    |     │
              │    ┌─┴────┴─┐   │
              │    │  COIL  │   │
              │    └────────┘   │
              │                 │
              │   GALVANOMETER  │
              └─────────────────┘

Figure 1.1: When a magnet is moved in/out of a coil, the galvanometer shows deflection indicating induced current.

1.2 Faraday's First Law

Statement: "Whenever a conductor cuts magnetic flux, an electromotive force (EMF) is induced in that conductor."

The conductor and magnetic field must have relative motion.
If the conductor is part of a closed circuit, induced current flows.
EMF exists only as long as the relative motion continues.

        📌 Key Point: The magnitude of induced EMF depends on:
        Strength of the magnetic field (flux density B)
Speed of relative motion
Length of the conductor
Number of turns in the coil

      

1.3 Faraday's Second Law

Statement: "The magnitude of the induced EMF in a coil is directly proportional to the rate of change of flux linkage."

ε = - N × (dΦ / dt)

Where:

ε = Induced EMF (Volts)
N = Number of turns in the coil
dΦ/dt = Rate of change of flux (Weber/second)
Negative sign (−) = Lenz's Law — induced EMF opposes the change that caused it.

Solved Example 1.1

A coil of 500 turns has a magnetic flux of 0.02 Wb linking with it. If this flux is reduced to zero in 0.1 seconds, calculate the average induced EMF.

ε = N × (ΔΦ / Δt) = 500 × (0.02 / 0.1) = 500 × 0.2 = 100 Volts

1.4 Lenz's Law

Statement: "The direction of the induced EMF is always such that it opposes the cause producing it."

This is why the negative sign appears in Faraday's equation. Lenz's Law ensures conservation of energy — you cannot get energy for free.

🧲 FIGURE 1.2: Lenz's Law Direction

        North pole approaching coil:
        
              N ──→  [  coil  ]   Induced current creates North pole
                           ↑      to repel approaching magnet.
                           |
                        (opposes motion)

Figure 1.2: Induced current always opposes the relative motion.

📘 CHAPTER 2: TRANSFORMER — CONSTRUCTION & WORKING

2.1 What is a Transformer?

Definition: A transformer is a static (no moving parts) electrical device that transfers electrical energy from one circuit to another at the same frequency, but with voltage and current levels changed.

Step-up Transformer: Increases voltage (V_s > V_p)
Step-down Transformer: Decreases voltage (V_s < V_p)
Isolation Transformer: V_s = V_p (used for electrical isolation)

2.2 Construction of a Transformer

A transformer consists of two main parts: Magnetic Core and Windings.

2.2.1 Magnetic Core

Made of silicon steel laminations (0.3 mm to 0.5 mm thick)
Laminations are insulated from each other by varnish or oxide layer
Purpose: To provide a low reluctance path for magnetic flux and reduce eddy current losses

2.2.2 Windings (Coils)

Primary Winding: Connected to AC supply
Secondary Winding: Connected to load
Made of copper (sometimes aluminum)
HV winding: thinner wire, more turns
LV winding: thicker wire, fewer turns

📐 FIGURE 2.1: Core-Type Transformer Construction

        ┌─────────────────────────────────────────┐
        │                ┌───┐                    │
        │    ════════════│   │═══════════════    │  ← HV Winding
        │    ════════════│   │═══════════════    │     (outer)
        │                │   │                    │
        │    ────────────│   │───────────────    │  ← LV Winding
        │    ────────────│   │───────────────    │     (inner)
        │                └───┘                    │
        │         LAMINATED CORE                  │
        └─────────────────────────────────────────┘

Figure 2.1: Core-type transformer — windings surround the core limbs.

2.3 Working Principle

Transformer works on the principle of Mutual Induction between two windings placed on a common magnetic core.

Step-by-Step Working:

Step 1: AC voltage V_p applied to primary winding.
Step 2: Alternating current I_p flows in primary.
Step 3: This current produces an alternating magnetic flux Φ in the core.
Step 4: The alternating flux links with both primary and secondary windings.
Step 5: According to Faraday's Law, EMF is induced in both windings:
E_p = -N_p × (dΦ/dt) and E_s = -N_s × (dΦ/dt)
Step 6: If secondary is connected to a load, current I_s flows.

🔄 FIGURE 2.2: Transformer Working Diagram

        AC Supply ──┐      ┌─────────┐      ┌── Load
                   │      │         │      │
                  ║│      │   CORE  │      │║
                  ║│      │   ███   │      │║
                  ║│      │   ███   │      │║
                   │      │         │      │
                  ─┴─     └─────────┘     ─┴─
                 Primary                 Secondary
                  Np                       Ns

Figure 2.2: AC supply on primary creates alternating flux, inducing EMF in secondary.

2.4 Why Transformer Does Not Work on DC?

❌ DC supply on transformer:

DC produces constant flux (not alternating)
No change in flux → no induced EMF in secondary (dΦ/dt = 0)
Primary winding has only resistance (no back EMF)
Very high current flows → overheating → transformer burns/smokes

✅ Conclusion: Transformers are designed for AC only.

2.5 Transformer Classification

Basis of Classification	Types
Voltage Level	Step-up, Step-down, Isolation
Core Construction	Core-type, Shell-type
Number of Phases	Single-phase, Three-phase
Cooling Method	Oil-filled (ONAN, ONAF), Dry-type (AN)
Application	Power, Distribution, Instrument (CT, PT), Auto-transformer

2.6 Transformer Ratings

kVA Rating: Apparent power = V × I / 1000 (single phase)
Voltage Ratio: V_p / V_s
Frequency: 50 Hz or 60 Hz
Impedance (%Z): Internal voltage drop at full load
Temperature Rise: Maximum allowable temperature (e.g., 55°C, 65°C)

Single Phase: kVA = (V × I) / 1000
Three Phase: kVA = (√3 × V_L × I_L) / 1000

📘 CHAPTER 3: EMF EQUATION OF A TRANSFORMER

3.1 Derivation

Let:

N₁ = Number of turns in primary winding
N₂ = Number of turns in secondary winding
Φ_m = Maximum flux in core (Weber)
f = Frequency of AC supply (Hz)

The flux varies sinusoidally: Φ = Φ_m sin(ωt)

By Faraday's Law: E = -N × (dΦ/dt)

After derivation:

E_rms = 4.44 × f × N × Φ_m

Therefore, for primary and secondary:

E₁ = 4.44 f N₁ Φ_m
E₂ = 4.44 f N₂ Φ_m

3.2 EMF Equation in Terms of Flux Density

E = 4.44 × f × N × B_m × A

Where: B_m = Maximum flux density (Tesla), A = Core cross-sectional area (m²)

3.3 Voltage Transformation Ratio (Turn Ratio)

K = N₂/N₁ = E₂/E₁ = V₂/V₁ = I₁/I₂

Type	K	Voltage	Current
Step-up	K > 1	V₂ > V₁	I₂ < I₁
Step-down	K < 1	V₂ < V₁	I₂ > I₁
Isolation	K = 1	V₂ = V₁	I₂ = I₁

3.4 Solved Example

Problem 3.1: A 50 Hz, single-phase transformer has a core with a cross-sectional area of 200 cm² and a maximum flux density of 1.2 T. If the primary has 500 turns, calculate the primary induced EMF.

Solution:
Φ_m = B_m × A = 1.2 × (200 × 10^-4) = 0.024 Wb
E₁ = 4.44 × f × N₁ × Φ_m
E₁ = 4.44 × 50 × 500 × 0.024
E₁ = 4.44 × 50 × 12 = 4.44 × 600 = 2664 Volts

📘 CHAPTER 4: TRANSFORMER LOSSES & EFFICIENCY

4.1 Types of Losses in a Transformer

4.1.1 Core Loss (Iron Loss) — Constant

Hysteresis Loss: Due to repeated magnetization and demagnetization of core.
P_h = η × B_m^1.6 × f × V
Eddy Current Loss: Due to circulating currents induced in the core.
P_e = k_e × B_m² × f² × t² × V

4.1.2 Copper Loss — Variable

P_cu = I₁²R₁ + I₂²R₂

Copper loss is proportional to the square of the load current (P_cu ∝ I² ∝ kVA²).

4.1.3 Stray Losses

Due to leakage flux — usually 10-15% of total losses.

4.2 Efficiency of Transformer

η = (Output Power / Input Power) × 100%

η = [V_sI_s cosφ / (V_sI_s cosφ + P_core + P_cu)] × 100%

Maximum Efficiency Condition: When Copper Loss = Core Loss

        📊 Typical Efficiency: Large power transformers operate at 95% to 98.5% efficiency.
      

📘 CHAPTER 5: OPEN CIRCUIT & SHORT CIRCUIT TESTS

5.1 Open Circuit (OC) Test

Purpose: To measure Core Loss and no-load current (I₀)
Connection: LV side energized with rated voltage, HV side open
Reading: Wattmeter measures core loss (copper loss negligible)

5.2 Short Circuit (SC) Test

Purpose: To measure Copper Loss and impedance (%Z)
Connection: HV side energized with low voltage (5-10%), LV side shorted
Reading: Wattmeter measures copper loss (core loss negligible)

Test	Winding Energized	Applied Voltage	Measured Loss
OC Test	LV (rated voltage)	Rated voltage	Core Loss
SC Test	HV (reduced voltage)	5-10% of rated	Copper Loss

📘 CHAPTER 6: DC GENERATOR — CONSTRUCTION & WORKING

6.1 Introduction

A DC Generator is a rotating electrical machine that converts mechanical energy into direct current (DC) electrical energy. It works on the principle of Faraday's Law of Electromagnetic Induction.

🔧 FIGURE 6.1: Basic DC Generator Principle

              N ─────────────────── S
                 │              │
                 │    ┌────┐    │
                 │    │    │    │
                 │    │ COIL│    │
                 │    │    │    │
                 │    └────┘    │
                 │              │
                 └───SLIP RINGS─┘
                       (AC Output)

Figure 6.1: When coil rotates in magnetic field, AC voltage is induced.

6.2 Construction of DC Generator

A DC generator consists of the following main parts:

📐 FIGURE 6.2: Cross-Section of a DC Generator

                    ┌─────────────────────┐
                    │       YOKE          │
                    │   ┌───────────┐     │
                    │   │  FIELD    │     │
                    │   │  POLES    │     │
                    │   │  ┌─────┐  │     │
                    │   │  │ARMA-│  │     │
                    │   │  │TURE │  │     │
                    │   │  └─────┘  │     │
                    │   │           │     │
                    │   └───────────┘     │
                    │    COMMUTATOR       │
                    │       │             │
                    │    BRUSHES          │
                    └─────────────────────┘

Figure 6.2: Main parts of a DC generator.

6.2.1 Yoke (Frame)

Material: Cast iron (for small machines) or cast steel (for large machines)
Function: Provides mechanical support for poles, protects internal parts, carries magnetic flux

6.2.2 Field Poles

Material: Cast steel laminations
Function: Produce magnetic flux. Consists of pole core and pole shoe
Field winding: Copper wire wound around pole core

6.2.3 Armature

Parts: Armature core + armature winding + commutator
Armature Core: Made of silicon steel laminations to reduce eddy current loss
Armature Winding: Copper conductors placed in slots
Function: Where EMF is induced

6.2.4 Commutator

Material: Copper segments insulated from each other by mica
Function: Converts alternating EMF (AC) into direct current (DC) — acts as a mechanical rectifier

6.2.5 Brushes

Material: Carbon or graphite
Function: Collect current from commutator and deliver to external load

6.3 Working Principle of DC Generator

🔄 FIGURE 6.3: EMF Generation in a Single Coil

        Position 1 (0°):        Position 2 (90°):        Position 3 (180°):
        ┌────┐                  ┌────┐                  ┌────┐
        │    │                  │    │                  │    │
      N ┼────┼ S              N ┼────┼ S              N ┼────┼ S
        │    │                  │    │                  │    │
        └────┘                  └────┘                  └────┘
        EMF = 0                 EMF = Max               EMF = 0

Figure 6.3: As coil rotates, induced EMF varies sinusoidally.

Step-by-Step Working:

Step 1: Field winding is energized with DC → produces constant magnetic flux.
Step 2: Armature is rotated by a prime mover (turbine, engine).
Step 3: Armature conductors cut the magnetic field.
Step 4: According to Faraday's Law, EMF is induced in the armature conductors.
Step 5: The induced EMF is alternating (AC) in nature.
Step 6: Commutator converts this AC into DC.
Step 7: Brushes collect this DC and deliver to the external load.

      📌 Key Point: The EMF induced in the armature is AC. The commutator acts as a mechanical rectifier to convert it to DC.
    

📘 CHAPTER 7: EMF EQUATION OF DC GENERATOR

7.1 Derivation

Let:

P = Number of poles
φ = Flux per pole (Weber)
N = Speed of armature (RPM)
Z = Total number of armature conductors
A = Number of parallel paths in armature

Flux cut by one conductor in one revolution = P × φ

Time for one revolution = 60/N seconds

Average EMF per conductor = (P × φ) / (60/N) = (P × φ × N) / 60

Total EMF = (EMF per conductor) × (Number of conductors in series per path)

Number of conductors in series per path = Z / A

E = (P × φ × N × Z) / (60 × A) (Volts)

7.2 For Different Windings

Lap Winding: A = P
E = (φ × N × Z) / 60
Wave Winding: A = 2
E = (P × φ × N × Z) / (120)

Solved Example 7.1

Problem: A 4-pole DC generator has a lap-wound armature with 500 conductors. The flux per pole is 0.02 Wb. Calculate the generated EMF when running at 1500 RPM.

Solution:
For lap winding: A = P = 4
E = (P × φ × N × Z) / (60 × A)
E = (4 × 0.02 × 1500 × 500) / (60 × 4)
E = (4 × 0.02 × 1500 × 500) / 240
E = (60,000) / 240 = 250 Volts

📘 CHAPTER 8: CLASSIFICATION OF DC GENERATORS

8.1 Types Based on Field Excitation

8.1.1 Separately Excited DC Generator

Field winding is energized from a separate external DC source
Output voltage depends only on speed and field current
Used where precise voltage control is required

🔌 FIGURE 8.1: Separately Excited DC Generator

        External DC ──┐
        Source        │
                     ║│
                     ║│  Field Winding
                     ║│
                      │
        Armature ─────┼───── Load
        (rotated)     │
                      └─────

8.1.2 Self-Excited DC Generator

Field winding is energized from the generator's own armature. Further classified into three types:

Type	Field Connection	Characteristics	Applications
Series Generator	Field in series with armature	Output voltage increases with load current	Boosters, series arc lighting
Shunt Generator	Field parallel (shunt) with armature	Nearly constant voltage	General lighting, battery charging
Compound Generator	Both series + shunt field	Good voltage regulation	Industrial power supply

8.1.3 Compound Generator Sub-Types

Cumulative Compound: Series field aids shunt field → voltage rises with load
Differential Compound: Series field opposes shunt field → voltage drops sharply
Over Compound: Full-load voltage > no-load voltage
Flat Compound: Full-load voltage = no-load voltage
Under Compound: Full-load voltage < no-load voltage

8.2 Voltage Build-Up in Shunt Generator

Conditions for voltage build-up:

There must be residual magnetism in the poles
Field winding should be connected such that flux aids residual flux
Field circuit resistance should be less than critical resistance
Speed should be above critical speed

📈 FIGURE 8.2: Voltage Build-Up Curve

        Voltage
          ↑
        V₀ ┤─────────────────
          │                ╱
          │              ╱
          │            ╱
          │          ╱
          │        ╱
          │      ╱
          │    ╱
          │  ╱
          │╱
          └──────────────────→ Time

Figure 8.2: Shunt generator voltage builds up gradually.

📘 CHAPTER 9: CHARACTERISTICS OF DC GENERATORS

9.1 Open Circuit Characteristic (OCC) or Magnetization Curve

Shows relationship between field current (I_f) and no-load voltage (E₀) at constant speed.

        E₀ (Volts)
          ↑
        V₀ ┤─────────────────
          │                ╱
          │              ╱
          │            ╱
          │          ╱
          │        ╱
          │      ╱
          │    ╱
          │  ╱
          │╱
          └──────────────────→ I_f (Amps)
                (Field Current)

Figure 9.1: Open Circuit Characteristic (Saturation Curve).

9.2 Internal (Total) Characteristic

Shows relationship between armature current (I_a) and generated EMF (E).

9.3 External Characteristic (Load Characteristic)

Shows relationship between load current (I_L) and terminal voltage (V_t). This is the most important characteristic for practical applications.

Generator Type	External Characteristic
Series Generator	Voltage rises with load (rises then falls at heavy load)
Shunt Generator	Voltage drops slightly with load
Compound Generator	Can be flat, rising, or dropping based on compounding

📘 CHAPTER 10: LOSSES & EFFICIENCY OF DC GENERATOR

10.1 Types of Losses

10.1.1 Copper Losses

Armature copper loss = I_a² × R_a
Field copper loss = I_f² × R_f

10.1.2 Iron Losses (Core Losses)

Hysteresis loss: P_h = η × B_m^1.6 × f × V
Eddy current loss: P_e = k_e × B_m² × f² × t² × V

10.1.3 Mechanical Losses

Friction losses (bearings, brushes)
Windage losses (air friction)

10.1.4 Stray Load Losses

Miscellaneous losses (usually 1% of output)

Total Losses = Copper Loss + Iron Loss + Mechanical Loss + Stray Loss

10.2 Efficiency

η = (Output Power / Input Power) × 100%

η = (V_t × I_L) / (V_t × I_L + Losses) × 100%

      📌 Maximum Efficiency Condition: Occurs when Variable Losses = Constant Losses

      i.e., Ia²Ra = Iron Loss + Mechanical Loss + Shunt Field Loss

📘 CHAPTER 11: APPLICATIONS OF DC GENERATORS

Generator Type	Applications
Separately Excited	Speed control systems, laboratory experiments
Series Generator	Boosters, arc lamps, series lighting
Shunt Generator	Battery charging, lighting, excitation of alternators
Cumulative Compound	Power supply for factories, lifts, elevators
Differential Compound	Welding generators (constant current)

📘 CHAPTER 12: KEY FORMULAS — QUICK REFERENCE

EMF Equation: E = (P × φ × N × Z) / (60 × A)

Lap Winding: A = P

Wave Winding: A = 2

Armature Current: I_a = I_L + I_sh (for shunt generator)

Terminal Voltage: V_t = E - I_aR_a (as generator)

Efficiency: η = Output / (Output + Losses) × 100%

📘 CHAPTER 13: DC MOTOR — INTRODUCTION & CONSTRUCTION

13.1 What is a DC Motor?

A DC Motor is a rotating electrical machine that converts direct current (DC) electrical energy into mechanical energy. It works on the principle that "when a current-carrying conductor is placed in a magnetic field, it experiences a mechanical force."

🔧 FIGURE 13.1: Basic DC Motor Principle

              N ─────────────────── S
                 │              │
                 │    ┌────┐    │
                 │    │    │    │
                 │    │ COIL│    │
                 │    │    │    │
                 │    │  ↓  │    │  ← Force (Motion)
                 │    └────┘    │
                 │      ↓       │
                 └───────────────┘

Figure 13.1: Current-carrying coil in magnetic field experiences torque.

13.2 Construction of DC Motor

The construction of a DC motor is identical to that of a DC generator. Same parts:

📐 FIGURE 13.2: DC Motor Cross-Section

                    ┌─────────────────────┐
                    │       YOKE          │
                    │   ┌───────────┐     │
                    │   │  FIELD    │     │
                    │   │  POLES    │     │
                    │   │  ┌─────┐  │     │
                    │   │  │ARMA-│  │     │
                    │   │  │TURE │  │     │
                    │   │  └─────┘  │     │
                    │   │           │     │
                    │   └───────────┘     │
                    │    COMMUTATOR       │
                    │       │             │
                    │    BRUSHES          │
                    └─────────────────────┘

Figure 13.2: Cross-section of a DC motor.

13.2.1 Main Parts Summary

Part	Material	Function
Yoke	Cast iron/steel	Provides mechanical support, carries flux
Field Poles	Cast steel laminations	Produces main magnetic flux
Field Winding	Copper wire	Carries field current
Armature Core	Silicon steel laminations	Holds armature conductors, reduces eddy currents
Armature Winding	Copper conductors	Carries armature current, where torque is produced
Commutator	Copper segments + mica insulation	Converts AC to DC (rectification), reverses current direction
Brushes	Carbon/Graphite	Collects current from commutator

📘 CHAPTER 14: WORKING PRINCIPLE OF DC MOTOR

14.1 Fleming's Left-Hand Rule (Motor Rule)

        📌 Fleming's Left Hand Rule:
        
              Force (Motion)
                  ↑
                  │
            ┌─────┴─────┐
            │  Thumb    │
            │           │
        ────┼───────────┼────→ Magnetic Field
            │  Index    │    (North to South)
            │    Finger │
            └───────────┘
                  │
                  ↓
              Current (Middle Finger)

Figure 14.1: Fleming's Left Hand Rule — used for motors.

Force on a conductor: F = B × I × L (Newtons)

Where:

B = Magnetic flux density (Tesla)
I = Current through conductor (Amperes)
L = Length of conductor (meters)

14.2 Step-by-Step Working

Step 1: DC supply is connected to the armature winding via brushes and commutator.
Step 2: Field winding is energized (produces main magnetic flux).
Step 3: Armature conductors carry current.
Step 4: Current-carrying conductors experience force (F = BIL).
Step 5: This force produces a torque that rotates the armature.
Step 6: Commutator reverses current direction every half-cycle to keep rotation continuous.

🔄 FIGURE 14.2: Continuous Rotation in DC Motor

        Position 1:          Position 2:          Position 3:
        ┌────┐              ┌────┐              ┌────┐
        │  ← │              │    │              │ →  │
      N ┼────┼ S          N ┼────┼ S          N ┼────┼ S
        │  → │              │    │              │ ←  │
        └────┘              └────┘              └────┘
        Torque CW           No torque           Torque CW
        (Commutator switches current direction to maintain torque)

Figure 14.2: Commutator reverses current to maintain unidirectional torque.

      📌 Key Point: The commutator in a DC motor acts as a mechanical inverter — it reverses the armature current direction every half-cycle so that torque always acts in the same direction.
    

📘 CHAPTER 15: BACK EMF (COUNTER EMF) IN DC MOTOR

15.1 What is Back EMF?

When the armature of a DC motor rotates, it also cuts the magnetic field. According to Faraday's Law, an EMF is induced in the armature. This induced EMF opposes the applied voltage — hence called Back EMF (E_b) or Counter EMF.

E_b = (P × φ × N × Z) / (60 × A) (Same as generator EMF formula)

V = E_b + I_aR_a

Where:

V = Applied voltage
E_b = Back EMF
I_a = Armature current
R_a = Armature resistance

15.2 Significance of Back EMF

Acts as a governor — automatically regulates armature current
When load increases → speed decreases → E_b decreases → I_a increases → torque increases
When load decreases → speed increases → E_b increases → I_a decreases → torque decreases
Makes DC motor self-regulating

      📌 Important: Back EMF is always less than applied voltage (V > Eb). If Eb becomes equal to V, Ia = 0 and motor stops.
    

📘 CHAPTER 16: TORQUE EQUATION OF DC MOTOR

16.1 Derivation

Let:

P = Number of poles
φ = Flux per pole (Weber)
Z = Total armature conductors
I_a = Armature current (Amperes)
A = Number of parallel paths

T_a = (P × φ × Z × I_a) / (2π × A) (Newton-meters)

T_a = 0.159 × φ × I_a × (P/A) (N-m)

T_a ∝ φ × I_a

16.2 Torque for Different Windings

Lap Winding (A = P): T_a ∝ φ × I_a (directly proportional)
Wave Winding (A = 2): T_a ∝ (P × φ × I_a) / 2

Solved Example 16.1

Problem: A 4-pole DC motor has a lap-wound armature with 800 conductors. The flux per pole is 0.025 Wb and armature current is 50 A. Calculate the torque developed.

Solution:
For lap winding: A = P = 4
T_a = (P × φ × Z × I_a) / (2π × A)
T_a = (4 × 0.025 × 800 × 50) / (2 × 3.14 × 4)
T_a = (4000) / (25.12) = 159.2 N-m

📘 CHAPTER 17: TYPES OF DC MOTORS

17.1 Classification Based on Field Connection

Type	Field Connection	Torque Characteristic	Speed Characteristic	Applications
Series Motor	Field in series with armature	Very high starting torque (T ∝ I_a²)	Variable speed, no-load dangerous	Cranes, hoists, electric trains, vacuum cleaners
Shunt Motor	Field parallel to armature	Medium starting torque (T ∝ I_a)	Almost constant speed (good regulation)	Lathes, fans, pumps, blowers, conveyors
Compound Motor	Series + Shunt both	High starting torque	Good speed regulation	Elevators, lifts, presses, rolling mills
Permanent Magnet Motor	Permanent magnet (no field winding)	Good starting torque	Good speed regulation	Small motors, toys, wipers, starters

17.2 Detailed Analysis — Series Motor

🔌 FIGURE 17.1: Series Motor Connection

        DC Supply (+)
            │
            ├─────┬─────┐
            │     │     │
           ║│     │     │
           ║│   ┌─┴─┐   │
           ║│   │   │   │
            │   └───┘   │
            │   Series  │
            │   Field   │
            │           │
            │   ┌───┐   │
            │   │   │   │
            │   │ A │   │
            │   │ R │   │
            │   │ M │   │
            │   │   │   │
            │   └───┘   │
            │           │
            └─────┬─────┘
                  │
               Load (-)

Starting Torque: Very high (T ∝ I²)
Speed Control: Poor — speed varies widely with load
Danger: NEVER run at no-load — speed becomes dangerously high (motor can self-destruct)

17.3 Detailed Analysis — Shunt Motor

🔌 FIGURE 17.2: Shunt Motor Connection

        DC Supply (+)
            │
            ├─────────────┐
            │             │
           ║│             │
           ║│   Field     │
           ║│   Winding   │
            │   (Shunt)   │
            │             │
            │   ┌───┐     │
            │   │   │     │
            └───┤ A ├─────┘
                │ R │
                │ M │
                │   │
                └───┘
                 │
              Load (-)

Speed Regulation: Excellent — nearly constant speed
Starting Torque: Medium (T ∝ I_a)
Application: Where constant speed is required regardless of load

17.4 Compound Motor — Cumulative vs Differential

Type	Series Field Effect	Result
Cumulative Compound	Aids shunt field	High starting torque + good speed regulation
Differential Compound	Opposes shunt field	Poor starting torque, unstable — rarely used

📘 CHAPTER 18: SPEED CONTROL OF DC MOTORS

18.1 Speed Equation

N ∝ (V - I_aR_a) / φ

From this equation, speed can be controlled by:

Varying armature voltage (V)
Varying field flux (φ)
Varying armature resistance (R_a)

18.2 Speed Control Methods

Method	How it works	Used for	Characteristics
Armature Voltage Control	Varying voltage applied to armature	Shunt motors	Speed below base speed, constant torque
Field Flux Control	Adding resistance in field circuit (weakening field)	Shunt motors	Speed above base speed, constant power
Armature Resistance Control	Adding resistance in series with armature	Series motors, cranes	Poor efficiency, speed below base speed
Ward-Leonard Method	Motor-generator set for variable voltage	Large motors, elevators	Wide range, good efficiency, expensive

      📌 Key Points for Speed Control:

      • Field weakening → Speed increases

      • Armature voltage reduction → Speed decreases

      • Armature resistance addition → Speed decreases (but wastes power)

📘 CHAPTER 19: LOSSES & EFFICIENCY OF DC MOTOR

19.1 Types of Losses (Same as Generator)

Copper Loss: I_a²R_a + I_f²R_f
Iron Loss: Hysteresis + Eddy current
Mechanical Loss: Friction + Windage
Stray Loss: Miscellaneous (≈ 1%)

19.2 Efficiency

η = (Output Mechanical Power / Input Electrical Power) × 100%

η = (2πNT / 60) / (V × I_L) × 100%

Where: N = Speed (RPM), T = Torque (N-m)

📘 CHAPTER 20: COMPARISON — DC MOTOR vs DC GENERATOR

Parameter	DC Generator	DC Motor
Energy Conversion	Mechanical → Electrical	Electrical → Mechanical
Working Principle	Faraday's Law (EMF induced)	Fleming's Left Hand Rule (Force on conductor)
Back EMF	Not applicable (EMF is generated)	Present (opposes applied voltage)
Power Flow	Prime mover → Armature → Load	Supply → Armature → Mechanical load
Current Relation	I_a = I_L + I_f (shunt)	I_L = I_a + I_f (shunt)
Voltage Relation	V = E - I_aR_a	V = E_b + I_aR_a

📘 CHAPTER 21: KEY FORMULAS — QUICK REFERENCE

Back EMF: E_b = (P × φ × N × Z) / (60 × A)

Voltage Equation: V = E_b + I_aR_a

Torque: T = (P × φ × Z × I_a) / (2π × A) (N-m)

Torque ∝ φ × I_a

Speed: N ∝ (V - I_aR_a) / φ

Mechanical Power Output: P_out = (2π × N × T) / 60 (Watts)

Efficiency: η = (Output Power / Input Power) × 100%

📘 CHAPTER 22: APPLICATIONS OF DC MOTORS

Motor Type	Applications
Series Motor	Electric trains, cranes, hoists, trolley buses, vacuum cleaners, sewing machines, electric drills
Shunt Motor	Lathe machines, drilling machines, fans, blowers, conveyors, centrifugal pumps, textile machinery
Cumulative Compound Motor	Elevators, lifts, rolling mills, presses, shears, punches, electric excavators
Permanent Magnet Motor	Automobile starters, windshield wipers, power windows, toys, robotics, small fans

📘 CHAPTER 23: COMPARISON — TRANSFORMER vs DC GENERATOR vs DC MOTOR

Parameter	Transformer	DC Generator	DC Motor
Energy Conversion	AC → AC (same frequency)	Mechanical → Electrical (DC)	Electrical (DC) → Mechanical
Moving Parts	No (static device)	Yes (rotating armature)	Yes (rotating armature)
Working Principle	Mutual Induction	Faraday's Law	Fleming's Left Hand Rule
Commutator	Not present	Yes (converts AC to DC)	Yes (reverses current for continuous rotation)
Back EMF	Not applicable	Not applicable	Present (opposes applied voltage)
Frequency Change	No (same frequency)	Not applicable	Not applicable
Efficiency	95-98.5%	85-95%	75-90%
Applications	Power transmission, distribution	Battery charging, lighting	Industrial drives, fans, pumps

📘 CHAPTER 24: COMPARISON — TYPES OF DC MOTORS

Characteristic	Series Motor	Shunt Motor	Compound Motor
Field Connection	Series with armature	Parallel to armature	Series + Shunt both
Starting Torque	Very High (T ∝ I²)	Medium (T ∝ I)	High
Speed Regulation	Poor (wide variation)	Excellent (5-10% drop)	Good (15-25% drop)
No-Load Operation	Dangerous (overspeed)	Safe	Safe
Speed Control	Difficult	Easy	Moderate
Torque-Speed Curve	Hyperbolic	Nearly flat	Slightly drooping
Applications	Cranes, trains, hoists	Lathe, fans, pumps	Elevators, presses

📘 CHAPTER 25: COMPARISON — LAP vs WAVE WINDING

Parameter	Lap Winding	Wave Winding
Number of Parallel Paths (A)	A = P (equal to poles)	A = 2 (always 2)
Number of Brushes	P (equal to poles)	2 (only 2 brushes)
Current Capacity	High (suitable for high current)	Low (suitable for high voltage)
Voltage Capacity	Low	High
EMF Equation	E = (φNZ)/60	E = (PφNZ)/120
Torque Equation	T ∝ φ × I_a	T ∝ (P × φ × I_a)/2
Applications	High current, low voltage generators	High voltage, low current generators

📘 CHAPTER 26: COMPARISON — CORE-TYPE vs SHELL-TYPE TRANSFORMER

Parameter	Core-Type	Shell-Type
Core Shape	Single rectangular core	Core surrounds windings
Windings	Around core limbs	Inside core (core surrounds windings)
Flux Path	2 magnetic paths	1 magnetic path (central limb)
Cooling	Better cooling (windings exposed)	Poor cooling (windings inside)
Mechanical Strength	Less	More (better for short circuits)
Applications	High voltage transformers	Low voltage, high current transformers

📘 CHAPTER 27: QUICK REVISION — ALL FORMULAS

TRANSFORMER FORMULAS

EMF Equation: E = 4.44 × f × N × φ_m

Turn Ratio: K = N₂/N₁ = V₂/V₁ = I₁/I₂

Efficiency: η = Output / (Output + Losses) × 100%

Voltage Regulation = (V_nl - V_fl) / V_nl × 100%

Copper Loss: P_cu = I₁²R₁ + I₂²R₂

kVA (Single Phase) = (V × I) / 1000

kVA (Three Phase) = (√3 × V_L × I_L) / 1000

DC GENERATOR FORMULAS

EMF Equation: E = (P × φ × N × Z) / (60 × A)

Lap Winding (A = P): E = (φ × N × Z) / 60

Wave Winding (A = 2): E = (P × φ × N × Z) / 120

Terminal Voltage: V = E - I_aR_a

DC MOTOR FORMULAS

Back EMF: E_b = (P × φ × N × Z) / (60 × A)

Voltage Equation: V = E_b + I_aR_a

Torque: T = (P × φ × Z × I_a) / (2π × A) (N-m)

Torque: T ∝ φ × I_a

Speed: N ∝ (V - I_aR_a) / φ

Force on Conductor: F = B × I × L (N)

Mechanical Power: P_m = (2π × N × T) / 60 (W)

📘 CHAPTER 28: IMPORTANT MCQs WITH EXPLANATIONS

Transformer MCQs

Q1. A transformer works on ________ supply.
A) DC B) AC C) Both D) None
Answer: B) AC
Explanation: Transformer works on electromagnetic induction which requires changing flux. DC produces constant flux, so no induction.

Q2. The core of a transformer is laminated to reduce ________.
A) Copper loss B) Hysteresis loss C) Eddy current loss D) All losses
Answer: C) Eddy current loss
Explanation: Laminations increase resistance to eddy currents, reducing eddy current loss.

Q3. Open circuit test on a transformer measures ________.
A) Copper loss B) Core loss C) Both D) Efficiency
Answer: B) Core loss
Explanation: OC test is done at rated voltage on LV side. Current is very small, so copper loss negligible. Power measured = core loss.

Q4. Short circuit test on a transformer measures ________.
A) Copper loss B) Core loss C) Iron loss D) Hysteresis loss
Answer: A) Copper loss
Explanation: SC test is done at reduced voltage. Flux is very low, so core loss negligible. Power measured = copper loss.

Q5. If a transformer primary is connected to DC supply, it will ________.
A) Work normally B) Work with low efficiency C) Work with high efficiency D) Burn
Answer: D) Burn
Explanation: DC causes constant flux, no back EMF, high current flows, overheating, and burning.

DC Generator MCQs

Q6. The function of a commutator in a DC generator is to ________.
A) Produce magnetic field B) Convert AC to DC C) Convert DC to AC D) Reduce sparking
Answer: B) Convert AC to DC
Explanation: Commutator acts as a mechanical rectifier, converting the AC induced in armature to DC output.

Q7. In a lap winding, the number of parallel paths (A) is equal to ________.
A) 2 B) Number of poles C) Number of conductors D) Number of coils
Answer: B) Number of poles
Explanation: In lap winding, A = P. In wave winding, A = 2.

Q8. The EMF generated in a DC generator depends on ________.
A) Flux per pole B) Speed C) Number of conductors D) All of the above
Answer: D) All of the above
Explanation: E = (PφNZ)/(60A) — depends on P, φ, N, Z, A.

DC Motor MCQs

Q9. Fleming's Left Hand Rule is used to find direction of ________.
A) Induced EMF B) Current C) Force on conductor D) Magnetic field
Answer: C) Force on conductor
Explanation: Left hand rule is for motors (force). Right hand rule is for generators (induced EMF).

Q10. Back EMF in a DC motor is maximum at ________.
A) Starting B) No-load C) Full load D) Half load
Answer: B) No-load
Explanation: At no-load, speed is maximum, so back EMF (E ∝ N) is maximum.

Q11. Which DC motor has the highest starting torque?
A) Shunt motor B) Series motor C) Compound motor D) PMDC motor
Answer: B) Series motor
Explanation: In series motor, T ∝ I², so starting torque is very high.

Q12. Which DC motor should never be started at no-load?
A) Shunt motor B) Series motor C) Compound motor D) All motors
Answer: B) Series motor
Explanation: At no-load, speed of series motor becomes dangerously high (overspeed), can damage motor.

📘 CHAPTER 29: QUICK REVISION — IMPORTANT POINTS

      📌 TRANSFORMER — Key Points:

      • Works on AC only (DC will burn it)

      • No frequency change (input = output frequency)

      • Core loss = constant, Copper loss ∝ (load)²

      • OC Test → Core loss, SC Test → Copper loss

      • Max efficiency when Copper loss = Core loss

      • Core material: Silicon steel (laminations)

      • Transformer oil: Cooling + Insulation

      • Capacitive load → Negative voltage regulation

      • Humming sound → Magnetostriction (core lamination vibration)

      📌 DC GENERATOR — Key Points:

      • Converts Mechanical → Electrical (DC)

      • Works on Faraday's Law (EMF induced)

      • Commutator converts AC to DC (mechanical rectifier)

      • EMF equation: E = (PφNZ)/(60A)

      • Lap winding: A = P (high current)

      • Wave winding: A = 2 (high voltage)

      • Residual magnetism needed for voltage build-up in shunt generator

      📌 DC MOTOR — Key Points:

      • Converts Electrical (DC) → Mechanical

      • Works on Fleming's Left Hand Rule (Force = BIL)

      • Back EMF opposes applied voltage (V = Eb + IaRa)

      • Torque equation: T ∝ φ × Ia

      • Series motor: High starting torque, no-load dangerous

      • Shunt motor: Constant speed, good speed regulation

      • Speed control: Field weakening (speed ↑), Armature voltage reduction (speed ↓)

📘 CHAPTER 30: KEY TERMINOLOGY — QUICK GLOSSARY

Term	Definition
EMF	Electromotive Force — voltage generated by induction
Back EMF	Voltage induced in motor armature opposing applied voltage
Flux (φ)	Magnetic field lines measured in Weber (Wb)
Flux Density (B)	Flux per unit area — Tesla (T)
Hysteresis Loss	Loss due to repeated magnetization and demagnetization
Eddy Current Loss	Loss due to circulating currents in core
Copper Loss	I²R loss in windings
Magnetostriction	Physical expansion/contraction of core in magnetic field — causes humming
Armature Reaction	Effect of armature flux on main field flux (demagnetizing + cross-magnetizing)
Commutator	Device that converts AC to DC (generator) or reverses current (motor)
Lap Winding	Winding where A = P — high current applications
Wave Winding	Winding where A = 2 — high voltage applications

📘 CHAPTER 31: TRANSFORMER — ADVANCED CONCEPTS

31.1 Transformer Polarity

Polarity indicates the instantaneous direction of induced EMF. Two types:

Additive Polarity: Voltage between H1-H2 + X1-X2 is sum (used for small transformers)
Subtractive Polarity: Voltage between H1-H2 + X1-X2 is difference (used for large transformers)

🔧 FIGURE 31.1: Polarity Marking

        Additive Polarity:          Subtractive Polarity:
        H1 ──┬── H2                H1 ──┬── H2
             │                           │
             ║                           ║
             │                           │
        X1 ──┴── X2                X1 ──┴── X2
        V(H1-H2 + X1-X2)          V(H1-H2 - X1-X2)

31.2 All-Day Efficiency (Energy Efficiency)

For distribution transformers, all-day efficiency is more important than commercial efficiency.

η_all-day = (Output energy in kWh) / (Input energy in kWh) × 100%

Distribution transformers run 24/7, so core loss is constant
All-day efficiency considers load variation over 24 hours
Core loss should be kept minimum for high all-day efficiency

31.3 Auto-Transformer

An auto-transformer has a single winding with a taping point. Part of the winding is common to both primary and secondary.

🔧 FIGURE 31.2: Auto-Transformer

        ┌─────────────────────────┐
        │   ┌─────────────────┐   │
        │   │                 │   │
        A ──┼─┼─────────────────┼─── C
        │   │       Coil       │   │
        │   │                 │   │
        B ──┼─┼─────────────────┼───
        │   └─────────────────┘   │
        └─────────────────────────┘
        
        A-B: Primary (full winding)
        B-C: Secondary (part of winding)

Voltage ratio: V₁/V₂ = N₁/N₂

Current ratio: I₁/I₂ = N₂/N₁

Advantages: Smaller size, less copper, higher efficiency, better voltage regulation

Disadvantage: No isolation between primary and secondary

31.4 Instrument Transformers

31.4.1 Current Transformer (CT)

Used to measure high currents
Primary winding connected in series with line (few turns)
Secondary winding connected to ammeter (many turns)
NEVER open CT secondary while primary is energized — high voltage may develop

CT Ratio = I_p / I_s = N_s / N_p

31.4.2 Potential Transformer (PT)

Used to measure high voltages
Primary winding connected across line (many turns)
Secondary winding connected to voltmeter (few turns)
Secondary should be kept short-circuited? NO — always keep open

PT Ratio = V_p / V_s = N_p / N_s

📘 CHAPTER 32: DC MACHINES — ADVANCED CONCEPTS

32.1 Armature Reaction in Detail

Armature reaction has two major effects:

📐 FIGURE 32.1: Armature Reaction

        Main Field Flux:        Armature Flux:          Resultant Flux:
        N ────────── S         N ────────── S         N ────────── S
          │       │              │   ↑   │              │   ↗   │
          │   →   │              │ ← │ → │              │  ╱│╲  │
          │       │              │   ↓   │              │ ↙ │ ↘ │
        ──┴───────┴──          ──┴───────┴──          ──┴───────┴──
        
        Effect: Flux is distorted (cross-magnetizing) and reduced (demagnetizing)

32.1.1 Compensating Windings

Placed in pole faces to neutralize armature reaction
Connected in series with armature
Used in large machines to reduce sparking

32.1.2 Interpoles (Commutating Poles)

Small poles placed between main poles
Wound with armature current
Improve commutation, reduce sparking

32.2 Losses in DC Machines — Detailed

Loss Type	Sub-Type	Percentage	Reduction Method
Copper Loss	Armature I²R	30-40%	Use thick copper wire
Copper Loss	Field I²R	20-25%	Optimize field current
Iron Loss	Hysteresis	15-20%	Silicon steel
Iron Loss	Eddy Current	10-15%	Laminations
Mechanical Loss		10-15%	Proper bearings, smooth surfaces
Stray Loss		~1%	Good design

32.3 Swinburne's Test

Indirect method to find efficiency of DC machines without loading.

Machine run as motor at no-load
Constant losses (iron + mechanical) measured
Efficiency can be predicted at any load

      📌 Advantage: No-load test, so power consumption is low

      Disadvantage: Cannot detect commutation problems

32.4 Hopkinson's Test (Back-to-Back Test)

Two identical DC machines coupled together
One runs as motor, other as generator
Both machines loaded equally
Most accurate method for large machines

📘 CHAPTER 33: NUMERICAL PROBLEMS WITH SOLUTIONS

Numerical 1: Transformer EMF

Problem: A 25 kVA, 2200/220 V, 50 Hz transformer has a core cross-section of 200 cm² and maximum flux density 1.2 T. Calculate primary and secondary turns.

Solution:
φ_m = B_m × A = 1.2 × (200 × 10⁻⁴) = 0.024 Wb
E = 4.44 × f × N × φ_m
N₁ = E₁ / (4.44 × f × φ_m) = 2200 / (4.44 × 50 × 0.024)
N₁ = 2200 / (4.44 × 1.2) = 2200 / 5.328 = 413 turns
N₂ = (V₂/V₁) × N₁ = (220/2200) × 413 = 41 turns

Numerical 2: DC Generator EMF

Problem: An 8-pole DC generator has 600 armature conductors, flux per pole 0.03 Wb, speed 500 RPM. Calculate EMF for lap and wave winding.

Solution:
Lap winding (A = P = 8):
E = (P × φ × N × Z) / (60 × A) = (8 × 0.03 × 500 × 600) / (60 × 8)
E = (8 × 0.03 × 500 × 600) / 480 = 72000 / 480 = 150 V

Wave winding (A = 2):
E = (8 × 0.03 × 500 × 600) / (60 × 2) = 72000 / 120 = 600 V

Numerical 3: DC Motor Torque

Problem: A 4-pole DC motor has lap winding with 500 conductors. Flux per pole is 0.02 Wb and armature current is 40 A. Calculate torque.

Solution:
T = (P × φ × Z × I_a) / (2π × A)
For lap winding: A = P = 4
T = (4 × 0.02 × 500 × 40) / (2 × 3.14 × 4)
T = 1600 / 25.12 = 63.7 N-m

Numerical 4: Transformer Efficiency

Problem: A 100 kVA transformer has iron loss 1000 W and full-load copper loss 2000 W. Calculate efficiency at full load, 0.8 pf lagging.

Solution:
Output power = kVA × pf = 100 × 1000 × 0.8 = 80,000 W
Total losses = Iron loss + Copper loss = 1000 + 2000 = 3000 W
Input power = Output + Losses = 80,000 + 3,000 = 83,000 W
η = (80,000 / 83,000) × 100 = 96.38%

Numerical 5: DC Shunt Generator

Problem: A 4-pole shunt generator with lap winding has 400 conductors. Flux per pole 0.05 Wb, speed 1000 RPM. Armature resistance 0.5 Ω, shunt field resistance 100 Ω. Calculate terminal voltage at no-load and at load current 50 A.

Solution:
E = (P × φ × N × Z) / (60 × A) = (4 × 0.05 × 1000 × 400) / (60 × 4)
E = 80,000 / 240 = 333.33 V (no-load terminal voltage)

At load: I_a = I_L + I_sh = 50 + (333/100) = 50 + 3.33 = 53.33 A
V = E - I_aR_a = 333.33 - (53.33 × 0.5) = 333.33 - 26.67 = 306.66 V

📘 CHAPTER 34: INTERVIEW QUESTIONS & ANSWERS

Q1. Why is the transformer core laminated?
Ans: To reduce eddy current loss. Laminations increase the resistance to circulating currents, minimizing eddy current loss.

Q2. What is the difference between power transformer and distribution transformer?
Ans: Power transformers are used at generating stations (high voltage, low current, high efficiency at full load). Distribution transformers are used near consumers (low voltage, high current, high all-day efficiency).

Q3. Why is a DC motor used for traction (electric trains)?
Ans: DC series motor provides very high starting torque (T ∝ I²), which is required for starting heavy loads like trains.

Q4. Why can't a transformer work on DC?
Ans: DC produces constant flux, so no induced EMF in secondary (dΦ/dt = 0). Primary draws excessive current due to low resistance, causing overheating and burning.

Q5. What is the function of a commutator?
Ans: In DC generator: converts AC to DC (mechanical rectifier). In DC motor: reverses current direction every half-cycle to maintain unidirectional torque.

Q6. What is back EMF? Why is it important?
Ans: Back EMF is the voltage induced in a DC motor armature that opposes the applied voltage. It acts as a governor — automatically regulating armature current based on load.

Q7. Why is silicon steel used for transformer cores?
Ans: Silicon steel has low hysteresis loss, high magnetic permeability, and high electrical resistivity (reduces eddy currents).

Q8. What happens if CT secondary is opened?
Ans: A very high voltage develops across the open secondary, which can be dangerous for personnel and can damage insulation.

📘 CHAPTER 35: ADDITIONAL MCQs — 50 QUESTIONS

Set 1: Transformer (1-15)

1. The primary and secondary windings of a transformer are ________ coupled.
A) Electrically B) Magnetically C) Mechanically D) Chemically
Answer: B) Magnetically

2. The efficiency of a transformer is maximum when ________.
A) Copper loss = Iron loss B) Copper loss = 0 C) Iron loss = 0 D) Both losses zero
Answer: A) Copper loss = Iron loss

3. Transformer oil provides ________.
A) Cooling B) Insulation C) Both A and B D) Lubrication
Answer: C) Both A and B

4. The humming sound in a transformer is due to ________.
A) Oil movement B) Core vibration C) Winding vibration D) Cooling fan
Answer: B) Core vibration (magnetostriction)

5. In a step-up transformer, the current in the secondary is ________ than primary.
A) Higher B) Lower C) Equal D) Double
Answer: B) Lower

6. The iron loss in a transformer depends on ________.
A) Load B) Voltage C) Frequency D) Both B and C
Answer: D) Both B and C

7. The voltage regulation of a transformer at unity power factor is ________.
A) Positive B) Negative C) Zero D) Infinity
Answer: A) Positive

8. The function of a breather in a transformer is to ________.
A) Cool the oil B) Absorb moisture C) Reduce noise D) Increase voltage
Answer: B) Absorb moisture

9. Conservator in a transformer is used for ________.
A) Oil expansion B) Cooling C) Insulation D) Protection
Answer: A) Oil expansion

10. Buchholz relay is used for ________ protection.
A) Overload B) Short circuit C) Internal fault D) Over voltage
Answer: C) Internal fault

11. The core of a transformer is made of ________.
A) Cast iron B) Silicon steel C) Copper D) Aluminum
Answer: B) Silicon steel

12. The mutual flux in a transformer links with ________.
A) Primary only B) Secondary only C) Both primary and secondary D) Core only
Answer: C) Both primary and secondary

13. The eddy current loss in a transformer is proportional to ________.
A) f B) f² C) 1/f D) 1/f²
Answer: B) f²

14. The hysteresis loss in a transformer is proportional to ________.
A) f B) f² C) 1/f D) f³
Answer: A) f

15. A transformer cannot step up ________.
A) Voltage B) Current C) Power D) Frequency
Answer: D) Frequency

Set 2: DC Generator (16-30)

16. The function of a yoke in a DC machine is to ________.
A) Provide mechanical support B) Carry magnetic flux C) Both A and B D) None
Answer: C) Both A and B

17. In a DC generator, the armature core is laminated to reduce ________.
A) Copper loss B) Hysteresis loss C) Eddy current loss D) Mechanical loss
Answer: C) Eddy current loss

18. The number of parallel paths in a lap winding is equal to ________.
A) 2 B) Number of poles C) Number of conductors D) Number of coils
Answer: B) Number of poles

19. For a DC generator, the generated EMF is maximum when the coil is at ________.
A) 0° B) 45° C) 90° D) 180°
Answer: C) 90° (cutting flux at maximum rate)

20. The commutator segments are insulated from each other by ________.
A) Paper B) Mica C) Rubber D) Plastic
Answer: B) Mica

21. The brushes in a DC machine are made of ________.
A) Copper B) Aluminum C) Carbon D) Silver
Answer: C) Carbon

22. A shunt generator builds up voltage only if ________.
A) Residual magnetism exists B) Field resistance is less than critical C) Speed is above critical D) All of these
Answer: D) All of these

23. The critical resistance of a DC shunt generator is ________.
A) Fixed B) Variable C) Zero D) Infinity
Answer: B) Variable (depends on speed)

24. In a DC generator, the iron loss occurs in ________.
A) Yoke B) Armature core C) Pole shoes D) All of these
Answer: B) Armature core

25. The terminal voltage of a shunt generator ________ with load.
A) Increases B) Decreases slightly C) Decreases sharply D) Remains constant
Answer: B) Decreases slightly

Set 3: DC Motor (31-45)

31. The direction of rotation of a DC motor can be reversed by ________.
A) Reversing armature connections B) Reversing field connections C) Either A or B D) None
Answer: C) Either A or B

32. The speed of a DC shunt motor is ________ proportional to armature current.
A) Directly B) Inversely C) Not D) Squarely
Answer: C) Not (speed is nearly constant)

33. The speed of a DC series motor is ________ proportional to armature current.
A) Directly B) Inversely C) Not D) Squarely
Answer: B) Inversely (speed decreases as load increases)

34. The starting torque of a DC series motor is ________ that of a shunt motor.
A) Lower than B) Higher than C) Equal to D) None
Answer: B) Higher than

35. For a DC shunt motor, the torque is proportional to ________.
A) I_a B) I_a² C) 1/I_a D) Constant
Answer: A) I_a

36. For a DC series motor, the torque is proportional to ________.
A) I_a B) I_a² C) 1/I_a D) Constant
Answer: B) I_a² (up to saturation)

37. The back EMF in a DC motor is maximum at ________.
A) Starting B) No-load C) Full load D) Half load
Answer: B) No-load

38. The back EMF in a DC motor is zero at ________.
A) Starting B) No-load C) Full load D) Half load
Answer: A) Starting

39. A starter is required for a DC motor because ________.
A) Starting current is high B) Back EMF is zero at start C) Both A and B D) None
Answer: C) Both A and B

40. The three-point starter has ________ terminals.
A) 2 B) 3 C) 4 D) 5
Answer: B) 3 (Line, Field, Armature)

📘 CHAPTER 36: FINAL FORMULA SHEET — PRINT READY

🔹 TRANSFORMER — EMF: E = 4.44 × f × N × φ_m

🔹 TRANSFORMER — Turn Ratio: K = N₂/N₁ = V₂/V₁ = I₁/I₂

🔹 TRANSFORMER — Efficiency: η = (Output) / (Output + Losses) × 100%

🔹 TRANSFORMER — Voltage Regulation = (V_nl - V_fl) / V_nl × 100%

🔹 TRANSFORMER — kVA (1φ) = V × I / 1000

🔹 TRANSFORMER — kVA (3φ) = √3 × V_L × I_L / 1000

🔹 DC GENERATOR — EMF: E = (P × φ × N × Z) / (60 × A)

🔹 LAP WINDING — A = P, E = (φ × N × Z) / 60

🔹 WAVE WINDING — A = 2, E = (P × φ × N × Z) / 120

🔹 DC GENERATOR — Terminal Voltage: V = E - I_aR_a

🔹 DC MOTOR — Back EMF: E_b = (P × φ × N × Z) / (60 × A)

🔹 DC MOTOR — Voltage Equation: V = E_b + I_aR_a

🔹 DC MOTOR — Torque: T = (P × φ × Z × I_a) / (2π × A) (N-m)

🔹 DC MOTOR — Torque Relation: T ∝ φ × I_a

🔹 DC MOTOR — Speed Relation: N ∝ (V - I_aR_a) / φ

🔹 DC MOTOR — Force: F = B × I × L (N)

🔹 DC MOTOR — Mechanical Power: P_m = (2π × N × T) / 60 (W)

🔹 LOSSES — Copper Loss = I²R

🔹 LOSSES — Hysteresis Loss ∝ B_m^1.6 × f

🔹 LOSSES — Eddy Current Loss ∝ B_m² × f² × t²

📚 Continue Learning:

← Previous: upda-3 Next: upda-5 →

📚 TRANSFORMERS & DC MACHINES

📖 TRANSFORMERS & DC MACHINES