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Overview

🌳 Concept tree

Lesson 4→The Electric Transformer and Electric Motor (DC Motor)

├── Principle and structure

├── Currents

├── Efficiency and energy losses

├── Power transmission

├── Principle

├── Structure

├── Action

├── Back emf

├── Comparisonmotor and galvanometer

├── Comparisongenerator and motor

├── Typical "give reasons" for this chapter

└── Summary of formulas

Lesson 4: The Electric Transformer and Electric Motor (DC Motor) · 1 / 14

Principle and structure

🧊 3D Isometric Model of Electric Transformer with Laminated Silicon-Steel Core

D Scientific Model→High-fidelity 3D structural model illustrating the laminated silicon-steel core engineered to suppress eddy current losses, primary coil connected to AC supply, secondary coil, and mutual electromagnetic induction flux lines obeying Vs/Vp = Ns/Np.

A transformer is a device based on mutual induction that steps an alternating voltage up or down.
It has a primary coil (N_p turns) and a secondary coil (N_s turns), both wound on a closed core of laminated soft iron (thin sheets insulated from each other, often silicon iron).
The core makes the flux lines of the primary pass through the secondary.
When AC is applied to the primary, the changing flux in the core links both coils and induces an alternating emf of the same frequency in the secondary.
For a negligible loss of flux (no stray lines), the same flux change links both coils:

Vs = Ns ΔΦ/Δt and Vp = Np ΔΦ/Δt, so

Vs / Vp = Ns / Np

Ns > Np step-up transformer (Vs > Vp). For example if Ns = 2Np then Vs = 2Vp.
Ns < Np step-down transformer (Vs < Vp).
A transformer does not work with DC (the flux does not change, so no emf is induced in the secondary).
Step-up transformers are used at power stations; step-down transformers are used in the distribution zones.

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Currents

For an ideal transformer (no energy losses), by conservation of energy the electrical energy supplied to the primary equals that delivered to the load in the secondary, so input power = output power:
V_p I_p = V_s I_s, so I_s / I_p = V_p / V_s = N_p / N_s

The current is inversely proportional to the number of turns the coil with more turns carries the smaller current (made of thinner wire); the coil with few turns carries the larger current (thick wire).
A step-up transformer steps current down.
If the secondary has twice the turns, its current is half that of the primary.

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Efficiency and energy losses

A 100%-efficient transformer is not available in practice.
efficiency η = (output power) / (input power) = (V_s I_s)/(V_p I_p) × 100%
An efficiency of 80% means that 20% of the input energy is lost; if the lost energy is 10% of the original energy the efficiency is 90%.
For a real transformer, Vs Is = η Vp Ip, so the current relation becomes Is/Ip = η Vp / Vs.
Causes of loss and remedies:

Comparison aspect	Main idea	Keywords
Heating of the copper wires	(I²R) → use thick wires of low resistance	thickresistance
Eddy currents in the core	→ laminated core of thin insulated sheets of silicon steel (high resistivity)	laminatedthininsulatedsheets
Hysteresis	(energy used in repeatedly magnetizing and demagnetizing the core turned into vibration of the magnetic molecules) → use soft iron whose molecules are easily magnetized and demagnetized	energyusedrepeatedlymagnetizing
•	Leakage of flux (not all the flux links both coils) → use a closed core	Leakagefluxlinksboth

Example 7.→A transformer has 400 primary turns and 100 secondary turns; V_p = 220 V, and the secondary current is 4.0 A. Assuming 100% efficiency: V_s = 220 × 100/400 = 55 V; I_p = I_s N_s/N_p = 4 × 100/400 = 1.0 A.

Comparison aspect	Main idea	Keywords
Example 8.	A transformer of 90% efficiency is connected to a 120 V supply and delivers 12 V at 3 A. P_out = 36 W; P_in = 36/0.9 = 40 W; I_p = 40/120 = 0.33 A.	transformerefficiencyconnectedsupply
Example 8a — step-down, ideal and real.	A transformer converts 2200 V to 220 V. The primary has 4000 turns, so N_s = 4000 × 220/2200 = 400 turns. The secondary supplies 10 A, so the output power is 2200 W. If the transformer is ideal, I_p = 2200/2200 = 1 A. If its efficiency is 88%, P_in = 2200/0.88 = 2500 W and I_p = 2500/2200 = 1.14 A.	transformerconverts2200primary
Example 8b — step-up.	A 12 V supply feeds a step-up transformer with 50 primary turns and a secondary of 1000 turns. V_s = 12 × 1000/50 = 240 V. If the secondary current is 0.5 A, the output power is 120 W. For an efficiency of 80% the input power is 150 W and the primary current is 150/12 = 12.5 A.	supplyfeedstransformerprimary

Example 8c — finding the efficiency.→A transformer takes 0.5 A from a 200 V source and delivers 0.9 A at 90 V. P_in = 200 × 0.5 = 100 W; P_out = 90 × 0.9 = 81 W; η = 81/100 = 81%. The lost power is 19 W.

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Power transmission

Power stations generate electricity at a few thousand volts. To transmit it over long distances:

A step-up transformer at the station raises the voltage to hundreds of kilovolts. For the same power P = VI, the current I is correspondingly smaller, and the power lost in the line, I²R, falls by the square of the voltage ratio. If the current is reduced to one-hundredth, the loss is reduced to one ten-thousandth.
Step-down transformers at the distribution areas reduce the voltage (finally to 220 V for homes).
Transformers are also found in doorbells, chargers, refrigerators and other appliances.
Example 9. A line of resistance 10 Ω transmits 100 kW. At 10 kV: I = 10 A, loss = 1 kW. At 100 kV: I = 1 A, loss = 10 W (a hundred times smaller).

Example 9a.→A power of 200 kW is sent through a line of total resistance 0.5 Ω. At 400 V the current is 500 A and the loss is 500² × 0.5 = 125 kW (63% of the power!). At 20 kV the current is 10 A and the loss is 10² × 0.5 = 50 W. The voltage was multiplied by 50, so the loss fell by 50² = 2500 times.

Lesson 4: The Electric Transformer and Electric Motor (DC Motor) · 5 / 14

Principle

A motor converts electrical energy into mechanical energy.
It works on the torque on a current-carrying coil in a magnetic field (the same principle as the moving-coil galvanometer), but must keep turning continuously in the same direction.

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Structure

Comparison aspect	Main idea	Keywords
Field magnet	a strong horseshoe permanent magnet or electromagnet	stronghorseshoepermanentelectromagnet
Armature	a rectangular coil abcd of many turns of insulated copper wire wound on a laminated soft-iron core (thin insulated sheets to reduce eddy currents)	rectangularcoilabcdmany
Commutator	a split cylinder (two halves x and y insulated from each other) joined to the ends of the coil. The plane separating the halves is perpendicular to the plane of the coil	splitcylinderhalvesinsulated
•	Two brushes F₁ and F₂ connected to a DC source (battery) touching the commutator halves the line joining them is parallel to the field lines	brushesconnectedsourcebattery

Lesson 4: The Electric Transformer and Electric Motor (DC Motor) · 7 / 14

Action

Start with the coil plane parallel to the field and brush F₁ (positive) on half x, F₂ on half y. The current flows d → c → b → a. By Fleming’s left-hand rule the force on ab is upward and the force on cd is downward: a couple makes the coil rotate.
The torque decreases as the coil turns and is zero when the plane is perpendicular to the field. The coil’s inertia carries it past this position; the commutator halves now exchange brushes, so the current in the coil reverses (now a → b → c → d). The forces on the two sides reverse too (ab down, cd up), and the torque again turns the coil in the same direction.
Thus the commutator reverses the current every half revolution, and the coil rotates continuously in one direction. The torque grows to a maximum when the plane is parallel to the field and falls to zero when it is perpendicular.
To increase the power and obtain smoother, nearly uniform rotation: use many coils at equal angles on the armature, each joined to opposite segments of a commutator with twice as many segments as coils (each pair of segments touches the brushes when its coil is in the position of largest torque), more turns, a stronger field, and a bigger current.

Lesson 4: The Electric Transformer and Electric Motor (DC Motor) · 8 / 14

Back emf

As the coil of a motor rotates in the field it cuts flux lines, so an emf is induced in it opposing the supply voltage (Lenz’s rule).
This is the back emf.
The current in the motor is I = (V − back emf)/R.
At start (no rotation) the back emf is zero and the current is largest; as the speed rises the back emf rises and the current falls.
Example 10. A DC motor of armature resistance 2 Ω is connected to a 220 V supply. At the moment of starting I = 220/2 = 110 A (hence the need for a starter resistance). At its running speed the back emf is 200 V, so I = (220 − 200)/2 = 10 A. The electrical power supplied is 220 × 10 = 2200 W, of which 200 × 10 = 2000 W becomes mechanical power and 10² × 2 = 200 W is heat in the windings.

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Comparison: motor and galvanometer

Both depend on the torque on a coil in a magnetic field.
The galvanometer rotates through a limited angle restrained by springs; the motor rotates continuously and so needs a commutator whose halves interchange positions relative to the brushes every half cycle.

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Comparison: generator and motor

Comparison aspect	AC generator	DC motor
Energy change	mechanical → electrical	electrical → mechanical
Principle	electromagnetic induction (Fleming’s right-hand rule)	force on a current in a field (left-hand rule)
Input	rotation of the coil by an external agent	current from a DC source
Output	emf induced in the coil	rotation of the coil
Contacts	slip rings (AC) or commutator (DC)	commutator

Lesson 4: The Electric Transformer and Electric Motor (DC Motor) · 11 / 14

Typical "give reasons" for this chapter

Comparison aspect	Main idea	Keywords
The core of a transformer is made of thin insulated sheets	to reduce eddy currents and the energy lost as heat	reduceeddycurrentsenergy
•	An iron bar inside a double-wound wire carrying current is not magnetized: the currents in adjacent turns are opposite the fields cancel	ironinsidedoublewound
A free wire in a field moves when current flows	there is a force on a current in a field (BIL)	force
A transformer cannot be used with DC	the flux is constant so no emf is induced in the secondary	fluxconstantinducedsecondary
The motor rotates (almost) uniformly	the commutator reverses the current every half turn and many coils at different angles keep the torque nearly constant	commutatorreversescurrentevery
A current dies out slowly in a coil with an iron core	the large self-induced emf opposes the fall	largeselfinducedopposes
•	The commutator of a DC generator is split in two halves insulated from each other: to reverse the connection to the external circuit each half turn so the output keeps one direction	commutatorgeneratorsplithalves
The current in a coil grows slowly when the switch is closed	the induced emf opposes the growth of the current (Lenz’s rule)	inducedopposesgrowthLenz

Lesson 4: The Electric Transformer and Electric Motor (DC Motor) · 12 / 14

Summary of formulas

Quantity	Formula
Faraday’s law	emf = −N ΔΦ_m/Δt
Moving wire	emf = B L v sin θ
Mutual induction	emf = −M ΔI_p/Δt (unit henry)
Self-induction	emf = −L ΔI/Δt
Generator	emf = N B A ω sin θ ; emf_max = 2π f N B A
Effective values	I_eff = 0.707 I_max ; V_eff = 0.707 V_max
Transformer	V_s/V_p = N_s/N_p ; I_s/I_p = N_p/N_s (ideal)
Efficiency	η = V_s I_s / (V_p I_p)
Motor	I = (V − back emf)/R

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Part 14

📖

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

Physics (Thanaweya Amma 2026/2027)

Lesson 4 – The Electric Transformer and Electric Motor (DC Motor)
The electric transformer changes alternating voltages via mutual induction between primary and secondary coils wound on a soft iron core.
The electric motor converts electrical energy into rotational kinetic energy based on magnetic deflecting torque.

🌳 Conceptual Hierarchy Tree

Transformers and Electric Motors

├── 1. Electric Transformer Operation

Comparison aspect	Main idea	Keywords
•	│ ├── Operates ONLY on Alternating Current (AC) Does NOT work on Direct Current (DC)	OperatesAlternatingCurrentDoes
│ ├── Ideal Transformer	Efficiency η = 100% (Pin = Pout)	EfficiencyPout
•	│ │ └── Voltage vs. Turns vs. Current: (Vs / Vp) = (Ns / Np) = (Ip / Is)	VoltageTurnsCurrent
│ ├── Step-Up Transformer	Ns > Np → Vs > Vp and Is < Ip (Used at power generation stations)	Usedpowergenerationstations
│ └── Step-Down Transformer	Ns < Np → Vs < Vp and Is > Ip (Used at residential distribution)	Usedresidentialdistribution

└── 2. Electric Motor (DC Motor)

├── PrincipleTorque on a current-carrying loop in magnetic field (τ = B·I·A·N·sin θ)

├── Split-ring commutatorReverses current direction in coil every half turn to maintain continuous rotation in same direction

└── Back-EMF in motorRegulates and stabilizes rotational speed

📋 Master Teacher's Blackboard Secrets

Comparison aspect	Main idea	Keywords
RULE 1	A transformer connected to a 12 V DC car battery produces ZERO voltage in the secondary coil (Vs = 0) because DC current creates a steady magnetic flux (ΔΦm/Δt = 0)	transformerconnectedbatteryproduces
RULE 2	Step-up transformer raises voltage but LOWERS current reducing power loss during long-distance transmission (Ploss = I² · Rwire)	Steptransformerraisesvoltage
RULE 3	Back-EMF in a motor acts as an automatic governor: If motor speed decreases back-EMF decreases allowing current to increase and accelerating the motor back to equilibrium speed	Backmotoractsautomatic
📊 Matrix	Comparative Analysis: Step-Up vs. Step-Down Transformers	ComparativeAnalysisStepDown

Feature	Step-Up Transformer	Step-Down Transformer
Secondary Turns vs. Primary	Ns > Np	Ns < Np
Secondary Voltage vs. Primary	Vs > Vp (Voltage amplifies)	Vs < Vp (Voltage steps down)
Secondary Current vs. Primary	Is < Ip (Current decreases)	Is > Ip (Current increases)
Wire Thickness	Secondary wire is thinner than primary	Secondary wire is thicker than primary
Key Installation Location	Power generation plants (Aswan Dam)	Consumer substations and household appliances

Lesson 4: The Electric Transformer and Electric Motor (DC Motor) · 14 / 14

Curriculum

Physics

Chapter 1 - Electric Current, Ohms Law and Kirchhoffs Laws

Chapter 2 - Magnetic Effects of Electric Current and Measuring Instruments

Chapter 3 - Electromagnetic Induction and AC Dynamo and Transformer

Chapter 4 - Alternating Current AC Circuits

Chapter 5 - Dual Nature of Wave and Particle

Chapter 6 - Atomic Spectra and X-Rays

Chapter 7 - Lasers and Principles

Chapter 8 - Modern Electronics and Semiconductors

More lessons

Lesson 4: The Electric Transformer and Electric Motor (DC Motor)