Motors, Generators, and Transformers
When you plug your cell phone into the outlet at home, you use electricity at 127 V. That energy may have been born in the alternator of a hydroelectric plant like Chicoasén, in Chiapas, and traveled hundreds of kilometers along high-voltage lines. Along the way it passed through several transformers. The motor in your blender, the generator at the power plant, and the transformer in your charger don't need any new physics: all three are based on electromagnetism. The motor takes advantage of the magnetic force on a current, and the generator and the transformer work by electromagnetic induction. Why does electricity travel at tens or hundreds of thousands of volts if at home we use only 127 V? We'll solve it at the end.
The electric motor: a pair of forces that rotates
If you connect a loop of copper wire to a battery and place it between the poles of a magnet, the loop gives a jolt and lines up toward the center. How do we get it not to stop there and to rotate continuously to run a blender, a fan, or an electric truck?
Back electromotive force
When the motor spins, its coils move inside the magnetic field. Therefore, their flux changes and, by Faraday's law, an emf is induced in the motor itself. By Lenz's law, that emf opposes the cause that produces it, that is, the current that makes it rotate. That is why it is called back electromotive force (back emf, $\varepsilon'$).
The rectangular loop of a DC motor is between the poles of a magnet and carries a current. a) Draw the magnetic forces on the two sides of the loop perpendicular to the…
The rectangular loop of a DC motor is between the poles of a magnet and carries a current.
- a) Draw the magnetic forces on the two sides of the loop perpendicular to the field in your notebook and explain why they form a couple that makes the loop rotate.
- b) What role does the split-ring commutator play? What would happen to the loop upon reaching the dead center if it did not exist?
The loop of a toy motor measures $10\ \text{cm}$ in length and $5\ \text{cm}$ in width, carries a current of $3\ \text{A}$ and is in a field of $0.4\ \text{T}$…
The loop of a toy motor measures $10\ \text{cm}$ in length and $5\ \text{cm}$ in width, carries a current of $3\ \text{A}$ and is in a field of $0.4\ \text{T}$ perpendicular to its axis of rotation, as shown in the figure.
- a) Calculate the magnetic force on each of the long sides of the loop, $F = I \times l \times B$.
- b) Explain why these two forces, acting in opposite directions, form a couple that rotates the loop instead of translating it.
- c) What happens to the couple when the plane of the loop is perpendicular to the field (dead center) and how does the split-ring commutator solve this?
When starting a $12\ \text{V}$ motor whose internal resistance is $0.5\ \Omega$, with the shaft still at rest, the initial current is $24\ \text{A}$. a) Verify this…
When starting a $12\ \text{V}$ motor whose internal resistance is $0.5\ \Omega$, with the shaft still at rest, the initial current is $24\ \text{A}$.
- a) Verify this value by applying Ohm's law and explain why, at startup, the motor behaves as a simple resistor.
- b) When the motor runs at full speed, the rotation of its coils in the field generates an induced emf of $10\ \text{V}$ (back electromotive force): calculate the current at full speed, $I = \frac{V - \varepsilon'}{R}$.
- c) Explain using Lenz's law why this induced emf has the opposite sign to the supply voltage.
- d) In light of the previous sections, explain why it is dangerous to stall the shaft of a running motor and what happens to its winding.
Generators: alternator and dynamo
Seen head-on, the loop "looks" wider or narrower depending on its position. That means the flux through it goes up and down. If the flux changes, an emf appears.
a) Explain the fundamental difference between an alternator and a dynamo: what type of current each produces and which connecting element (slip rings or split-ring…
- a) Explain the fundamental difference between an alternator and a dynamo: what type of current each produces and which connecting element (slip rings or split-ring commutator) determines it.
- b) What physical phenomenon are both devices based on?
In a bicycle with a dynamo, the light turns on when pedalling. a) Explain how the dynamo generates the current that powers the bulb. b) According to Lenz's law, the…
In a bicycle with a dynamo, the light turns on when pedalling.
- a) Explain how the dynamo generates the current that powers the bulb.
- b) According to Lenz's law, the induced current opposes the rotation of the dynamo: explain where the energy consumed by the bulb ultimately comes from, and why more resistance is felt when pedalling with the dynamo connected.
The alternator of a small power plant is equivalent to a coil of $200$ turns of $100\ \text{cm}^2$ rotating at a frequency of $50\ \text{Hz}$ in a field of…
The alternator of a small power plant is equivalent to a coil of $200$ turns of $100\ \text{cm}^2$ rotating at a frequency of $50\ \text{Hz}$ in a field of $0.5\ \text{T}$ (take $\pi \approx 3.14$).
- a) Calculate the angular speed of rotation, $\omega = 2 \times \pi \times f$.
- b) Calculate the maximum generated emf, $\varepsilon_0 = N \times B \times S \times \omega$.
- c) Write the expression $\varepsilon(t)$ assuming that at $t = 0$ the flux is maximum and the emf is zero.
Alternating current and rms value
A light bulb plugged into the outlet at your home lights up at "127 V." But the voltage of an alternating current changes all the time, even passing through zero. So what does that 127 mean?
When a loop rotates at a constant angular speed in a uniform magnetic field, an alternating current is generated, as shown in the figure. a) Explain, based on how the…
When a loop rotates at a constant angular speed in a uniform magnetic field, an alternating current is generated, as shown in the figure.
- a) Explain, based on how the magnetic flux through the loop varies, why the emf continually changes value.
- b) Reason why the emf changes sign every half turn.
The domestic voltage of $230\ \text{V}$ is the rms value of an alternating current of $50\ \text{Hz}$ whose maximum value is $325\ \text{V}$ (take $\pi \approx 3.14$ and…
The domestic voltage of $230\ \text{V}$ is the rms value of an alternating current of $50\ \text{Hz}$ whose maximum value is $325\ \text{V}$ (take $\pi \approx 3.14$ and $\sqrt{2} \approx 1.41$). The graph of voltage against time is shown in the figure.
- a) Calculate the angular frequency $\omega$ and write the expression for emf as a function of time, $\varepsilon(t) = \varepsilon_0 \times \sen(\omega \times t)$.
- b) Verify that the rms value satisfies the relationship $\varepsilon_{\text{ef}} = \frac{\varepsilon_0}{\sqrt{2}}$.
- c) If the power station alternator works with a field of $0.6\ \text{T}$ and turns of $0.05\ \text{m}^2$, calculate how many turns its winding needs to generate that maximum emf, $\varepsilon_0 = N \times B \times S \times \omega$.
Induction in everyday life
You brake in your electric car and the battery recharges. You pluck a string on an electric guitar and sound comes out of the speaker. You put your cell phone on a wireless charger and it charges without a cable. What do they have in common?
Applications
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Bicycle dynamo. You pedal and spin a magnet near a coil. The energy for the light comes from your legs. That is why you notice more resistance: the induced current, by Lenz's law, slows the rotation.
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Regenerative braking. When you release the accelerator, the car's motor works as a generator. Part of the kinetic energy goes to the battery and the induced current slows the car.
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Guitar pickup. The steel string vibrates in front of a magnet with a coil, magnetizes the metal, and changes the field in the coil. That flux oscillates at the frequency of the note and an emf is induced with that same frequency. That is why the string must be made of a material that can be magnetized (steel), not nylon.
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Wireless charging. A transmitting coil with alternating current creates a changing field that induces an emf in the phone's coil. It is a transformer without an iron core, and it uses high frequencies because the emf depends on how fast the flux changes.
A $1500\ \text{kg}$ electric car travels at $72\ \text{km/h}$. On releasing the accelerator, its motor acts as a generator and recharges the battery (regenerative…
A $1500\ \text{kg}$ electric car travels at $72\ \text{km/h}$. On releasing the accelerator, its motor acts as a generator and recharges the battery (regenerative braking).
- a) Express the speed in $\text{m/s}$ and calculate the kinetic energy of the car.
- b) If the system recovers $60\%$ of that energy, how much energy returns to the battery?
- c) During braking, the motor winding ($300$ turns) undergoes a flux variation of $0.4\ \text{Wb/s}$ in each turn: calculate the generated emf.
- d) Explain using Lenz's law why the induced current slows down the car instead of accelerating it.
The pickup of an electric guitar is a small magnet surrounded by a coil of $2000$ turns of $1\times10^{-5}\ \text{m}^2$ each. When the metal string is plucked (A note,…
The pickup of an electric guitar is a small magnet surrounded by a coil of $2000$ turns of $1\times10^{-5}\ \text{m}^2$ each. When the metal string is plucked (A note, $440\ \text{Hz}$), the field through the coil oscillates, with a variation of $1\times10^{-3}\ \text{T}$ in each quarter period. Observe the diagram:
- a) Calculate the duration of a quarter of a period of the vibration.
- b) Calculate the average induced emf during that interval, in millivolts.
- c) Explain qualitatively how the electrical signal reaching the amplifier is generated and reason why the string must be made of a ferromagnetic material.
A wireless charger has a transmitter coil of $20$ turns powered at $12\ \text{V}$ with alternating current of $100\ \text{kHz}$; the phone's coil has $15$ turns. Observe…
A wireless charger has a transmitter coil of $20$ turns powered at $12\ \text{V}$ with alternating current of $100\ \text{kHz}$; the phone's coil has $15$ turns. Observe the diagram:
- a) Calculate the induced voltage in the phone's coil assuming ideal coupling, using the relationship $\frac{V_1}{N_1} = \frac{V_2}{N_2}$.
- b) Explain how this system resembles a transformer and what fundamental difference there is regarding the core compared to a conventional one.
- c) Reason, based on Faraday's law, why such a high frequency is used instead of the $50\ \text{Hz}$ mains frequency: what happens to $\frac{\Delta \Phi}{\Delta t}$ and to the required size of the coils?
The transformer
When you wind two independent coils on the same iron ring and connect one to alternating current, the second lights a bulb with no electrical contact between the two. How does the energy travel, and what does the voltage that the bulb receives depend on?
A transformer has two coils wound on the same iron core. Explain, based on Faraday's law, why connecting the primary to a constant direct current produces no voltage in…
A transformer has two coils wound on the same iron core. Explain, based on Faraday's law, why connecting the primary to a constant direct current produces no voltage in the secondary, whereas connecting it to alternating current does.
The primary of a transformer has $N_1 = 1000$ turns and is connected to $V_1 = 230\ \text{V}$. The secondary has $N_2 = 50$ turns. a) Write the turns ratio,…
The primary of a transformer has $N_1 = 1000$ turns and is connected to $V_1 = 230\ \text{V}$. The secondary has $N_2 = 50$ turns.
- a) Write the turns ratio, $\frac{V_1}{N_1} = \frac{V_2}{N_2}$.
- b) Calculate the secondary voltage and state whether the transformer is step-up or step-down.
Mains voltage, $230\ \text{V}$, needs to be transformed to $12\ \text{V}$ to power a model circuit. a) Calculate the required turns ratio, $\frac{N_1}{N_2}$. b) If the…
Mains voltage, $230\ \text{V}$, needs to be transformed to $12\ \text{V}$ to power a model circuit.
- a) Calculate the required turns ratio, $\frac{N_1}{N_2}$.
- b) If the secondary has $30$ turns, how many turns must the primary have?
A toy electric train operates at $18\ \text{V}$ with a current of $2.5\ \text{A}$, using an ideal transformer connected to the $230\ \text{V}$ mains. a) Calculate the…
A toy electric train operates at $18\ \text{V}$ with a current of $2.5\ \text{A}$, using an ideal transformer connected to the $230\ \text{V}$ mains.
- a) Calculate the turns ratio of the transformer.
- b) Calculate the power consumed by the train.
- c) Calculate the current drawn by the transformer from the mains.
Efficiency and losses in motors and transformers
Your charger gets warm, and so does the motor of a drill. If all the electrical energy were converted into useful work, they wouldn't heat up. Where does the rest go?
Where energy is lost
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Joule heating in the windings: the wire has resistance, and $I^2 R$ is dissipated as heat.
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Eddy currents in the core: the changing flux induces circular currents in the iron itself and heats it. That is why the core is built of thin laminations, stacked and insulated from each other, to cut off their path.
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Magnetic hysteresis: the iron wastes energy as it is magnetized and demagnetized in each cycle.
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Mechanical friction (in motors): bearings, brushes, and ventilation.
A drill connected to $230\ \text{V}$ draws a current of $2\ \text{A}$ and delivers a useful mechanical power of $400\ \text{W}$. a) Calculate the electrical power it…
A drill connected to $230\ \text{V}$ draws a current of $2\ \text{A}$ and delivers a useful mechanical power of $400\ \text{W}$.
- a) Calculate the electrical power it absorbs from the mains.
- b) Calculate its efficiency, as a percentage.
- c) Name two mechanisms by which a real motor loses energy and briefly explain them.
An ideal transformer has $N_1 = 460$ turns in the primary and $N_2 = 40$ in the secondary. The primary is connected to $230\ \text{V}$. Observe the diagram: a) Calculate…
An ideal transformer has $N_1 = 460$ turns in the primary and $N_2 = 40$ in the secondary. The primary is connected to $230\ \text{V}$. Observe the diagram:
- a) Calculate the secondary voltage.
- b) A current of $2.3\ \text{A}$ flows through the secondary circuit: calculate the power delivered by the transformer.
- c) In an ideal transformer, the primary power is equal to the secondary power: calculate the current drawn by the transformer from the mains.
A laptop charger delivers $19\ \text{V}$ and $38\ \text{W}$ to the laptop, and its transformer has an efficiency of $95\%$. The charger is connected to the…
A laptop charger delivers $19\ \text{V}$ and $38\ \text{W}$ to the laptop, and its transformer has an efficiency of $95\%$. The charger is connected to the $230\ \text{V}$ mains.
- a) Calculate the current flowing through the secondary.
- b) Calculate the power drawn by the charger from the mains and the corresponding primary current.
- c) Calculate the power lost inside the charger.
- d) Explain the two main causes of losses in a real transformer: eddy currents in the core (which is why it is made of stacked, insulated laminations) and the Joule effect in the windings.
Transmitting energy: why at high voltage
CFE's power plants are often far from the cities, and the energy travels hundreds of kilometers. What physical trick makes it possible to do this without losing almost everything along the way?
That is why at the power plant the voltage is stepped up before transmission (step-up transformer) and stepped down near where it is consumed (step-down transformer). The combination of alternating current and transformers is what makes it feasible to transmit electricity hundreds of kilometers with reasonable losses. With direct current you could not easily change the voltage, because transformers do not work with it.
Electricity companies transport energy at voltages of hundreds of kilovolts. a) Write the relationship between power, voltage and current, $P = V \times I$, and explain…
Electricity companies transport energy at voltages of hundreds of kilovolts.
- a) Write the relationship between power, voltage and current, $P = V \times I$, and explain why, to transport the same power, a high voltage implies a low current.
- b) The power lost due to the Joule effect in the cables is $I^2 \times R$: explain why it is so important for the transported current to be small.
A town consumes $50\ \text{kW}$ and the supply line feeding it has a total resistance of $2\ \Omega$. a) If transmission were carried out at $1000\ \text{V}$, calculate…
A town consumes $50\ \text{kW}$ and the supply line feeding it has a total resistance of $2\ \Omega$.
- a) If transmission were carried out at $1000\ \text{V}$, calculate the current through the line and the power lost in it, and express this loss as a percentage of the transmitted power.
- b) Repeat the calculations assuming transmission is carried out at $10\ \text{kV}$.
- c) Draw a conclusion as to why it is advantageous to transmit energy at high voltage.
A wind farm generates $10\ \text{MW}$ and transmits it at $220\ \text{kV}$ over a line with a total resistance of $40\ \Omega$. a) Calculate the current flowing through…
A wind farm generates $10\ \text{MW}$ and transmits it at $220\ \text{kV}$ over a line with a total resistance of $40\ \Omega$.
- a) Calculate the current flowing through the line.
- b) Calculate the power dissipated in the line and express it as a percentage of the generated power.
- c) Repeat the calculations assuming transmission were carried out at $25\ \text{kV}$.
- d) Explain the dual role played by transformers (and therefore alternating current) in making long-distance electrical power transmission viable.
From the power plant to your home: costs and benefits
Where does the energy that drives the alternator come from? Each source has its advantages and its problems. "Producing" electricity means converting stored energy (water in a dam, wind, heat from the Earth, nuclear fuel) into electrical energy, and concentrating it where and when it is needed.
In a hydroelectric plant, the energy flows like this:
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The impounded water has gravitational potential energy.
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As it falls, it is converted into kinetic energy of the water.
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The water drives the turbine (rotational kinetic energy).
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The turbine spins the rotor of the alternator and, by induction, electrical energy appears.
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Its voltage is stepped up at the step-up transformer, it travels along the line, and it is stepped down at the step-down transformer.
At each step there are losses in the form of heat, so the overall efficiency is always less than $100\ \%$. Energy is not destroyed, but it is transformed into less useful forms.
Every way of generating electricity has economic, social, and environmental costs and benefits, in the short and long term:
| Source | Benefits | Costs and problems |
|---|---|---|
| Hydroelectric (Chicoasén) | Economic: cheap energy once built, with a service life of decades and no fuel to buy. Environmental: emits no $\text{CO}_2$ during operation. Social: the reservoir also serves for flood control and irrigation | Economic: very high initial investment. Social: flooding land and relocating communities. Environmental: changes the river's ecosystem and depends on rainfall |
| Wind (Isthmus of Tehuantepec) | Economic: no fuel expense, low cost per kWh, and jobs in construction and maintenance. Environmental: emits no $\text{CO}_2$ during operation | Economic: requires backup from other plants when there is no wind. Social: conflicts over the use of communal lands and over rent payments. Environmental: impact on birds |
| Geothermal (Cerro Prieto) | Economic: runs day and night, without depending on imported fuels. Environmental: takes up little space | Economic: drilling wells is expensive and is only possible in areas with underground heat. Environmental: extraction of fluids and gases |
| Nuclear (Laguna Verde) | Economic: a lot of energy from little fuel. Environmental: no $\text{CO}_2$ during operation | Economic: very high construction and decommissioning costs. Environmental and social: radioactive waste and the need for strict safety |
In a hydroelectric power plant, water rotates the turbine, which drives the alternator rotor at a frequency of $25\ \text{Hz}$ (take $\pi \approx 3.14$). The alternator…
In a hydroelectric power plant, water rotates the turbine, which drives the alternator rotor at a frequency of $25\ \text{Hz}$ (take $\pi \approx 3.14$). The alternator winding is equivalent to $100$ turns of $0.2\ \text{m}^2$ rotating in a field of $0.8\ \text{T}$.
- a) Calculate the angular speed of the rotor.
- b) Calculate the maximum generated emf, $\varepsilon_0 = N \times B \times S \times \omega$.
- c) Describe the chain of energy transformations from the dammed water to the electrical energy leaving the power plant.
A power station generates electricity at $6\ \text{kV}$. A step-up transformer increases it to $132\ \text{kV}$ for transmission and, in the city, another transformer…
A power station generates electricity at $6\ \text{kV}$. A step-up transformer increases it to $132\ \text{kV}$ for transmission and, in the city, another transformer steps it down to $230\ \text{V}$ for consumption. Observe the diagram:
- a) Calculate the turns ratio $\frac{N_2}{N_1}$ of the step-up transformer.
- b) Calculate the turns ratio $\frac{N_1}{N_2}$ of the city step-down transformer.
- c) Explain why the step-up transformer is installed next to the power station and the step-down transformer next to the points of consumption, and not the other way round.
A popular science article proposes "transporting electricity directly at $230\ \text{V}$ to save the cost of transformers". A city requires a power of $1\ \text{MW}$ and…
A popular science article proposes "transporting electricity directly at $230\ \text{V}$ to save the cost of transformers". A city requires a power of $1\ \text{MW}$ and the line would have a resistance of $1\ \Omega$.
- a) Calculate the current that would have to be transmitted.
- b) Calculate the power that would be lost in the line and compare it with the power intended for transmission.
- c) In light of the result, explain why the proposal is unfeasible and how the combination of alternating current and transformers solves the problem.