Back to the on-screen lesson ·

Generators and transformers

Motional emf $BLv$ and the work it costs, generators and alternating current, transformers and why power lines run at high voltage.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to follow energy through induction and analyze generators, transformers and transmission lines.

2. What you already have

From the last lesson you know Faraday's law, $\mathcal{E} = -N\,\Delta\Phi/\Delta t$, and Lenz's law. From the magnetic force lesson you know $F = ILB$ and that a coil in a field feels a torque. This lesson follows the energy through induction and builds the machines that power the grid.

3. Words for this lesson

TermWhat it means
Motional emf$\mathcal{E} = BLv$ across a conductor of length $L$ moving at $v$ through a field $B$.
GeneratorA spinning coil in a field; peak emf $NBA\omega$.
Alternating currentCurrent that reverses periodically; $60$ Hz on the American grid.
TransformerTwo coils sharing a core; $V_2/V_1 = N_2/N_1$.
Step-up, step-downRaising or lowering voltage with more or fewer secondary turns.
Transmission loss$I^2R$ heating of power lines.

4. Induction turns work into electrical energy

Move a conductor through a magnetic field and the free charges in it feel a force $qvB$ along it, piling up at the ends: an emf of

$$\mathcal{E} = BLv.$$

Connect the ends in a circuit and a current flows. That current now feels a force $BIL$ from the field, pointing against the motion, by Lenz's law. To keep the conductor moving, something must push, and the work it does becomes electrical energy: mechanical power $Fv$ equals electrical power $\mathcal{E}I$. A coil spinning in a field is a generator. Two coils sharing a changing flux make a transformer, which trades voltage for current: $V_2/V_1 = N_2/N_1$.

Another way: picture

Picture pedaling a bicycle with a generator lamp. With the lamp off, pedaling is easy. Switch it on and the pedals push back a little: the current in the generator's coil feels a force opposing its motion. Your legs do the extra work, and it comes out as light. A power station does the same with steam or water instead of legs.

Another way: steps

  1. For a moving rod: $\mathcal{E} = BLv$, current $I = \mathcal{E}/R$, opposing force $BIL$.
  2. For a spinning coil: peak emf $NBA\omega$, with $\omega = 2\pi f$.
  3. For a transformer: $V_2 = V_1N_2/N_1$ and, if ideal, $V_1I_1 = V_2I_2$.
  4. For power lines: $I = P/V$ and loss $= I^2R$.
  5. Check: energy out never exceeds work in.

5. The sliding rod

Picture a metal rod lying across two parallel rails, the rails joined at one end, all in a magnetic field pointing into the page. Push the rod along the rails and the circuit's area grows, so the flux grows, and Faraday's law gives $\mathcal{E} = B\,\Delta A/\Delta t = BLv$. The same answer comes from the force on the charges in the moving rod.

The current that flows feels a force $BIL$ opposing the rod's motion. Stop pushing and the rod slows and stops, its kinetic energy turned to heat in the resistance. Keep pushing at a steady speed, and your power $Fv$ exactly equals the electrical power $\mathcal{E}I$.

6. Generators

A generator spins a coil in a magnetic field, or spins magnets past fixed coils. The flux through the coil swings back and forth, $\Phi = BA\cos\omega t$, and the induced emf follows $\mathcal{E} = NBA\omega\sin\omega t$, reaching its peak $NBA\omega$ as the coil passes edge-on, where the flux changes fastest.

The result is alternating current. In the United States, generators spin at rates that give $60$ cycles per second, so every clock on the grid can keep time by it. Whether the spinning comes from steam, falling water, wind or a gasoline engine, the electrical principle is Faraday's.

7. Transformers

A transformer wraps two coils around one iron core. Alternating current in the primary coil makes a changing flux that the core carries through the secondary. Every turn of either coil sees the same flux change and so the same emf, so the voltages are in the ratio of the turns: $V_2/V_1 = N_2/N_1$.

An ideal transformer wastes no energy, so $V_1I_1 = V_2I_2$: stepping the voltage down by ten steps the current up by ten. Real transformers are about ninety-eight to ninety-nine percent efficient. They need alternating current; a steady current makes a steady flux and induces nothing.

8. Why the grid runs at high voltage

Power lines have resistance, and they lose power $I^2R$ as heat. For a given power $P = IV$, raising the voltage lowers the current, and the loss falls as the square. Sending $100$ MW at $345$ kV instead of $138$ kV cuts the current by a factor of $2.5$ and the loss by more than six.

So generators' output, about $20$ kV, is stepped up to $115$ to $765$ kV for long-distance lines, then stepped down at substations to $7$ to $35$ kV for neighborhood lines, and finally by pole or pad transformers to $240$ V for homes. Transformers are why alternating current won the battle over Edison's direct current.

9. The method, step by step, and how to check it

  1. Identify the device: moving rod, spinning coil or transformer.
  2. Find the emf: $BLv$, $NBA\omega$, or $V_1N_2/N_1$.
  3. Find currents with Ohm's law or power balance.
  4. Check energy: mechanical work in must equal electrical energy out, plus losses.

Checking an answer. The force on a moving rod must oppose its motion. A step-down transformer must raise current as it lowers voltage. Power line losses must fall as voltage rises.

10. Why each step is allowed

The motional emf follows either from the magnetic force on charges in the moving rod or from the rate of change of flux as the circuit's area grows; both give $BLv$, as they must. Energy conservation ties the mechanical and electrical sides together exactly.

The transformer ratio assumes all the flux from one coil passes through the other and that the coils' resistance is negligible. Real transformers leak a little flux and heat a little, but the ideal model is accurate to a percent or two.

11. Hoover Dam and wind farms

At Hoover Dam on the Colorado River, water falling about $150$ m spins seventeen turbines, each driving a generator whose rotor carries electromagnets past fixed coils. Together they produce about $2000$ MW, supplying Nevada, Arizona and California.

A modern wind turbine in Iowa or Texas does the same with wind: blades turn a shaft, a gearbox or direct-drive design spins the generator, and electronics convert its output to the grid's $60$ Hz. Wind now supplies over a tenth of American electricity, and every kilowatt-hour comes from Faraday's law.

12. Regenerative braking

An electric or hybrid car's motor is also a generator. When the driver brakes, the car's motion spins the motor, which now induces a current that charges the battery. By Lenz's law, that current makes a force opposing the motion, slowing the car.

Regenerative braking recovers a substantial share of the energy that friction brakes would waste as heat, which is why electric cars get better efficiency in stop-and-go city traffic than on the highway. Subway trains in New York and many other cities feed braking energy back into the rails for other trains to use.

13. Chargers and adapters

The block on a laptop or phone charger contains a transformer, though modern designs first convert the $60$ Hz supply to tens of kilohertz, which lets a much smaller transformer do the job. It steps $120$ V down to $5$ to $20$ V, and rectifiers turn the result into direct current.

Older adapters with heavy iron transformers ran at $60$ Hz and hummed; their bulk came from the large core needed at low frequency. The faster the flux changes, the smaller the core needed for the same emf, one of Faraday's law's quieter gifts to portable electronics.

14. Power on the pole

The gray cylinders on American utility poles are distribution transformers. A typical one takes $7200$ V from the line and steps it down to the $240$ V that enters a house, split into two $120$ V halves by a center tap on the secondary coil. With a turns ratio of thirty to one, a household current of $100$ A draws only about $3.3$ A from the line.

Each transformer serves a handful of homes. On hot summer evenings, when every air conditioner runs, they work near their limits, and the hum you hear from them is the core vibrating as the flux reverses $120$ times a second.

15. Faraday's disk and the dynamo

The first generator was Faraday's own: in 1831 he spun a copper disk between the poles of a magnet and drew a steady current from sliding contacts at its rim and axle. Each radius of the spinning disk is a rod moving through the field, with a motional emf along it. The output was tiny, but it showed that motion could be turned into electricity continuously.

Fifty years later, engineers such as Werner von Siemens and Thomas Edison built dynamos with electromagnets and many-turn coils that produced real power. Edison's Pearl Street Station in lower Manhattan began supplying electricity to paying customers in 1882, lighting about four hundred lamps. Every power plant since, from Niagara Falls in 1895 to today's gas turbines, has been a larger and better version of Faraday's spinning disk.

16. In the world: high-voltage transmission

American utilities move power from plants to cities over lines at $115$ to $765$ kV. Sending $100$ MW at $345$ kV requires about $290$ A; through a line with $10$ Ω of resistance, the loss is $I^2R$, about $840$ kW, under one percent. At $138$ kV the current would be $725$ A and the loss over $5$ MW.

That is why the tall steel towers carry such high voltages, and why transformers sit at both ends of every line. The Energy Information Administration estimates that about five percent of American electricity is lost in transmission and distribution, mostly in the lower-voltage lines near homes. Utilities keep building higher-voltage lines to bring wind and solar power from the Great Plains and the desert Southwest to distant cities.

17. In the world: the pole transformer

A pole-mounted transformer on an American street steps the neighborhood line's $7200$ V down to the $240$ V that enters houses. If it serves homes drawing a total of $50$ A at $240$ V, that is $12$ kW, and on the primary side it draws only about $1.7$ A from the line, since an ideal transformer passes the same power at thirty times the voltage.

Line workers treat these transformers with great respect: the primary side is lethal, and a transformer can back-feed a line thought to be dead if a home's generator is connected improperly during an outage. That is why electrical codes require transfer switches for home backup generators, so the transformer cannot step their $240$ V up to $7200$ V on the line.

18. Generators convert energy; they do not make it

It is natural to think a generator produces electrical energy on its own. It only converts: the induced current's force opposes the motion, so whatever turns the generator must do work equal to the electrical energy delivered, plus losses. That is why power plants burn fuel or need falling water or wind.

A related error is to think a transformer that raises voltage also raises power. It raises voltage by lowering current in the same proportion; the power out can never exceed the power in.

19. A rod on rails

  1. A $0.50$ m rod slides at $4.0$ m/s through a $0.60$ T field on rails with $3.0$ Ω total resistance. Find the emf.

    $\mathcal{E} = 0.60 \times 0.50 \times 4.0 = 1.2\ \text{V}$

    Motional emf.

  2. Find the current.

    $I = \dfrac{1.2}{3.0} = 0.40\ \text{A}$

    Ohm's law.

  3. Find the magnetic force on the rod.

    $F = 0.60 \times 0.40 \times 0.50 = 0.12\ \text{N}$

    Opposing the motion.

  4. Find the mechanical power needed.

    $P = Fv = 0.12 \times 4.0 = 0.48\ \text{W}$

    Pushing at steady speed.

  5. Check the electrical power.

    $\mathcal{E}I = 1.2 \times 0.40 = 0.48\ \text{W}$

    Energy conserved.

20. A hand-crank generator

  1. A hand-crank flashlight's coil has $500$ turns of area $4.0$ cm² in a $0.20$ T field, turned $5.0$ times a second. Find the angular speed.

    $\omega = 2\pi \times 5.0 = 31.4\ \text{rad/s}$

    Five turns a second.

  2. Find the peak emf.

    $\mathcal{E}_{\max} = 500 \times 0.20 \times 4.0 \times 10^{-4} \times 31.4 = 1.26\ \text{V}$

    $NBA\omega$.

  3. Crank twice as fast. Find the peak emf.

    $\mathcal{E}_{\max} = 2.51\ \text{V}$

    Proportional to $\omega$.

  4. Describe the output.

    $\text{alternating, reversing each half turn}$

    A diode converts it for the battery.

  5. The LED draws $0.10$ A at $2.5$ V. Find the power.

    $P = 2.5 \times 0.10 = 0.25\ \text{W}$

    Electrical power out.

  6. Explain why cranking is harder with the light on.

    $\text{current in the coil opposes its turning}$

    Lenz's law.

21. Stepping down for a doorbell

  1. A doorbell transformer steps $120$ V down to $16$ V. Its primary has $900$ turns. Find the secondary turns.

    $N_2 = 900 \times \dfrac{16}{120} = 120$

    Turns ratio.

  2. The chime draws $0.75$ A. Find the power.

    $P = 16 \times 0.75 = 12\ \text{W}$

    Secondary side.

  3. Find the primary current, if ideal.

    $I_1 = \dfrac{12}{120} = 0.10\ \text{A}$

    Same power.

  4. Compare the current ratio with the turns ratio.

    $\dfrac{I_2}{I_1} = 7.5 = \dfrac{N_1}{N_2}$

    Currents go inversely with turns.

  5. Find the power if $95\%$ efficient.

    $P_1 = \dfrac{12}{0.95} = 12.6\ \text{W}$

    A little lost as heat.

  6. Explain why it works only on AC.

    $\text{DC gives steady flux, no induction}$

    Faraday's law needs change.

22. Your turn: a transformer has $400$ primary turns and $100$ secondary turns on $120$ V. What is the secondary voltage?

  1. Write the turns ratio rule.

    $V_2 = V_1\dfrac{N_2}{N_1}$

    Voltage per turn is shared.

  2. Substitute the values.

    $V_2 = 120 \times \dfrac{100}{400}$

    A quarter.

  3. Your turn: work this step out. Its working is at the end of the packet.

    Evaluate the voltage.

23. Guided practice

A metal rod $0.8$ m long moves at $4$ m/s through a uniform $0.3$ T field, with the rod, its velocity and the field all mutually perpendicular. What emf appears between its ends?

24. Guided practice

Complete the worked solution: an ideal transformer has $3000$ primary turns on $7200$ V AC and $50$ secondary turns. The secondary delivers $100$ A to a load. Find the secondary voltage in V, the power delivered in W, and the primary current in A.

  1. Find the secondary voltage.

    $V_2 = V_1\dfrac{N_2}{N_1} =$ o

    Turns ratio.

  2. Find the power delivered.

    $P = V_2I_2 =$ w

    To the load.

  3. Find the primary current.

    $I_1 = \dfrac{P}{V_1} =$ j

    Same power in as out.

  4. Compare the currents.

    $\text{stepping voltage down steps current up}$

    Their product is fixed.

25. Guided practice

Match each device to the principle it uses.

turns mechanical motion into an emfchanges an alternating voltage by a ratio of turnsturns current in a field into torqueuses induced currents to slow a moving conductor
generator
transformer
electric motor
magnetic brake

26. Practice

A rod $0.3$ m long slides at a steady $4$ m/s along metal rails through a $1.2$ T field; the circuit's resistance is $3$ Ω. Fill in the emf in V, the current in A, and the magnetic force on the rod in N, which a hand must match to keep it moving.

value
emf (V)
current (A)
force on the rod (N)

27. Practice

An ideal transformer's primary coil has $3600$ turns and is connected to $7200$ V AC. Write the secondary voltage, in V, as a function of the number of secondary turns $N$.

Answer:

28. Practice

A generator coil of $100$ turns, each of area $50$ cm², spins $60$ times a second in a $0.8$ T field. What is its peak emf, in V?

Answer: V

29. Somewhere new

A utility sends $50$ MW from a power plant to a city over a line of total resistance $5$ Ω at $69$ kV. How much power is lost as heat in the line, in kW?

Answer: kW

30. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

31. Test question

An ideal transformer's primary coil has $1000$ turns and is connected to $12000$ V AC. Write the secondary voltage, in V, as a function of the number of secondary turns $N$.

Answer:

32. What you can do now

You can analyze induction devices. Explain to someone why power lines carry such high voltages.

Working for the steps left to you

22. Your turn: a transformer has $400$ primary turns and $100$ secondary turns on $120$ V. What is the secondary voltage?, step 3

$V_2 = 30\ \text{V}$

Stepped down.