Grade 10

Advanced

10 min

Lesson 12 of 12

Motors, Generators, and Transformers

Not Started

Applies the principles of electromagnetism to real-world technology and power distribution.

Have you ever wondered why your phone charger gets warm, or how a giant dam can power a city hundreds of miles away without losing all its energy along the wires?

Learning Objectives

1.Distinguish between the energy conversions in electric motors and generators
2.Calculate primary and secondary voltages using the transformer turns ratio equation
3.Analyze how high-voltage AC transmission minimizes energy loss in the power grid

Motors vs. Generators: Two Sides of a Coin

At their core, motors and generators are nearly identical in construction but opposite in function. An Electric Motor converts electrical energy into mechanical energy. It uses the Lorentz Force: when current flows through a wire in a magnetic field, the field pushes the wire, causing rotation. Conversely, an Electric Generator converts mechanical energy into electrical energy. It operates on Faraday’s Law of Induction, where spinning a coil inside a magnetic field creates a changing magnetic flux, which induces a voltage (EMF) in the wire. One uses electricity to create motion; the other uses motion to create electricity.

Quick Check

If you manually spin the shaft of a simple DC motor, what does it effectively become?

Think about the direction of energy conversion.

Answer

It becomes a generator, producing an electric current from the mechanical motion.

Transformers and Mutual Induction

A transformer changes the voltage of Alternating Current (AC) using mutual induction. It consists of two coils—the primary and secondary—wrapped around an iron core. When AC flows through the primary coil, it creates a fluctuating magnetic field that 'links' to the secondary coil, inducing a new voltage. The ratio of the voltages is directly proportional to the number of turns (

N

) in each coil:

\frac{V_p}{V_s} = \frac{N_p}{N_s}

. A step-up transformer increases voltage (

N_s > N_p

), while a step-down transformer decreases it (

N_s < N_p

A phone charger uses a step-down transformer to convert $240V$ from a wall socket to $12V$ . If the primary coil has $1000$ turns, how many turns are in the secondary coil?

1. Identify knowns: $V_p = 240V$ , $V_s = 12V$ , $N_p = 1000$ . 2. Use the formula: $\frac{240}{12} = \frac{1000}{N_s}$ . 3. Simplify: $20 = \frac{1000}{N_s}$ . 4. Solve for $N_s$ : $N_s = \frac{1000}{20} = 50$ turns.

Quick Check

In an ideal transformer, if the voltage is stepped up by a factor of 10, what happens to the current?

Recall the Law of Conservation of Energy.

Answer

The current is reduced by a factor of 10 to conserve energy ( $P = VI$ ).

The Power Grid: Why High Voltage?

When electricity travels through long wires, some energy is lost as heat. This is known as Joule heating. The power lost is calculated by $P_{loss} = I^2R$ , where $I$ is current and $R$ is resistance. Because the loss depends on the square of the current, even a small reduction in current leads to a massive reduction in energy loss. By using transformers to step up the voltage to hundreds of thousands of volts, we can transmit the same amount of power ( $P = VI$ ) with a very low current, making long-distance distribution efficient.

Compare the power lost in a $10\Omega$ wire when transmitting $10,000W$ at $1,000V$ versus $10,000V$ .

1. At $1,000V$ : $I = \frac{P}{V} = \frac{10,000}{1,000} = 10A$ . Loss: $P_{loss} = (10^2) \times 10 = 1,000W$ . 2. At $10,000V$ : $I = \frac{10,000}{10,000} = 1A$ . Loss: $P_{loss} = (1^2) \times 10 = 10W$ . 3. Result: Increasing voltage by 10x reduced energy loss by 100x!

A power plant generates $50MW$ ( $50 \times 10^6 W$ ) of power. It is stepped up from $20kV$ to $500kV$ for transmission over a line with a total resistance of $20\Omega$ . Calculate the percentage of power lost during transmission.

1. Calculate transmission current: $I = \frac{P}{V} = \frac{50,000,000}{500,000} = 100A$ . 2. Calculate power loss: $P_{loss} = I^2R = 100^2 \times 20 = 200,000W$ (or $0.2MW$ ). 3. Calculate percentage loss: $\frac{0.2MW}{50MW} \times 100 = 0.4\%$ . 4. Note: If we hadn't stepped up the voltage, the loss would have exceeded the power generated!

Key Takeaways

1Motors convert electrical to mechanical energy; Generators convert mechanical to electrical energy.
2The transformer equation $\frac{V_p}{V_s} = \frac{N_p}{N_s}$ shows that voltage change depends on the ratio of coil turns.
3High-voltage transmission is essential because it lowers current, which exponentially reduces power loss ( $I^2R$ ).
4Transformers only work with Alternating Current (AC) because they require a changing magnetic field.

Quick Check

Multiple Choice

Which principle explains how a generator produces electricity?

Multiple Choice

A transformer has 200 primary turns and 800 secondary turns. If the input voltage is $110V$ , what is the output?

True/False

Power loss in transmission lines is reduced by increasing the current.

What's Next?

Review Tomorrow

In 24 hours, try to explain to someone why transformers don't work with Direct Current (DC) and sketch the $I^2R$ relationship.

Practice Activity

Look at the 'Input' and 'Output' labels on a laptop power brick. Calculate the turns ratio ( $N_p/N_s$ ) based on the voltages listed.

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Motors, Generators, and Transformers

Learning Objectives

Motors vs. Generators: Two Sides of a Coin

Transformers and Mutual Induction

The Power Grid: Why High Voltage?

Key Takeaways

Quick Check

What's Next?

Motors, Generators, and Transformers

Learning Objectives

Motors vs. Generators: Two Sides of a Coin

Transformers and Mutual Induction

The Power Grid: Why High Voltage?

Key Takeaways

Quick Check

What's Next?