Topic 4: Motors and Generators Where do we see motors in everyday

life?

What makes your phone vibrate?

Important Terms to Know • Current: the movement of electrons in a

circuit

• Electrical Energy: Energy due to the movement of charge (electrons)

• Mechanical Energy: energy due to movement or position of an object

• Chemical potential energy: energy stored in the bonds of a molecule (ex. Battery acid)

Energy can be transformed from one form to another to so that it is in a form we can use. For example, solar energy is

transformed into chemical potential energy in plants. Motors and generators

are similar to this, but have different energy transformations.

Electrons flowing through a metal wire produces a magnetic field, just like a magnet

Motors and generators work based on this idea!

Bill Nye explains induction! (5:06 – 6:20)

Current (Electrical Field)

Magnet(Magnetic Field)

Motion/ Movement

When we have 2 of the above in a triangle, it produces the third

What is the relationship between current, magnetic

fields and motion?

This is an important concept to

know!

Try this applet to see it happen

Simple Electric Motor

What is the purpose of having a motor? Motors convert electrical energy into mechanical energy

All motors consist of a few key components:

Magnets

Coil of wire

Power source

How does a motor work? 1. As current flows through the coil, a magnetic field is generated (induced magnetic field)

2. The induced magnetic field generated by the coil will interact with the permanent magnet causing turning of the coil

3. The armature can be connected to blades that turn in a blow dryer

NN SS

brushes

Split ring commutator

NN SS

NN SS

NN SS

1. In which diagram is there current flowing through the coil?

Step 1 and 3

2. When there is current flowing through the coil, what else is generated?

Magnetic field

3. Based on what you know about motion, what keeps objects in motion?

Momentum

Color code the split ring commutator

NN SS

brushes

Split ring commutator

NN SS

NN SS

NN

NN

NN SS

Step 1 and 3: Current flow through the coil and a ___________ field is generated. This ___________ field interacts with the permanent magnet, causing the coil to _______.

magneticmagnetic

rotate

Step 2 and 4: When there is no current, there is no ____________ field. The coil will continue to spin due to momentum.

magnetic

Magnetic Fields can be intensified!!!

• How are stronger motors built?1. Increasing the number of coils 2. Increase the strength of the

magnets 3. Changing the shape of the magnets 4. Increasing the voltage supplied5. By wrapping current carrying wire

around a piece of metal like iron

We are going to build this in class!We are going to build this in class!

From Motors to Generator• Generators have the opposite

energy conversions as a motor and converts mechanical energy into electric energy

• Electrical induction (producing a current) occurs when a coil moves next to a magnet or a magnet moves within a coil

• In the diagram to the right, the coil is moving within a magnet and electricity is generated

Watch it happen!

Motor

Generator

Mechanical energy

Electrical energy

DC and AC Generators• Both DC and AC generators exist, what is the difference?

Listen to this!Bill Nye shows us DC vs. AC (14:55 – 15:37)

The shapes of the graphs are important!

Direct current (DC): Electrons flow in one direction in a circuitExamples: batteries and solar panels

Alternating current (AC): Electrons flow in both direction in a circuit Examples: anything that plugs into the wall

Types of Circuits

Series Circuit •There is one single loop (one pathway for current to flow)•Example: old Christmas lights•When one burns out, they all go out!

Parallel Circuit• There are loops within a loop

(multiple pathways for current to flow)

• Example: Car & home, power bar • When one burns out, the others

still work!

Current and circuits What is a circuit and where do you see them?

• Amount of charge or electrons that passes a point per second

• Electrons move from negative to positive (not the same as an electric field)

• Current (I) is measured in Amperes (A) using an ammeter connected in series in a circuit

Important Term - Current

What does high current mean?

Energy Source - Batteries• Batteries are usually represented by • Batteries create voltage, which is the amount of electrical

potential energy per unit charge available for doing work• Voltage is what causes the movement of electrons and creates a

current• Measured in volts (V) with a voltmeter which is connected parallel in

the circuit.

Practice Questions:

Which circuit diagram below correctly shows the connection of ammeter and voltmeter?

#4, since the ammeter is in series and voltmeter is in parallel

In the circuit diagram below, label which (1,2,3 and 4) are ammeters and which are voltmeters?

In the circuit diagram above, where should an ammeter be located to correctly measure the total current? ______________

In the circuit diagram above, where should an ammeter be located to correctly measure the current on the shorter path? _____________

ammeter

ammeter

voltmetervoltmeter

#1

#2

Connecting batteries in series

• When batteries are connected in SERIES….– Total Voltage (VT) = V1 + V2 + V3

– Example: VT = 4V + 4V + 4V = 12V

** This allows for increased energy output!**

(a brighter light bulb!)

4V 4V 4V

Connecting batteries in parallel

•When batteries are connected in PARALLEL….– Total Voltage (VT) = V1 = V2 = V3

– Example: VT = 4V

** This reduces energy output, but cells last longer !**

4V

4V

4V

VT = V1 + V2 + V3 + V4

VT = 1.5V + 1.5V + 1.5V +1.5V VT= 6V

6.0 V

Resistor• A resistor is any device in a circuit that disrupts electron

flow, generating heat (thermal energy) or light – Ex. a light bulb, toaster, lights, stereo and even a wire!

• Resistance is measured in Ohms (Ω) with an ohmmeter • The symbol for resistance is

Calculating Resistance

• In series These are on page 3 of the data booklet!

RT = R1 + R2 + R3

• In parallel 1 = 1 + 1 + 1 RT R1 R2 R3

3 Ω

2 Ω

4 Ω

Calculate the total resistance in the following circuits:

4 Ω 2 Ω 3 Ω

4Ω 2Ω 3Ω

Example 1: Calculate the total resistance in the following circuits

3 Ω

2 Ω

4 Ω

RT = R1 + R2 + R3

= 4Ω + 2Ω + 3Ω = 9Ω

1 = 1 + 1 + 1 RT R1 R2 R3

1 = 1 + 1 + 1 RT 4Ω 2Ω 3Ω

1 = 13 RT 12

RT = 12 or 0.92Ω 13

Which circuit has greater resistance and why?

Series, because the electrons have to pass through more resistors!

Example 2: What is the equivalent (total) resistance in a series circuit containing a 16Ω light bulb, a 27Ω heater and a 12Ω motor? (Answer: 55Ω)

Example 3: A string of 8 lights connected in series has a total resistance of 120Ω. If the lights are identical, what is the resistance of each bulb? (Answer: 15.0Ω)

RT = ?R1 = 16ΩR2 = 27ΩR3 = 12Ω

RT = R1 + R2 + R3

RT = 16Ω + 27Ω + 12Ω

RT = 55Ω

RT = 120ΩR1 = ?But all the bulbs are the same

RT = 8R1

120Ω = 8(R1)

R1= 120/8 = 15.0Ω

4, 3, 1, 2For this question, you could put in

values for each resistor and do calculations or think about the number of resistors an electron

will flow through

Practice: Draw a circuit diagram with 2 cells (batteries) in series and 2 resistors in parallel. Don't forget voltmeters (2), ammeters (3) and switch that turns off the entire circuit.

Don’t forget that voltmeters are placed in parallel and ammeters are placed in series!

Series Circuit Parallel Circuit

Current through each resistor

General diagram

Voltage through each component Formula to calculate total resistance

What happens when one resistor fails?

Total Resistance through the circuit

Is the same through every resistor Varies depending on the resistor

Varies depending on the resistor Is the same through every resistor branch

RT = R1 + R2 + …..+ RN 1 = 1 + 1 + 1 RT R1 R2 R3

Higher than in parallel Lower than in series

All the others go out The rest still work

10V

6V

4V 10V 10V 10V

Ohm’s Law

• Current is directly proportional to voltage – as I goes up, V goes ____________

• Current is inversely proportional to resistance – as R goes up, I goes ____________

V = IRV

Ω

A

Equation is found on page 3 of the data booklet

up

down

Voltage (_________)

Current (_________)

Resistance (_________)

Ohm’s Law Graphically

Volts

Current Resistance

Current

Current is inverselyproportional toresistance

x

x

÷

÷

Practice converting units using page 1 of your data booklet

1. Convert 16 A to milliamps

16 A = ___________mA (we need to divide by 10-3)16 000

2. Convert 500 kV to volts

500 kV = ___________V (we need to multiply by 103)500 000

Ohm’s Law Example 1: A headlight in an automobile draws a current of 5.0A with a 2.4Ω resistance. Is the current passing through the headlamp AC or DC? How do you know?

____________________________________________________________________________ What is the voltage that is drawn from the car battery? (Answer: 12V)

DC, since batteries produce direct current

R = 2.4Ω I = 5.0 A

V = ?

V = IRV = (5.0A)(2.4Ω)V = 12 V

12 V

Example 2: You must rearrange the formula to solve for R.

Example 3: You must rearrange the formula to solve for I.

Ohm’s Law Example 4:

R1 = 435 R3 = 1380 R2 = 1125 Is this a parallel or series circuit? __________________

Does the current stay the same throughout the circuit? ____

How about voltage? ________

series

yes

no

RT = R1 + R2 + R3

I =?V = 17.26V

RT = 435Ω + 1125Ω + 1380Ω

RT = 2940Ω

I = V R

I = 17.26V 2940Ω

I =0.0058707A

I =5.87 mA

To convert A to mA we divide by 10-3

V = ?

RT = R2 + R3 RT = 1125Ω + 1380Ω

RT = 2505Ω

I =0.0058707A

V = IRV = (0.00587…A)(2505Ω)

V = 14.70622 V

V = 14.7V

Click to watch a tutorial on #5 and #6

Protection from Electricity!

• When your body becomes part of the circuit for electricity…..you get zapped!!

• This can be dangerous!• Fuses, Circuit Breakers and polarized plugs are

designed to protect us

Fuses• Fuses are placed in series in circuit• If current is too high, fuse conductor

overheats & melts• Circuit is then broken• Fuses use materials with

– low melting point– narrow cross section– less conductive material

Is pencil lead a conductor?

Circuit Breakers• Opens circuits automatically if current

is too high• Found in fuse boxes in basements of

homes• If current rises above a safe level,

breaker trips and circuit is broken• Without breaker, wire could

overheat!!!

Click to watch a review of circuits (Bozeman Science)

Energy and Power• Power can be defined as the rate of doing work or

transforming energy • Power Is measured in Watts, W or J/s• Equations for calculating power:

P=IV or P = I2RW

Ω

A

V

This is on page 3 of the data booklet!

Depending of what information is given, we need to determine which equation to use

Power (_____)

Current (_____)

Voltage (_____)

Resistance (_____)

Practice Problem: A speaker has a resistance of 4.0 and allows 2.00A of current to flow through. Calculate the power consumed by the speaker. (Answers: 16W)

R = 4.0Ω I = 2.00A

P = ?

P = IV (but we don’t have V, so we use Ohm’s Law)

V = IRV = (2.00A)(4.0Ω)V = 8.0V

P = IVP = (2.00A)(8.0V)P = 16W

16W

Practice Problem: Calculate the current running through a 100W light bulb with a resistance of 15. (Answers: 2.6A)

P = 100WI = ?

R = 15Ω

P = I2R2.6A

I2 = P R

I = 2.5819888A

• Using power and energy consumed, utility costs can be calculated.

• First how do we calculate energy?

• Energy can be expressed as Joules (J) or kilowatt hours (kWh)• Depending on what the question is asking for, we need to convert

our variables into the correct units

Energy and Power

E = PtW

J

s

This is on page 3 of the data booklet!

Energy (____)

Power (____)Time (____)

Energy in kWh Energy in Joules

Don’t forget to use your basic skills intro and page 1 of the

data booklet to help with conversions!

What to do when the question asks for energy in kWh or when the question asks for energy in Joules?

E = Pt

Power (kW)

Energy (kWh)Time (h)

E = Pt

Power (W)

Energy (J)Time (s)

Complete the practice conversion

questions

Practice converting units using page 1 of your data booklet

x

x

÷

÷

1. Convert 25 W to kilowatts

1. Convert 2.4 kW to watts

1. Convert 14.5 J to megajoules

1. Convert 320 000 MJ to joules

5.4.5 hours to seconds

5.312 400 s to hours

5.2.4 days to seconds

25 W = ___________mA (divide by 103)0.025

2.4 kW = ___________W (multiply by 103)2400

14.5 J = ___________MJ (divide by 10-6)1.45 x 10-5

320 000 MJ = ___________J (multiply by 10-6)3.2 x 1011

4.5 x 3600 = 16 200s

312 400 / 3600 = 86.6h

2.4 x 86 400 = 207 360s

Power Problem 1: If a 40 W bulb is on for 2 hours, how much energy (J) does it use? (Answers: 2.9 x 105 J)

P = 40Wt = 2 hours

E = Pt

Joules (J)

To convert hours to seconds we multiply hours by 3600.

E = ?

Since the desired units is Joules (J), we need power in watts (W) and time in seconds (s)

E = (40W)(2 hours x 3600s)

E = 288 000J 2.9 x 105J

Power Problem 2: A light bulb operates on a 100 V circuit and draws 0.91 A. Assume you turn on the lights for 12 hours a day, what will it cost per day, if utility rate is $0.07/kWh? (Answers: $0.08)

V = 100VI = 0.91 A

E = Pt (but we don’t have P, so we use P = IV first)

Kilowatt-hours(kWh)

To calculate cost of electricity, we first calculate the energy used, then multiply by the cost of electricity

t = 12 hours

Since the desired units is kilowatt hours (kWh), we need power in kilowatts (kW) and time in hours (h)

P = IV

P = (0.91A)(100V)

$0.08

Cost = $0.07/kWhP = 91W

E = PtE = (91/103)(12h)E = 1.092 kWh

To convert W to kW divide by 103

Cost = (1.092 kWh)($0.07/kWh)Cost = $0.07644

Power Problem 3: Did you know most models of TVs and DVD players use electrical energy even when they are turned off? This stand-by power is used to run clocks and to provide an “instant on” feature, allowing electronics to become operational with a click of the remote control. Average values for stand-by power is about 8.0 W. Since this power is required 24 h a day, the electrical energy consumption is significant. a. Determine the electrical energy required to supply 8.0 W of stand-by power for both a TV and a DVD player during one year. Express your answer in kilowatt-hours. (Answers: 70kWh)

P = 8.0Wt = 1 year

E = Pt

Assume 1 year has 365 days

E = ?

E = (8.0W/103)(1 year x 365 days x 24 hours)

E = 70.08 kWh

70 kWh

Kilowatt-hours(kWh)

Since the desired units is kilowatt hours (kWh), we need power in kilowatts (kW) and time in hours (h)

b. If the price of electricity is 9.3¢/kWh, determine the cost in dollars of providing stand-by power to the DVD and TV for 365 days (one year). (Answers: $6.51)

c. There are about 2.0 million TVs and DVDs that operate with stand-by power in Alberta. Use this fact to estimate the total annual cost of maintaining stand-by power for all of these devices in Alberta. (Answers: $13 million)

Cost = (70.08 kWh)(9.3cents/kWh)Cost = 651.744 cents

Cost = $6.52

$6.52 x 2 000 000 = $13 034 880 = ~$13 million dollars

Power Problem 4: Calculate the cost of running an 80 W bulb for 2.0 days if the utility cost is $10/GJ. (Answers: $0.14)

P = 80Wt = 2 days

E = Pt

Gigajoules (GJ)

To convert Joules to Gigajoules, we divide by 109 (according to page 1 of the data booklet)

E = ?

Since the desired units is Gigajoules (GJ), we to first calculate energy in Joules (J) then covert to gigajoules (GJ)

E = (80W)(2 days x 24 hours x 3600s)

E = 13 824 000J $0.14

E = 13 824 000J/109

E = 0.013824 GJ

Cost = $10/GJ

Cost = (0.013824 GJ)($10/GJ)Cost = $0.13824

Electrical energy generated at a power generating station must be transmitted to distribution stations before it arrives at our houses

Transmitting Electrical Energy

Why? In order to conserve power, but reduce energy loss over long distances, we want to DECREASE resistance and current (electrons create heat) and INCREASE voltage during

transmission

The current passing through the transmission

lines is usually AC

Label this picture in your workbooks

Transformers

• The voltage carried in power lines must be

high (550kV) to be efficient but is too high for most house hold appliances (240V/120V)

• In order to change voltage (and current) of the electricity being delivered to our homes, we use transformers – Transformers may be step up - increasing

voltage (at generating plant) or step down – decreasing voltage (at home)

Key Components of a Transformer

• Transformers are made of 2 coils of insulated wire wrapped around an iron core

• The primary coil receives an input voltage and the secondary coil supplies the output voltage

Electrons do not flow through the transformer

corePrimary coil Primary coil Secondary coil Secondary coil

Review: Circuits, Charges and FieldsRecall: • When electrons move in a circuit this creates a electric field

and with a moving/changing magnetic field around a coil of wire creates a(n) magnetic field

Try this animation!

Current (Electrical Field)

Magnet(Magnetic Field)

Motion/ Movement

Review: Circuits, Charges and Fields

Direct current – electrons move in one direction, producing a static magnetic field

Alternating current – electrons move in both directions, producing a changing magnetic field

How does current travel from the primary coil to the secondary coil?

1. An alternating current (AC) entering the primary coil will produce a changing magnetic field

2. This changing magnetic field produced by the primary coil is concentrated by the iron core

3. The changing magnetic field induces an alternating current (AC) in the second coil

• Electrons DO NOT travel through the iron core!

4. The principle used by transformers is electromagnetic induction!

AnimationWatch it happen!

Where is electromagnetic induction used?

Where is electromagnetic induction used?

How does wireless charging work? Watch

this!

2 Types of Transformers

Step up• Voltage is increased• There are more secondary

coils

Step down• Voltage is decreased• There are more primary coils

Both conserve power!

fewer more higher lowerStays the

same

more fewer lower higherStays the

same

Primary Coil(input voltage)

Secondary Coil(output voltage)

Iron Core

Np = Vp Ns Vs

Np = Is Ns Ip

Vp = Is Vs Ip

Look on page 3 of the data booklet to find the 3 formulas

used in transformer calculations

N = number of turns, P = primary and s = secondary

Transformer Problem 1: Calculate the voltage produced by the transformer if the number of primary coils is 368, the number of secondary coils is 878 and the voltage in the primary coil is 34.45 volts. Is this a step up or down transformer? (Answer: 82.2V, step up)

Ns = 878

Np = 368

Vp = 34.45V

Vs = ?

Cross multiply

Vs = 82.1932 V

Vs = 82.2 V

This is a step up transformer because voltage increases!

Transformer Problem 2: Calculate the current produced by the transformer if the number of primary coils is 246, the number of secondary coils is 466 and the current in the primary coil is 10.5 A. Is this a step up or down transformer (Answer: 5.54A, step up)

Ns = 466

Np = 246

Ip = 10.5A

Is = ?

Cross multiply

Is = 5.5429 A

Is = 5.54 V

This is a step up transformer because the number of secondary turns is higher!

Transformer Problem 5: Calculate the voltage input into a transformer with a current of 25A in the primary coil, 45A in the secondary coil and a voltage of 110 V in the secondary coil. Identify the transformer type as step up or step down. (Answer: 2.0 x 102 V/ Step down)

Is = 45 A

Ip = 25 A

Vs = 110 V

Vp = ?

Cross multiply

Vp = 198 V

Vp = 2.0 x 102V