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.

Motors and generators work based on this idea! 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 What is the relationship between current, magnetic fields and motion? This is an important concept to know! Current (Electrical Field) Magnet (Magnetic Field) Motion/ Movement When we have 2 of the above in a triangle, it produces the third 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 Motors convert electrical energy into mechanical energy 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

Color code the split ring commutator 1. In which diagram is there current flowing through the coil? 2. When there is current flowing through the coil, what else is generated? N S Magnetic field Step 1 and 3 Split ring commutator brushes N S N S Color code the split ring commutator N S 3. Based on what you know about motion, what keeps objects in motion? Momentum

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 _______. N magnetic magnetic N S rotate Split ring commutator brushes N S N S N N S 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? Increasing the number of coils Increase the strength of the magnets Changing the shape of the magnets Increasing the voltage supplied By wrapping current carrying wire around a piece of metal like iron 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 http://www.youtube.com/watch?v=Pv8WxPsJf4A&feature=youtu.be Watch it happen!

Generator Mechanical energy Electrical energy Motor

The shapes of the graphs are important! DC and AC Generators The shapes of the graphs are important! Both DC and AC generators exist, what is the difference? Direct current (DC): Electrons flow in one direction in a circuit Examples: batteries and solar panels Alternating current (AC): Electrons flow in both direction in a circuit Examples: anything that plugs into the wall We will study 2 types of current Direct Current or DC current flows in one direction only - electrons move from - to + Constant amount of current flowing Static magnetic field generated Ex. Batteries, solar cells Alternating Current or AC 2 way current electrons move back and forth and reverse directions at regular intervals Changing magnetic field Ex. Lamp, anything that plugs into the wall, all house hold circuitry Listen to this! Bill Nye shows us DC vs. AC (14:55 – 15:37)

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

Important Term - Current 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 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? _____________ #1 #2

Connecting batteries in series 4V 4V 4V 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!)

Connecting batteries in parallel 4V 4V 4V When batteries are connected in PARALLEL…. Total Voltage (VT) = V1 = V2 = V3 Example: VT = 4V ** This reduces energy output, but cells last longer !**

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 4 Ω 2 Ω 3 Ω Calculate the total resistance in the following circuits:

Example 1: Calculate the total resistance in the following circuits 4Ω 2Ω 3Ω 3 Ω 2 Ω 4 Ω 1 = 1 + 1 + 1 RT R1 R2 R3 RT = R1 + R2 + R3 = 4Ω + 2Ω + 3Ω = 9Ω 1 = 1 + 1 + 1 RT 4Ω 2Ω 3Ω 1 = 13 RT 12 Which circuit has greater resistance and why? Series, because the electrons have to pass through more resistors! RT = 12 or 0.92Ω 13

RT = ? RT = R1 + R2 + R3 R1 = 16Ω RT = 16Ω + 27Ω + 12Ω R2 = 27Ω 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) RT = ? RT = R1 + R2 + R3 R1 = 16Ω RT = 16Ω + 27Ω + 12Ω R2 = 27Ω RT = 55Ω R3 = 12Ω 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 = 120Ω RT = 8R1 R1 = ? 120Ω = 8(R1) But all the bulbs are the same R1= 120/8 = 15.0Ω

4, 3, 1, 2 For 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 General diagram 6V General diagram 10V 4V 10V 10V 10V Current through each resistor Is the same through every resistor Varies depending on the resistor Voltage through each component Varies depending on the resistor Is the same through every resistor branch Formula to calculate total resistance 1 = 1 + 1 + 1 RT R1 R2 R3 RT = R1 + R2 + …..+ RN Total Resistance through the circuit Higher than in parallel Lower than in series What happens when one resistor fails? All the others go out The rest still work

Ohm’s Law V = IR Current is directly proportional to voltage Equation is found on page 3 of the data booklet A Current (_________) V = IR Ω Resistance (_________) Voltage (_________) V Current is directly proportional to voltage as I goes up, V goes ____________ Current is inversely proportional to resistance as R goes up, I goes ____________ up down

Ohm’s Law Graphically Volts Current Resistance Current Current is inversely proportional to resistance

x ÷ x ÷ Practice converting units using page 1 of your data booklet 1. Convert 16 A to milliamps 16 000 16 A = ___________mA (we need to divide by 10-3) 2. Convert 500 kV to volts 500 kV = ___________V (we need to multiply by 103) 500 000

DC, since batteries produce direct current 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 I = 5.0 A V = IR R = 2.4Ω V = (5.0A)(2.4Ω) 12 V V = ? 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.

Click to watch a tutorial on #5 and #6 Ohm’s Law Example 4: Is this a parallel or series circuit? __________________ Does the current stay the same throughout the circuit? ____ How about voltage? ________ series yes no V = IR V = ? I = V R I =? V = (0.00587…A)(2505Ω) RT = R2 + R3 V = 17.26V V = 14.70622 V RT = 1125Ω + 1380Ω RT = R1 + R2 + R3 I = 17.26V 2940Ω V = 14.7V RT = 435Ω + 1125Ω + 1380Ω RT = 2505Ω RT = 2940Ω I =0.0058707A I =0.0058707A I =5.87 mA To convert A to mA we divide by 10-3

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 The wider prong on the polarized plug will permit it to be plugged in only with the correct polarity. The narrower prong is the "hot" lead and the switch to the appliance is placed in that lead, gauranteeing that no voltage will reach the appliance when it is switched off. A non-polarized plug may have the switch in the neutral leg and thus be a shock hazard even when it is switched off.

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?

Click to watch a review of circuits 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)

P=IV or P = I2R 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: Current (_____) A This is on page 3 of the data booklet! P=IV or P = I2R Resistance (_____) Ω W V Power (_____) Voltage (_____) Depending of what information is given, we need to determine which equation to use

Practice Problem: A speaker has a resistance of 4. 0  and allows 2 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 P = IV (but we don’t have V, so we use Ohm’s Law) I = 2.00A V = IR P = IV 16W P = ? V = (2.00A)(4.0Ω) P = (2.00A)(8.0V) V = 8.0V P = 16W

Practice Problem: Calculate the current running through a 100W light bulb with a resistance of 15. (Answers: 2.6A) I = ? P = I2R   I = 2.5819888A P = 100W 2.6A I2 = P R R = 15Ω  

Energy and Power Using power and energy consumed, utility costs can be calculated. First how do we calculate energy? J Energy (____) E = Pt Time (____) s Power (____) W This is on page 3 of the data booklet! 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

Complete the practice conversion questions What to do when the question asks for energy in kWh or when the question asks for energy in Joules? Energy in kWh Energy in Joules E = Pt E = Pt Power (W) Power (kW) Energy (J) Energy (kWh) Time (s) Time (h)   Don’t forget to use your basic skills intro and page 1 of the data booklet to help with conversions! Complete the practice conversion questions

x ÷ x ÷ Practice converting units using page 1 of your data booklet 4.5 hours to seconds 312 400 s to hours 2.4 days to seconds  Convert 25 W to kilowatts Convert 2.4 kW to watts Convert 14.5 J to megajoules Convert 320 000 MJ to joules   4.5 x 3600 = 16 200s 25 W = ___________mA (divide by 103) 0.025 2.4 kW = ___________W (multiply by 103) 2400 312 400 / 3600 = 86.6h 14.5 J = ___________MJ (divide by 10-6) 1.45 x 10-5 2.4 x 86 400 = 207 360s 320 000 MJ = ___________J (multiply by 10-6) 3.2 x 1011

To convert hours to seconds we multiply hours by 3600. 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 = 40W E = Pt t = 2 hours E = (40W)(2 hours x 3600s) E = ? 2.9 x 105J E = 288 000J Joules (J) Since the desired units is Joules (J), we need power in watts (W) and time in seconds (s) To convert hours to seconds we multiply hours by 3600.

E = Pt (but we don’t have P, so we use P = IV first) I = 0.91 A P = IV 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 = 100V E = Pt (but we don’t have P, so we use P = IV first) I = 0.91 A P = IV $0.08 E = Pt t = 12 hours P = (0.91A)(100V) E = (91/103)(12h) Cost = $0.07/kWh E = 1.092 kWh P = 91W Cost = (1.092 kWh)($0.07/kWh) Kilowatt-hours (kWh) Cost = $0.07644 To calculate cost of electricity, we first calculate the energy used, then multiply by the cost of electricity Since the desired units is kilowatt hours (kWh), we need power in kilowatts (kW) and time in hours (h) To convert W to kW divide by 103

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) E = ? E = Pt P = 8.0W E = (8.0W/103)(1 year x 365 days x 24 hours) 70 kWh t = 1 year E = 70.08 kWh Kilowatt-hours (kWh) Since the desired units is kilowatt hours (kWh), we need power in kilowatts (kW) and time in hours (h) Assume 1 year has 365 days

Cost = (70.08 kWh)(9.3cents/kWh) 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) Cost = (70.08 kWh)(9.3cents/kWh) Cost = 651.744 cents Cost = $6.52 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) $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 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 = 80W E = Pt t = 2 days E = (80W)(2 days x 24 hours x 3600s) E = ? $0.14 E = 13 824 000J Cost = $10/GJ E = 13 824 000J/109 E = 0.013824 GJ Cost = (0.013824 GJ)($10/GJ) Gigajoules (GJ) Cost = $0.13824 Since the desired units is Gigajoules (GJ), we to first calculate energy in Joules (J) then covert to gigajoules (GJ) To convert Joules to Gigajoules, we divide by 109 (according to page 1 of the data booklet)

Transmitting Electrical Energy Electrical energy generated at a power generating station must be transmitted to distribution stations before it arrives at our houses 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 Electrons do not flow through the transformer core Primary coil Secondary coil 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

Review: Circuits, Charges and Fields Recall: 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 Current (Electrical Field) Magnet (Magnetic Field) Motion/ Movement Try this animation!

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? Animation Watch it happen! An alternating current (AC) entering the primary coil will produce a changing magnetic field This changing magnetic field produced by the primary coil is concentrated by the iron core 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!

Where is electromagnetic induction used?

Where is electromagnetic induction used? How does wireless charging work? Watch this!

Step up Step down Both conserve power! 2 Types of Transformers Voltage is increased There are more secondary coils Step down Voltage is decreased There are more primary coils Both conserve power!

fewer more higher lower more fewer lower higher Stays the same

Np = Vp Ns Vs Np = Is Ns Ip Vp = Is Vs Ip Primary Coil Secondary Coil (input voltage) Secondary Coil (output voltage) Look on page 3 of the data booklet to find the 3 formulas used in transformer calculations Iron Core Np = Vp Ns Vs Np = Is Ns Ip Vp = Is Vs Ip N = number of turns, P = primary and s = secondary

This is a step up transformer because voltage increases! 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) Np = 368 Cross multiply Ns = 878 Vs = 82.2 V Vp = 34.45V Vs = ? Vs = 82.1932 V This is a step up transformer because voltage increases!

Np = 246 Ns = 466 Is = 5.54 V Ip = 10.5A Is = ? Is = 5.5429 A 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) Np = 246 Cross multiply Ns = 466 Is = 5.54 V Ip = 10.5A Is = ? Is = 5.5429 A 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) Ip = 25 A Cross multiply Is = 45 A Vp = 2.0 x 102V Vs = 110 V Vp = ? Vp = 198 V