Voltage, Current and Resistance

Similar documents
Regents Physics Mr. Mellon Based on Chapter 22 and 23

Resistance and Ohm s Law

Unit 6 ~ Learning Guide Name:

Activity Electrical Circuits Simulation

Resistance and Ohm s law

Activity Electrical Circuits Simulation

Chapter 12 Electric Circuits

1 V = IR P = IV R eq. 1 R i. = R i. = R eq. V = Energy Q. I = Q t

South Pasadena A.P. Physics Chapter Electric Current & DC Circuits Date / / Period Electricity Practice Test

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in

OHM'S LAW AND RESISTANCE NETWORKS OBJECT

Forces and Electrical Charges

Electromagnetism Unit- Current Sub-Unit

These are samples of learning materials and may not necessarily be exactly the same as those in the actual course. Contents 1.

INTRODUCTION TO CIRCUITS NOTES

Chapter 13. Electric Circuits

AP Physics - Problem Drill 14: Electric Circuits

Unit 3. Electrical Circuits

ENGINEERING COUNCIL CERTIFICATE LEVEL ENGINEERING SCIENCE C103

Electric Circuits. Physics 6 th Six Weeks

Circuitry II. Name: Date: Section C D F. Mr. Alex Rawson Physics

Electricity Practice Test 1

Why it s important: Electrical circuits are the basis of every electrical device, from electric lights to microwave ovens to computers.

Syllabus OP49 Test electrical conduction in a variety of materials, and classify each material as a conductor or insulator

DATE: NAME: CLASS: Drawing Circuit Diagrams

Resistance and Ohm s Law

Closed circuit complete path for electrons follow. Open circuit no charge flow and no current.

The following symbols are used in electric circuits:

ELECTRICAL CIRCUITS. All you need to be an inventor is a good imagination and a pile of junk. -Thomas Edison

Final Reg Current and Circuits Review SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Electric Current & DC Circuits

Electrical Measurements

RESISTANCE & OHM S LAW (PART I

Ohm s Law and Electrical Circuits

18-3 Circuit Analogies, and Kirchoff s Rules

Duration of resource: 23 Minutes. Year of Production: Stock code: VEA12041

2 Which arrangement of identical resistors would draw the most current when connected to the same potential difference?

Vocabulary. Electric Current. Electric Circuit. Open Circuit. Conductors. Insulators. Ohm s Law Current. Voltage. Resistance.

The answer is R= 471 ohms. So we can use a 470 ohm or the next higher one, a 560 ohm.

ENGR 1181 Lab 3: Circuits

Downloaded from

Chapters 34: Ohm s Law

Lightbulbs and Dimmer Switches: DC Circuits

Lab 5: Real DC Circuits

SUBELEMENT T5 Electrical principles: math for electronics; electronic principles; Ohm s Law 4 Exam Questions - 4 Groups

ELECTRIC Circuits Test

Wallace Hall Academy Physics Department. Electricity. Pupil Notes Name:

Experiment 3. Ohm s Law. Become familiar with the use of a digital voltmeter and a digital ammeter to measure DC voltage and current.

Experiment 2. Ohm s Law. Become familiar with the use of a digital voltmeter and a digital ammeter to measure DC voltage and current.

Chapter 23 Electric Current (cont.)

ELEXBO. Electrical - Experimentation Box

Series and Parallel Circuits Basics 1

Electric Circuits. Alternate Units. V volt (V) 1 V = 1 J/C V = E P /q V = W/q. Current I ampere (A) 1 A = 1 C/s V = IR I = Δq/Δt

Section A. Two resistors of 10 Ω and 15 Ω are connected in series to a battery of 6V. How can the values of current passing through them be compared?

EK307 Introduction to the Lab

Resistors in Series or in Parallel

Lab 3 DC CIRCUITS AND OHM'S LAW

HANDS-ON ACTIVITY 4 BUILDING SERIES AND PARALLEL CIRCUITS BACKGROUND WIRING DIRECTIONS

Any path along which electrons can flow is a circuit A Battery and a Bulb

Pre-Lab for Batteries and Bulbs

AC/DC ELECTRICAL SYSTEMS

Exercise 2: Ohm s Law Circuit Current

Electric Circuits Vocabulary

Book page Syllabus 2.8, 2.9, Series and parallel circuits

YAL. 12 Electricity. Assignments in Science Class X (Term I) IMPORTANT NOTES

CBSE TEST PAPER-01 CLASS - X Science (Electricity and its Effects)

Example 25 1: A total charge of 25 C passes through a wire every 5 seconds. What is the current in this wire?

Combined Series and Parallel Circuits

+ A Supply B. C Load D

ELECTRIC CIRCUIT PROBLEMS 12 AUGUST 2014

8) Name three more types of circuits that we will not study in this class.

Electric Circuits Notes 1 Circuits

charge time Electric Current and Circuits Current HEAT will flow if there is a difference in temperature

Chapters 35: Electric Circuits

Period 12 Activity Sheet Solutions: Electric Circuits

In this section you will learn about Ohm's Law as applied to a single resistor circuit. Phillips Textbook pp including some maths on notation.

Current, resistance, and Ohm s law

DC CIRCUITS AND OHM'S LAW

Module 1, Lesson 2 Introduction to electricity. Student. 45 minutes

Ohm's Law and the Measurement of Resistance

ELECTRIC CIRCUITS. 1. Which one of the following situations results in a conventional electric current that flows westward?

V (in volts) = voltage applied to the circuit, I (in amperes) = current flowing in the circuit, R (in ohms) = resistance of the circuit.

Resistive components in circuits

Unit 4: Electricity (Part 1)

The Discussion of this exercise covers the following points:

Radio Teacher Technician Test Subelement T4 Notes

SCRIPT. Voltage Dividers

Resistance and Ohm s Law R V I. 1 ohm = 1 volt ampere

Experiment 1: Circuits Experiment Board

Electricity Transition Questions Applied General in Science

A resistor adds resistance to a circuit. Describe what the effect of adding resistance would have on the current flowing in the circuit.

Section 4. Ohm s Law: Putting up a Resistance. What Do You See? What Do You Think? Investigate

Exercise 3: Voltage in a Series Resistive Circuit

Series and Parallel DC Circuits

Exercise 3: Ohm s Law Circuit Voltage

Practical 2.1 BASIC ELECTRICAL MEASUREMENTS AND DATA PROCESSING

Lab #1 Help Document. This lab will be completed in room 335 CTB. You will need to partner up for this lab in groups of two.

Conceptual Physics. Chapter 23: ELECTRIC CURRENT

Unit 3.C Electrical Theory, Circuits Essential Fundamentals of Electrical Theory, Circuits

Survival Skills for Circuit Analysis

Transcription:

Voltage, Current and Resistance Foundations in Engineering WV Curriculum, 2002 Foundations in Engineering Content Standards and Objectives 2436.8.3 Explain the relationship between current, voltage, and resistance, and use appropriate meters to measure each. Voltage, Current, and Resistance Over the last few days, you have learned about three different terms used in electricity and electronics: voltage, current, and resistance. Voltage is the pressure that pushes electrons through a conductor and is measured in volts (V). It results from the potential difference between a positive and a negative terminal. Voltage can be measured with a voltmeter and is always the measurement of the difference between two different parts of a circuit. The individual voltage measurements across all parts of a series circuit add up to the voltage across the power source. V1 V1 = V2 + V3 V3 The voltage measurements in a parallel circuit across each branch will be the same as the voltage across the power source. V2 V1 V1 = V2 = V3 V2 V3

Current is the quantity of electrons that flow through a conductor. Current is measured in amperes (A). In microelectronics, you often deal with very small current values expressed as milliamperes (ma), which equals 1/1000th of an ampere. Current is affected by the voltage between two points, but it is also affected by the resistance from different materials. Resistance is the opposition to the flow of electrons through a conductor. The type of material affects its resistance, as well as the length of the conductor and its temperature. Resistance is measured in ohms (R). It is often necessary to reduce the current flow in a circuit to protect components. Voltage divider is the term used to describe the method used for reducing the supply voltage from the battery. Electricians use specially constructed components called resistors that withstand the heat they produce. Each resistor has a power rating consisting of a coded set of colored bands that tells the electrician the value of the resistor. (see Transparency TM 10-3-1) To determine the value of a resistor, the numbers represented by the first two colored bands are multiplied by the number represented by the third band. Orange = 3, blue = 6 and red = 100 36 x 100 = 3600 R Resistors can be combined to produce a needed resistance. If they are combined in series, the total resistance equals the sum of all of the individual resistances. RT = R1 + R2 + R3 50 + 100 + 1000 = 1150 R If they are combined in parallel, the total resistance is determined by the following formula: 1 1 1 1 RT = R1 + R2 + R3 1 1 1 20 10 1 31 1 50 + 100 + 1000 = 1000 + 1000 + 1000 = 1000 = 32.2 R Often a circuit calls for different levels of voltage for different situations. A variable resistor, or potentiometer, is designed to adjust the flow of electricity

much like water is controlled through a tap. A 200 ohm variable resistor can be adjusted from 0 ohms to 200 ohms. Ohm s Law The units of volts, amperes, and ohms are mathematically related in Ohm s Law: 1 volt of pressure through 1 ohm of resistance equals 1 ampere of current. This mathematical law is used to calculate an unknown value of voltage, current, or resistance when the other two values are known. There are three ways of stating the relationship, and the graphic below will help you to remember the formulas: V V = Voltage in volts I = Current in amperes R = Resistance in ohms I R V = I x R I = V/R R = V/I Example: Find the current in an automobile light bulb that has a resistance of 16 ohms and operates on 12 volts. Choose the correct formula: Substitute known values: Current (I) = Voltage (V)/Resistance (R) I = 12/16 = 3/4 =.75 amps Activity 1. Have 10 Tronix kits, 10 multimeters, and 10 calculators for student group use during this activity. 2. Discuss voltage and current and how to calculate voltage measurements within a circuit. 3. Have students complete Handout 10-3-1. 4. Discuss resistance, how to determine the value of a resistor, (Transparency TM 10-3-1) and how to calculate the resistance in both series and parallel circuits. 5. Have students complete Handout 10-3-2. 6. Demonstrate how to use a multimeter to measure voltage and current. Remember that voltage is measured ACROSS the load being tested, and

current is measured IN the circuit. 7. Hand out the Tronix kits and multimeters for groups of students to use. Have students complete Handout 10-3-3, experimenting with resistors, variable resistors and multimeters. Review activity. 8. Discuss Ohm s Law and how to determine an unknown value of voltage, current or resistance when the other two values are known. (Transparency TM 10-3-2) A great website with a student tutorial and a lot of applets for students to do on-line experiments is at www.mste.uiuc.edu/users/noemi/ohm/menu.html 9. Have students complete Handout 10-3-4 and review in class.

Resistor Color Coding Transparency TM 10-3-1 4 BAND 25K 5% 5 BAND 460K 1% 6 BAND 276K 5%.01 Sil 10%.1 Gold 5% 0 0 0 1 1 1 1 10 1% 100 2 2 2 100 2% 50 3 3 3 1K 25 4 4 4 10K 15 5 5 5 100K.5% 6 6 6 1M.25% 7 7 7 10M.1% 8 8 8 9 9 9 1 ST DIGIT 2 ND DIGIT 3 RD DIGIT MULTIPLIER TOLERANCE TEMP COEF

Ohm s Law Transparency TM 10-3-2 There are three ways of stating Ohm s Law, and the graphic below will help you to remember the formulas: V = Voltage in volts I = Current in amperes R = Resistance in ohms V I R V = I x R I = V/R R = V/I

Figuring Voltage Handout 10-3-1 Calculate the voltage measurement in each of the following circuits. Make Sure to show your work. Remember: For Series circuits - the voltage is added VT = V1 + V2 + V3 For Parallel circuits - each branch remains the same VT = V1 = V2 = V3 1. 2. V? 36V 6V 3V 3V V? Lamps = 3 V each 3. 4. 36 V 24 V V 6 V 3 V V? V?

Figuring Voltage Handout 10-3-1 Answer Sheet Calculate the voltage measurement in each of the following circuits. Make Sure to show your work. Remember: For Series circuits - the voltage is added VT = V1 + V2 + V3 For Parallel circuits - each branch remains the same VT = V1 = V2 = V3 1. V4 3 V1 6 V2 2. V5 V1 36 V4 V3 V2 V3 3 V1 = V2 + V3 + V4 V1 = 6 + 3 + 3 V1 = 12 Lamps = 3 V each V1 = V2 + V3 + V4 + V5 36 = 3 + 3 + 3 + V5 36 = 9 + V5 V5 = 27 3. V1 36 V VV2 4. V4 6 V V1 24 V V2 2 V V3 V3 V1 = V2 = V3 36 = V2 36 = V3 V1 = V2 + V3 + V4 24 = 2 + V3 + 6 24 = 8 + V3 V3 = 16 V5 V3 = V5 16 = V5

Working with Resistors Handout 10-3-2 1-4. Figure the value of each resistor below from their color code. Make sure to show your work. R G O P B G R Bl Br Y W B R 5. Are the currents in both circuits the same? Why or why not? 60 R 100 R 10 V 40 R 10V 6. a. What is the total resistance of this circuit? b. If you wanted to be able to vary the resistance in this circuit from 0 ohms to the answer in a. above, what would you have to replace the current resistors with? 50R 30 R 10 V 20 R

7. Are the currents in both circuits the same? Why or why not? 12 V 12 V 30 R 60 R 30 R 8. What size resistor would you need to replace these 12 V resistors in this circuit? 400 R 400 R 200 R 9. Figure the value of this series combination of resistors below from their color code. Make sure to show your work. R G R Y P B G O R 10. Figure the value of this parallel combination of resistors below from their color code. Make sure to show your work. R G R G B R Br R G Br

Challenge Yourself! Any network of resistors in a circuit can be replaced by a single resistor. Its value is called the network s equivalent resistance. You can find the equivalent resistance of networks by simplifying the circuit step by step. See if you can figure out the equivalent resistance for the circuit below. It helps to draw a new circuit for each step. Show all work. 300 9V 500 400 400 Step 1. Simplify the 2 parallel 400 ohm resistors. Step 2. Simplify the series circuit. Step 3. Simplify the parallel circuit.

Working with Resistors Handout 10-3-2 Answer Sheet 1-4. Figure the value of each resistor below from their color code. Make sure to show your work. 25 x 1K = 25,000 705 x 100 = 70,500 or 25K 61 x 10K = 61,000 90 x 100 = 9,000 or 61K or 9K 5. Are the currents in both circuits the same? Why or why not? Yes, because in a series circuit, the values of the resistors is added together 40 + 60 = 100 60 R 100 R 10 V 40 R 10V 6. a. What is the total resistance of this circuit? 100 ohms b. If you wanted to be able to vary the resistance in this circuit from 0 ohms to the answer in a. above, what would you have to replace the current 50R resistors with? 100 ohm variable resistor (potentiometer) 30 R 10 V 20 R

7. Are the currents in both circuits the same? Why or why not? No, because in a parallel circuit, the inverse of the values of the resistors are added together. 30 R 12 V 12 V 3/60 = 1/20 R should equal 20 60 R 30 R 8. What size resistor would you need to replace these resistors in this circuit? 100 ohms 1/400 + 1/400 + 1/200 = 4/400 = 1/100 12 V 400 R 400 R 200 R 8. Figure the value of this series combination of resistors below from their color code. Make sure to show your work. R G R Y P B G O R (25 x 100)+(47 x 1)+(53 x 100) = 2500 + 47 + 5300 = 7847 R 10. Figure the value of this parallel combination of resistors below from their color code. Make sure to show your work. R G R G B R Br R G Br 1/R = 1/2500 + 1/ 5000 + 1/1250 = 2/5000 + 1/5000 + 4/5000 1/R = 7/5000 = 1/714.2 R = 714.2

Challenge Yourself! Any network of resistors in a circuit can be replaced by a single resistor. Its value is called the network s equivalent resistance. You can find the equivalent resistance of networks by simplifying the circuit step by step. See if you can figure out the equivalent resistance for the circuit below. It helps to draw a new circuit for each step. Show all work. 300 9V 500 400 400 Step 1. The 400 R and 400 R resistors are in parallel. They can be replaced by a single 200 R resistor. 1/R = 1/400 + 1/400 = 2/400 = 1/200 R = 200 9 500 300 200 Step 2. The 300 R and 200 R resistors are in series. They can be replaced by a single 500 R resistor. R = 300 + 200 = 500 9 500 500 Step 3. You are now left with a circuit with two 500 R resistors in parallel. Again, parallel 1/RT = 1/R1 + 1/R2 1/R = 1/500 + 1/500 = 1/R = 2/500 1/R = 1/250 R = 250 9 250

Materials and Equipment 2 6-volt bulbs 1 6-volt battery 1 0 to 10 volt voltmeter 1 1K variable resistor 1 2K variable resistor 1 1K resistor connecting wires baseboard Procedures Variable Resistors Handout 10-3-3 1. Assemble the circuit shown, using 2 6-volt bulbs. Measure the voltage across Bulb A using a voltmeter set at 10V or above. Then measure the voltage across Bulb B. What is happening in the circuit? A B Voltage A = Voltage B = 2. Replace Bulb A with a 200-ohm variable resistor. Adjust the variable resistor and record how dim the bulb goes, and how bright it goes. 3. Replace the 200-ohm variable resistor with a 100-ohm variable resistor. What difference does it make compared to the task above? 4. Adjust the variable resistor until the bulbs are just glowing. Now measure the voltage across Bulb B and Variable Resistor A. What did you find out? 5. Now replace Bulb B with a 1K resistor. Record the voltage range across 1K resistor B when Variable Resistor A is at either end of its movement by adjusting the control knob. Can you come up with a statement about the relationship of voltage to resistance?

Ohm s Law Handout 10-3-4 Ohm s Law V = I * R 1. What current would flow through the resistor in the circuit if the voltage across it were 5V? 500 2. What voltage would be needed to make 5 ma flow through the resistor? (Hint: Ohm s Law expresses current in Amps.) 3. To what value would you have to set the resistor at to make the current equal to 10 ma, if the voltage were 8 V? 4. The lamp is designed to work at a current of 60 ma. Its resistance is 100 ohms. What voltage would you have to put across the lamp to make it work at its working current? How much voltage would be needed if another lamp was added in series?

5. An LED uses only 2V and can take only 10mA without being damaged. Find the value of the resistor that will be needed in a series circuit if a 6-volt battery is used. 6V 6. Find the value of the resistor that will be needed in a series circuit if two 9-volt batteries are used. Express value in kilo-ohms. Critical Thinking 7. Now think about the relationship between the current, voltage and resistance. Show what would happen to the voltage if the current was doubled. What would the resistance be if the current was doubled? What would the current be if the voltage were doubled? (Hint: substitute these figures into your equations: voltage = 6, current = 3 and resistance = 2).

8. Below is a simplified schematic of the heater blower motor in a car. The switch can connect the circuit to each of the three pathways. The low pathway contains all three resistors in series. The medium pathway contains just two of the resistors. The high pathway only passes through one resistor. Use what you know about Ohm s law to explain why the heater blower motor runs at a different power on each of these settings. 9. Household voltage varies around the world. In the US we have a standard voltage of 110 volts. Some other countries don t even have a standard voltage. What would happen if you plugged a 110-volt hair dryer from the US into a 240-volt outlet in England? Explain your answer using Ohm s Law.

Ohm s Law Handout 10-3-4 Answer Sheet Ohm s Law V = I * R 1. What current would flow through the resistor in the circuit if the voltage across it were 5V? Looking for current (I), so formula is I = V/R I = 5/500 = 1/100 A =.01A = 10 ma (multiple by 1000) 500 2. What voltage would be needed to make 5 ma flow through the resistor? (Hint: Ohm s Law expresses current in Amps.) Looking for voltage (V), so formula is V = I * R Change 5 ma to A by dividing by 1000 =.005 A V =.005 A * 500 = 2.500 V 3. To what value would you have to set the resistor at to make the current equal to 10 ma, if the voltage were 8 V? Looking for resistance (R), so formula is R = V/I Change 10 ma to A by dividing by 1000 =.010 A R = 8/.010 = 800R 4. The lamp is designed to work at a current of 60 ma. Its resistance is 100 ohms. What voltage would you have to put across the lamp to make it work at its working current? How much voltage would be needed if another lamp was added in series? Looking for voltage (V), so formula is V = I * R Change 60 ma to A by dividing by 1000 =.06 A V =.06 A * 100 = 6 V a. Change 10 ma to A 10/1000 =.01 A b. R = V/I R = 4/.01 = 400 ohms

5. An LED uses only 2V and can take only 10mA without being damaged. Find the value of the resistor that will be needed in a series circuit if a 6-volt battery is used. c. Find voltage across resistor. VT = V1+ V2 6 = V1 + 2 Voltage across resistor = 4 volts 6V a. Find voltage across resistor. VT = V1+ V2 18 = V1 + 2 Voltage across resistor = 16 volts b. Change 10 ma to A 10/1000 =.01 A c. R = V/I R = 16/.01 = 1600 ohms or 16K ohms 6. Find the value of the resistor that will be needed in a series circuit if two 9-volt batteries are used. Express value in kilo-ohms. 7. Critical Thinking Now think about the relationship between the current, voltage and resistance. Show what would happen to the voltage if the current was doubled. What would the resistance be if the current was doubled? What would the current be if the voltage were doubled? (Hint: substitute these figures into your equations: voltage = 6, current = 3 and resistance = 2). Original Double Current What happens? V = I * R 6 = 3 * 2 V = 6 * 2 = 12 current doubled voltage doubles R = V/I 2 = 6/3 R = 6/6 = 1 current doubled resistance halved I = V/R 3 = 6/2 I = 12/2 = 6 voltage doubled current doubled

8. Below is a simplified schematic of the heater blower motor in a car. The switch can connect the circuit to each of the three pathways. The low pathway contains all three resistors in series. The medium pathway contains just two of the resistors. The high pathway only passes through one resistor. Use what you know about Ohm s law to explain why the heater blower motor runs at a different power on each of these settings. As the resistance is cut down, the current increases. Therefore, because the low path has the most resistance, it will run the slowest. As the resistance is made lower, the current will be higher, and the blower will run faster. 9. Household voltage varies around the world. In the US we have a standard voltage of 110 volts. Some other countries don t even have a standard voltage. What would happen if you plugged a 110 volt hair dryer from the US into a 240 volt outlet in England? Explain your answer using Ohm s Law. As the voltage increases, so does the current. Too much current through the hair dryer would burn up the components.