Math-in-CTE Lesson Planarchives.lessoncorner.com/8578a7d9bbaede1c3.pdfPA GOVERNOR’S INSTITUTE--2005 Ohm’s Law, Series Circuits 1 Math-in-CTE Lesson Plan Lesson Title: Using Ohm’s

PA GOVERNOR’S INSTITUTE--2005 Ohm’s Law, Series Circuits 1

Math-in-CTE Lesson Plan

Lesson Title: Using Ohm’s Law in Telephone Circuits Lesson Number: 20

Occupational Area: Information Technology CTE Concept(s): Use Ohm’s Law to find missing circuit values. Math Concepts: Solving 3-variable linear equations for the missing variable when two variables are given.

Lesson Objective: Students will demonstrate a working knowledge of solving problems involving formulas with specified symbols (math concept) and its application in telecommunications (technical field), while recognizing it in other contexts.

Supplies Needed: Telecommunications Trainer Lab Manual, RSR Electronics, 2000 Fill-in-the –blank worksheets with Ohm’s Law terms Crossword puzzles to reinforce Ohm’s law formulas and calculations Multi-meter and Telecommunications Trainer set up for a phone circuit to measure voltage, resistance, and current. Calculator to compute missing values of the circuit. Comparison of measured circuit current (with multi-meter) to calculated current (using Ohm’s Law with calculator) for accuracy.


THE "7 ELEMENTS" TEACHER NOTES (and answer key)

1. Introduce the CTE lesson. When your lights dim during a storm what is really happening? Why would you want to be able to figure out the effects of varying resistance on a power line of a fixed voltage? So, how do you figure out the amount of current in the power lines? What two Ohm’s Law factors determine current? What Ohm’s Law formula can be used to determine the value of current?

This serves as an explanation to the dimming of the lights in your house during an electrical storm. The electricity in the air affects the resistance of the power line which affects the current flow to your house. This can be visualized by flickering lights during a storm. Dimming lights are caused by reduced current caused by increased resistance caused by electrical storms near the power lines that supply your house.

You would need to know the voltage and the resistance of the power line to determine the current. This Ohm’s Law relationship can be expressed using the formula:

I=E/R, where I=current, E=voltage, and R=resistance.

2. Assess students’ math awareness as it relates to the CTE lesson. The same sort of change happens when the resistance varies in a telephone circuit. Instead of flickering lights we experience static on the line or absence of dial tone altogether. Provide students with a hands-on example. Have students connect a circuit on the Telecommunications Trainer

When the resistance is increased, the current is decreased (as a result, the light will dim from less current through it). When the resistance is decreased, the current is increased (as a result, the light will glow brighter from more current through it).


typical of a circuit found in a telephone. Have students connect several different values of resistors into the circuit. Have students measure and record each of the resistor values with a multi-meter. Have students connect a fixed voltage source to the circuit. Have students measure and record current through different resistors with the multi-meter. Have students compute Ohm’s Law calculations and compare these calculations with their measured values. Repeat with higher and lower value resistors. What relationship exists between current when resistance values are changed?

3. Work through the math example embedded in the CTE lesson. What is the current in a phone circuit with 12 volts applied to a resistance of 12 ohms? So, if the resistance in the phone circuit were increased to 1200 ohms due to a faulty condition, what would the new current be? Would you predict it to be higher or lower than the functioning circuit?

I=E/R

=12/12 =1ampere

I=E/R

=12/1200 =.01amperes

The current is now 100 times less in the faulty circuit than in the functioning circuit.


And if the resistance in the phone circuit were decreased to 1 ohm due to a faulty condition, what would the new current be? Would you predict it to be higher or lower than the functioning circuit? What would be a mathematical name for this Ohm’s Law formula? Using this same formula, how do we determine unknown voltage when current and resistance are given as 5 amperes and 100 ohms respectively?

I=E/R

=12/1 =12amperes

The current is now 12 times greater than the circuit in normal conditions. Remember, the higher the current in the circuit, the more heat produced and the greater the chance for a fire.

I=E/R is an algebraic equation with 3 variables, 2 of which are given in the problem. You have to divide to find the missing third variable (I) in this equation.

Start with the original equation: I=E/R

If we multiply both sides of the equation by ‘R’ we will not change its balance.

(R)(I) = (E/R)(R)

On the right side the R’s cancel out. Our new equation is E=IR

=5(100) =500 volts


How do we determine unknown resistance when current is 5 amperes and voltage is 60 volts?

We can also determine resistance from this formula if we divide both sides by I,

E=IR

E/I=(IR)/I

R=E/I Any variable can be solved for as long as we are given the values of the other two variables.

4. Work through related, contextual math-in-CTE examples.

1. Find the current through a 1200 ohm resistive circuit when 24 volts is applied.

2. Find the resistance of a circuit that draws .05 amperes with 12 volts applied.

3. Find the applied voltage of a circuit that draws 0.2 amperes through a 4800 ohm resistance.

1. I=E/R = 24/1200= .02 amperes 2. R=E.I = 12/.05 = 240 ohms

3. E = IR = (0.2)(4800) = 960 volts

5. Work through traditional math examples. 1. Given the equation x=yz, where y=2 and z=5, solve for x. 2. Given the equation x=yz, where x=2 and z=10, solve for y. 3. Given the equation x=yz, where x=6 and y=4, solve for z.

1. x=yz = (2)(5) = 10 2. y=x/z = 2/10 = 1/5 or 0.2

3. z=x/y = 6/4 = 3/2 or 1.5


6. Students demonstrate their understanding. So how would you explain to a customer how a dimmer works?

Begin with class discussion on the Ohm’s Law relationship between which two variables?

Students should be able to verbally describe the inverse relationship between resistance and current, how the dimmer alters circuit resistance, and how current is affected by resistance and its effect on light brightness.

Students could use the attached worksheet for homework, to be filed in student portfolio.

7. Formal assessment. Possible test questions.

1. Find the applied voltage of a telephone circuit that draws .017amperes through a resistance of 15,000 ohms.

2. Given the equation a=bc, solve for c where a=7 and b=7/3

1. E=IR = (.017)(15000) = 255 volts

2. c=a/b = 7/(7/3) = 3


Adaptations for special needs students. Teacher Notes: Worksheets and puzzles are pre-designed with small chunks of information for special needs students.

Students can complete these in class or as homework. Students can also go to resource room for help with these worksheets.

Math Standards and Assessment Anchors addressed with this lesson. M.11.A.2.1.3 M11.D.2.1.3 M11.D.2.1.4 M11.D.3.1.1 M11.D.3.1.2

References. Telecommunications Trainer Lab Manual RSR Electronics, 2000

Author(s): Position: School: Andrew Jacobs

Telecommunications Instructor

CAT Pickering

Dr. Ronald Husband Principal CAT Pickering Peggy LaRosa Math Teacher CAT Pickering


Worksheet

Answer problems 1-3 using the following information: The formula for the area of a rectangle is A=LW, where A=area, L= length, and W=width.

1. Find the area of a 4 foot long, 2 foot wide rectangle. 2. Find the width of a rectangle with an area of 26 square feet and a length of 13 feet. 3. Find the length of a 36 square foot, 9 foot wide rectangle.

Answer problems 4-6 using the following information: The distance formula is D=RT, where T=time, R=rate, and D=distance.

4. How long does it take to walk two miles at a rate of 4 miles per hour? 5. At what rate of speed would it take you to walk 8 miles in 2 hours? 6. What is the distance walked in one and a half hours at a rate of 4 miles per hour?


Homework Solve each equation for the given variable.

1. d=rt; solve for t

2. d=rt; solve for r

3. A=bh; solve for b

4. A=bh; solve for h

5. A=LW; solve for L

6. A=LW; solve for W

7. Find the area of a parallelogram with a base of 7 inches and a height of 8 inches.

8. Find the base of a 120 inch square inch parallelogram with a height of 6 inches?

9. Find the height of a 240 square inch parallelogram with a base of 48 inches?


Worksheet Answer Key:

1. A=LW = 4(2) = 8 feet 2. W=A/L = 26/13 = 2 feet 3. L=A/W = 36/9 = 4 feet 4. T=D/R = 2/4 = ½ hour 5. R=D/T = 8/2 = 4 miles/hour 6. D=RT = 1.5(4) = 6 miles

Homework Answer Key:

1. t=d/r 2. r=d/t 3. b=A/h 4. h=A/b 5. L=A/W 6. W=A/L 7. A=bh = 7(8) = 56 square inches 8. b=A/h = 120/6 = 20 inches 9. h=A/b = 240/48 = 5 inches

Documents

Math-in-CTE Lesson Planarchives.lessoncorner.com/8578a7d9bbaede1c3.pdfPA GOVERNOR’S INSTITUTE--2005 Ohm’s Law, Series Circuits 1 Math-in-CTE Lesson Plan Lesson Title: Using Ohm’s