Download Using Ohms Law in Telephone Circuits

PA GOVERNOR’S INSTITUTE--2005
Ohm’s Law, Series Circuits
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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:
Supplies Needed:
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.
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.
PA GOVERNOR’S INSTITUTE--2005
THE "7 ELEMENTS"
Ohm’s Law, Series Circuits
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?
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.
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?
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).
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PA GOVERNOR’S INSTITUTE--2005
Ohm’s Law, Series Circuits
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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?
I=E/R
=12/12
=1ampere
So, if the resistance in the phone circuit were increased to 1200 ohms
due to a faulty condition, what would the new current be?
I=E/R
=12/1200
=.01amperes
Would you predict it to be higher or lower than the functioning circuit?
The current is now 100 times less in the faulty circuit
than in the functioning circuit.
PA GOVERNOR’S INSTITUTE--2005
And if the resistance in the phone circuit were decreased to 1 ohm due
to a faulty condition, what would the new current be?
Ohm’s Law, Series Circuits
I=E/R
=12/1
=12amperes
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?
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
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PA GOVERNOR’S INSTITUTE--2005
How do we determine unknown resistance when current is 5 amperes
and voltage is 60 volts?
Ohm’s Law, Series Circuits
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.
1. x=yz = (2)(5) = 10
2. Given the equation x=yz, where x=2 and z=10, solve for y.
2. y=x/z = 2/10 = 1/5 or 0.2
3. Given the equation x=yz, where x=6 and y=4, solve for z.
3. z=x/y = 6/4 = 3/2 or 1.5
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PA GOVERNOR’S INSTITUTE--2005
Ohm’s Law, Series Circuits
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
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PA GOVERNOR’S INSTITUTE--2005
Ohm’s Law, Series Circuits
Adaptations for special needs students.
Worksheets and puzzles are pre-designed with small chunks of
information for special needs students.
Teacher Notes:
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):
Andrew Jacobs
Dr. Ronald Husband
Peggy LaRosa
Position:
Telecommunications
Instructor
Principal
Math Teacher
School:
CAT Pickering
CAT Pickering
CAT Pickering
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PA GOVERNOR’S INSTITUTE--2005
Ohm’s Law, Series Circuits
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?
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PA GOVERNOR’S INSTITUTE--2005
Ohm’s Law, Series Circuits
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?
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PA GOVERNOR’S INSTITUTE--2005
Worksheet Answer Key:
1.
2.
3.
4.
5.
6.
A=LW = 4(2) = 8 feet
W=A/L = 26/13 = 2 feet
L=A/W = 36/9 = 4 feet
T=D/R = 2/4 = ½ hour
R=D/T = 8/2 = 4 miles/hour
D=RT = 1.5(4) = 6 miles
Homework Answer Key:
1.
2.
3.
4.
5.
6.
7.
8.
9.
t=d/r
r=d/t
b=A/h
h=A/b
L=A/W
W=A/L
A=bh = 7(8) = 56 square inches
b=A/h = 120/6 = 20 inches
h=A/b = 240/48 = 5 inches
Ohm’s Law, Series Circuits
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