Problem Solving Unit 9 Ohms Law

1
Problem SolvingUnit 9Ohms Law
GE184 Problem Solving

• John Elberfeld
• JElberfeld_at_itt-tech.edu
• 518 872 2082

2
Electrical Charges

• People have seen the effects of electricity for
  thousands of years
• Lightning, electric shocks, static cling
• Only two types of electric charges exist
  positive ( ) and negative ( - )

3
Atomic Structure

• Nucleus has positive protons and neutral neutrons
• Very light negative electrons revolve in orbit
  around the nucleus
• Only electrons move away from the atom and make
  up electric current

4
Electricity

• Electric current is the flow of electrons
• Electrons flow easily through some materials.
• Other materials resist the flow of electrons

5
Conductors and Insulators

• Current electricity is the movement of electrons
• Conductors let electrons move easily and so offer
  low resistance to current flow
• Insulators oppose the flow of electrons and so
  offer high resistance to current flow
• Semiconductors offer moderate, controllable
  resistance
• Resistance to current flow depends on the atomic
  structure of the material

6
Conductors

• Conductors contain a large number of free
  electrons.
• High conductance means the material is a good
  electricity conductor (low resistance)

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Insulators

• Insulators prevent the flow of electricity
• A material with high insulation (resistance) is a
  poor conductor of electricity

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Current Flow

• Electron current flow occurs as electrons move
  away from the negative terminal towards the
  positive terminal.

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Analogy for Current Flow
10
Charges

• Electrons have a negative charge
• Protons have a positive charge
• The charge on an electron is equal in magnitude
  but opposite in sign to a proton
• The charge on an electron is the smallest unit of
  electrical charge that exists in nature

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Real World Measurements

• We usually deal with billions and billions of
  electrons
• 1 coulomb 6.25 x 1018 electrons
• The charge on a object depends on how many EXTRA
  charged particles are there
• Usually the number of electrons balances the
  number of protons, so the net charge is 0

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Charges

• When an object has a negative charge, it has
  more electrons than protons
• Charge is determined by how many more electrons
  it has
• When an object has a positive charge, it is
  missing electrons and has more protons than
  electrons
• Protons do NOT move only electrons move to and
  from an object
• A positively charged objects has had electrons
  removed from it

13
Large and Small Numbers

• In electricity, you will work with very large and
  very small numbers
• You must use engineering notation in your work
  and in your answers
• 736 400 736.4 x 103
• Engineering notation is like scientific notation,
  but the exponent is always a multiple of three
  (3) 3, 6, 9. -3, -6

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Powers of Ten


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Positive Powers of 10

• If you move the decimal point in the number to
  the left, you are making the number smaller.
• To keep the same value, you must multiply by a
  bigger power of 10

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Negative Powers of 10

• If you move the decimal point in the number to
  the right, you are making the number bigger.
• To keep the same value, you must multiply by a
  negative power of 10

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Engineering Notation

• Write the number using powers of ten.
• Move the decimal point left while increasing the
  exponent or right while decreasing the exponent.
• The final exponent must be zero or a number that
  is evenly divisible by three.
• The number itself must be greater than one and
  less than 1000

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Practice Exercises NOW!

• Convert the following numbers into engineering
  notation
• 1 2 3 4 5 6 7. 1.234567 x 106
• .0 0 1 2 3 1.23 x 10-3
• 1 2 3 . 4 x 105 12.34 x 106
• 1. 2 3 x 10-7 123. x 10-9

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Set Your Calculator

• Set your calculator to Engineering mode
• If you are an electronics student, and you dont
  have a good calculator with Engineering mode, buy
  one as soon as possible (Casio is good)
• In engineering mode, the calculator always shows
  numbers as a power of ten with an exponent a
  multiple of three

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Practice Exercises NOW!

• Enter the following numbers into your calculator
  and compare results when you choose engineering
  mode
• 1 234 567
• .00123
• 12.3 x 105
• 1.23 x 10-7
• .0123 x 108
• .00123 x 10-5
• 123.5

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Calculations with Exponents

• When multiplying, you add exponents
• (10x10x10)x(10x10) 10x10x10x10x10 105
• 103 x 102 1032 105
• 6 x 104 x 2 x 102 12 x 106
• Use your calculator in Engineering notation to
  find
• 3.21 x 105 x 12.98 x 108
• 3.21 EXP 5 x 12.98 EXP 8 416.7 x 1012

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More Calculations

• When dividing, you subtract exponents
• 10x10x10x10x10 10x10 10210x10x10
• 105 /103 105-3 102
• 10x10 1
  10-310x10x10x10x10 10x10x10
• 102 / 105 102-5 10-3

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Example

• Fact 1 Coulomb (C) 6.25 x 1018 electrons
• A capacitor stores a charge of 1.5 x 10-5 C. How
  many extra electrons are in the capacitor?
• 93.75 x 1012 electrons
• This is a simple unit conversion problem

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Example

• Fact 1 C 6.25 x 1018 electrons
• How many coulombs in 3.0 x 1013 electrons?
• 4.8 x10-6 C

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On your own

• Fact 1 C 6.25 x 1018 electrons
• If you have 25.2 x 1026 extra electrons, what is
  the charge in coulombs in engineering notation?
• If you have 428 x 10-6 coulomb, what number of
  extra electrons do you have?

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Prefixes

• Engineers usually replace the power of ten with a
  prefix
• Because all powers of 10 are multiples of 3 in
  Engineering Notation, we dont have so many to
  remember
• Some prefixes use Greek letters as symbols
• Procedure to use a prefix
• 1. Write the quantity in engineering notation.
• 2. Replace the power of ten with its prefix.

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Prefixes and Symbols

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Prefixes and Symbols

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Practice
106M103k10-3 m10-6 µ10-9 n

• Enter the following numbers in your calculator
  set in Engineering notation, hit the enter key,
  and use the proper prefix
• 12 x 104 C 120x103 C 120 kC
• .066 x 10-4 C
• 2345 mC
• .03 µC

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Electric Current

• Current
• Movement of electrons from negatively charged
  atoms to positively charged atoms.
• Represented as I.
• Coulomb
• 6.25 x 1018 electrons.
• Represented as C.

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Electric Current

• Electric current is the motion of free electrons
  through a material.
• Current is measured in Amperes
• 1 Ampere of current has 1 coulomb of electrons
  flowing past a given point in the wire in just 1
  second
• 1 A 1 C/s

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Current Symbol

• The symbol for current is the letter I.
• I Q/T
• C is already used for capacitors and coulombs, so
  I is less confusing for current, and Q for charge
• Units are Coulombs/second Ampere
• Batteries are a source of electric current
• Electrons flow from the negative terminal through
  the wire to the positive terminal

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Conventional Current

• Before scientist knew about electrons, they
  guessed that electricity was the flow of positive
  particles, but they were wrong
• However, many books and most electrical symbols
  are drawn to show the direction this positive
  current would flow.

Electrons
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Current

• Current formula

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Current Calculations
106M103k10-3 m10-6 µ10-9 n

• Fact I Q / t
• What is the current if a charge (Q) of 12 mC
  passes a point in time (t) of .023 s?
• I 12 mC / .023 s 522 mA
• What is the current if 76 kC passes a point 834
  sec?
• I

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Voltage

• Voltage is the push behind the moving electrons.
• Voltage gives the electrons energy and gets them
  to move from a negative area to a positive area

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Resistance

• The property of a material where an opposition to
  the flow of electrons is offered is known as
  resistance.
• Unit of resistance is Ohm and is denoted by the
  Greek letter omega, O.

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Resistance

• Not every material will conduct electricity
• Copper and aluminum conduct electricity easily
  and are labeled conductors and have low
  resistance
• Plastic and rubber do not conduct electricity
  easily and are labeled insulators and have high
  resistance

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Measuring Resistance

• Resistance is a measure of how much a material
  opposes the flow of electricity through it
• A material with a resistance of 1 ohm requires 1
  volt of energy to get 1 coulomb of charge
  through it per second
• 1 ohm 1 O 1 V / A

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Resistance Ranges

• Conductors have resistances of a few ohms or less
• Insulators have resistances in range of millions
  of ohms or more
• There is no absolute insulator or perfect
  conductor

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Ohms Law

• MEMORIZE V I R
• Ohms Law
• If you increase the voltage, you increase the
  current proportionally
• 3 times the voltage gives you three times the
  current
• Resistance (ohms) is the proportionality constant
  and depends on the atomic structure of the
  material conducting the current

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Experiment
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Graph of Data
x
V
x
x
x
x
I
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Reasoning
V I R

• Ohms Law V I R
• High voltage produces high current for a given
  resistance
• Low voltage produces low current for a given
  resistance
• For a given voltage, a high resistance produces a
  low current
• For a given voltage, a low resistance produces a
  high current

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Electronic Circuit

• A battery with the voltage V pushes a current I
  through a resistor R

V I R
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Ohms Law
V I R

• This is the BIG IDEA for the day (year)!
• V I R
• What if we divide both sides by R?
• V I R R R
• But R/R 1, so we dont need to write it down
• I V I V / R R

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Ohms Law
V I R

• V I R
• What if we divide both sides by I?
• V I R I I
• But I / I 1, so we dont need to write it down
• R V R V / I I

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Ohms Law

• Memorize V I R
• Use algebra to find
• I V / R
• R V / I
• If you can, learn all three variations, but you
  can get by if you memorize
• V I R

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Practice

• V I R
• What voltage (V) is needed to push a current of
  10 Amperes (I) through a resistance of 7 Ohms (R)
  ?

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Practice

• V I R
• What voltage (V) is needed to push a current of
  10 Amperes (I) through a resistance of 7 Ohms (R)
  ?
• V I R
• V 10 A x 7 O
• V 70 V

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Practice

• V I R
• What current (I) flows through a resistance of 8
  ohms when the resistor is connect to a 24 volt
  battery?

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Practice
V I R

• What current (I) flows through a resistance of 8
  ohms when the resistor is connect to a 24 volt
  battery?
• V I R
• 24 V I x 8 O
• I 24 V / 8 O
• I 3 A

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Practice
V I R

• What size resistor allows 4 amperes of current
  through it when it is connected to a 12 Volt
  power supply?

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Practice
V I R

• What size resistor allows 4 amperes of current
  through it when it is connected to a 12 Volt
  power supply?
• V I R
• 12 V 4 A x R
• R 12 V / 4 A
• R 3 O

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Now we add prefixes

• Very few electronics labs deal with simple volts,
  amps, and ohms.
• There is almost always a prefix, like millivolts,
  microamperes, and kilo ohms
• To work with real electric data, you have to be
  able to handle these units on your calculator

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Examples

• Ohms Law V I R k 103 µ 10-6 m10-3
• How much voltage must be connected across a 1.2 k
  O resistor to cause 575 µA of current to flow?
• V I R
• V 575 µA 1.2 k O
• V .69V 690 x 10-3V 690 mV

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Examples
103k10-3 m10-6 µ

• Ohms Law V I R
• How much current flow through a 25 O resistor
  with 10 V across it?
• V I R I V / R
• 10 V I 25 O
• I 10 V / 25 O
• I .4 A or 400 x 10-3A 400 mA

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Examples
103k10-3 m10-6 µ

• Ohms Law V I R
• If a certain resistor allows 250 mA to flow when
  35 V are across it, what is the resistance?
• V I R R V / I
• 35 V 250 mA R
• R 35 V / 250 ma
• R 140 O

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Examples
103k10-3 m10-6 µ

• Ohms Law V I R
• How much current flows through a 3.3k O resistor
  with 4.5 mV across it?
• V I R I V / R
• 4.5 mV I 3.3k O
• I 4.5 mV / 3.3k O
• I 1.36 µ A

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Series Circuit

• A series circuit has one path for current flow.
• The current is the same at any point in the
  circuit.

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SERIES CIRCUITS

• Total resistance in a series circuit is the
  summation of the individual resistor values
• RT R1 R2 R3 . Rn
• Where n the number of resistors

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OHMS LAW

• Total resistance in a series circuit is equal to
  the total circuit voltage divided by the total
  current Remember, IT is the same at every point
  in the circuit

VT ITRT IT VT / RT
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Single Resistor Circuit

• Single resistor example
• V I R
• 10 V 2 A x 5 O

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OHMS LAW EXAMPLES

• Voltage 25 V
• Resistance 25 ?
• I _____

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OHMS LAW EXAMPLES

• Voltage 15 V
• Resistance 1 k?
• I _______

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OHMS LAW EXAMPLES

• Resistance 50 ?
• Current 2.5 A
• Voltage _______

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OHMS LAW EXAMPLES

• Resistance 10 k?
• Current 4 mA
• Voltage ______

V ? V
I 4mA
R 10kO
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OHMS LAW EXAMPLES

• Current 15 mA
• Voltage 15 V
• Resistance _______

V 15V
I 15mA
R ?O
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OHMS LAW EXAMPLES

• Current 4 mA
• Voltage 60 V
• Resistance _______

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BASIC CIRCUIT EXAMPLES

• R 10 k?, I 10 mA, V _____
• V 100 V
• V 50 V, I 2 mA, R ______
• R 25 k?
• V 75 V, R 10 k?, I ______
• I 7.5 mA

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Example

• Series circuit comprised of resistors
• To calculate out the total resistance we use the
  formula RT R1 R2 R3 etc.
• The total resistance for the figure is
• RT 10O 20O 30O 60 OHMS

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Replace The Resistors
IT VT/RT 120V / 60kO 2 mA
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Summary

• In a pure series circuit, the current in every
  resistor is the same the same as the current
  that leaves the battery
• The effective resistance of the pure series
  circuit is the sum of all resistors in series

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Replace The Resistors II
IT VT/RT 9V / 30kO 300 µA
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Replace The Resistors III
-
-
V 11V
I ?mA
R 4.27kO
IT VT/RT 11V / 4.27kO 2.58mA
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Series Circuit Review

• There is just one path for the current
• The current is the same in every component in a
  series circuit
• Total resistance is the sum of the resistors

77
Parallel Circuits

• There are two or more paths for current flow.
• The voltage is the same across all parallel
  branches.
• The total current entering any point is equal to
  the current leaving the point

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Parallel Combination of Resistances

• When all the resistors in a circuit are connected
  in parallel, the circuit is called a parallel
  circuit.

79
Voltage Drop in Parallel Resistive Circuit

• The voltage drop across each component in a
  parallel circuit is the same, since all the
  resistances are connected in parallel. Thus, the
  voltage across each connection is equal to the
  supply voltage.
• The voltage drop in a parallel circuit is
  where, V is the
  supply voltage.

80
A Practical Example

• Car circuits (and house circuits) are all
  parallel
• You can run the radio without the horn being
  turned on an advantage

81
OHMS LAW EXAMPLES

• Voltage 25 V
• Find the total Current_______
• Find the equivalent Resistance ______

V 25V
R2 50O I2 ?
R1 25O I1 ?
82
Calculations

• I1 V1/R1 25V/25O 1 A
• I2 V2/R2 25V/50O .5A (500mA)
• IT I1I2 1 A 500 mA 1.5 A
• RT VT/IT 25V/ 1.5 A 16.7 O
• Notice that the total resistance is LESS than any
  one resistor in the circuit.

83
TOTAL RESISTANCE

• Ohms law method

84
Parallel Circuit Current

• Current leaving the () terminal is the same
  current entering the () terminal.
• This is referred to as total current (IT).
• Individual electrons may follow a variety of
  paths, but every electron that leaves the battery
  is balanced by an electron entering the battery

85
Parallel Circuit Current

• Since the total current in the individual
  resistors is equal to the current supplied by the
  source, the total current can be stated as
• IT IR1 IR2 IRn

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Sample Problems
R12kO
R36kO
R24kO
VT 36V
87
Sample Problems
R12kO
R36kO
R24kO
VT 36V
88
Finding Total Series Resistance

• Finding the total resistance for resistors in
  SERIES is easy
• In SERIES, just add up the all the resistors

89
Finding Total Parallel Resistance

• Every resistor you add in parallel actually
  REDUCES the equivalent resistance in the circuit
• Each resistor you add in parallel increases the
  current, which makes the resistance appear lower
• Simple addition does NOT work for resistors in
  parallel

90
Finding Total Parallel Resistance

• 1/RT 1/R1 1/R2 1/R3 1/R4

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Equation

• The reciprocal of the total resistance is equal
  to the sum of the reciprocals of the resistors in
  parallel
• 1/RT 1/1kO 1/2kO 1/3kO

R11kO
R33kO
R22kO
VT 24V
A
B
C
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Calculations

• 1/RT 1/1kO 1/2kO 1/3kO
• 1/1 1/2 1/3 6/6 3/6 2/6 11/6 1/RT
• RT 6/11 .545 kO 545 O
• RT 545 O
• To check on a calculator
• 1 EXP 3 INV 2 EXP 3 INV 3 EXP 3 INV
  EQUALS INV

93
Practice
R12kO
R36kO
R24kO
VT 36V
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Series-Parallel Combination of Resistances

• This type of circuit consists of a combination of
  resistances in series and parallel.
• Series-parallel combination of resistances can be
  classified as
• Simple series-parallel circuits
• Complex series-parallel circuits

95
Parallel Resistors

• For resistors to be in parallel, both ends of
  each resistor must be in direct electrical
  contact with each other with no resistance
  between the ends
• Parallel Not Parallel (blue)

96
Recognizing Series Components
97
Series-Parallel Circuits

• Pure series-parallel circuits can be reduced to a
  single equivalent resistor
• Example

98
Example

• Another simple series-parallel circuit

99
Example - Simple
100
Voltage Drop in Series-Parallel Resistive Circuit

• To find the equivalent resistance of a
  series-parallel resistive circuit
• Identify the resistance combinations as either
  series or parallel.
• Simplify, combine and recombine series and
  parallel combinations
• Find the value of the total resistance of the
  circuit.
• Once you know the equivalent resistance, you can
  calculate voltage or current

101
Circuit Analysis

• Combine the two parallel resistors to one
  equivalent resistor
• 1/R23 1/20O 1/30O 3/60O 2/60O
• 1/R23 5/60O
• R23 12 O

R1 10O
R1 10O
V 25V
R3 30O
R23 12O
V 25V
R2 20O
102
Analysis

• Combine the series resistors
• R123 10 O 12 O 22 O

R123 22O
R1 10O
V 25V
V 25V
R23 12O
103
Analysis

• V 25 V, RT 22O,
• IT V/R 25V/22 O
• IT 1.14 A

R1 10O
R3 30O
V 25V
R2 20O
104
Class Activity

• What is the equivalent resistance of the
  following circuit?

R1 5.6kO
V 30V
R2 10kO
R5 3.3kO
R3 4.7kO
R4 2.2kO
105
Current Analysis

• Combine resistors in series
• R23 10kO 4.7kO 14.7kO

R1 5.6kO
V 30V
R5 3.3kO
R23 14.7kO
R4 2.2kO
106
Current Analysis

• Combine parallel resistors
• 1/R235 1/14.7 kO 1/3.3kO
• (Use your calculator!)
• R235 2.70 kO

V 30V
R1 5.6kO
R235 2.7 kO
R4 2.2kO
107
Current Analysis

• Combine series resistors
• R12345 5.6 kO 2.7kO 2.2 kO 10.5 kO

R12345 10.5kO
V 30V
108
More Analysis

• IT VT/RT 30V / 10.5 KO
• IT 2.85 mA

R1 5.6kO
V 30V
R2 10kO
R3 4.7kO
R4 2.2kO
109
Ready for Thought Projects

• You are now ready to tackle Electronics Projects
  1 2 on pages 135 - 146