This document provides an overview of basic electrical concepts for corrosion control. It defines key terms like voltage, current, resistance, and conductance. It explains the differences between alternating and direct current, and how electrical circuits work. Ohm's Law is discussed, along with different circuit types like series and parallel. Measurement units are defined, such as volts, amps, ohms, and coulombs. The importance of these concepts for understanding corrosion is emphasized.

1. NACE International, 10,000
Lakes Pipeline Corrosion Control
Seminar
BASIC CORROSION SERIES, ROOM
C
TUESDAY, FEBRUARY 4, 2020
Basic Electricity, 9:00AM – 9:50AM
Eric Langelund
Piping & Corrosion Specialties, Inc.

2. BASIC ELECTRICITY
 Introduction
 Electrical Fundamentals
◦ Physical Matter
◦ The Two General Types of Electricity
 Basic Terms
 Ohm’s Law
 The Basic Electrical Circuit
 The Series Electrical Circuit
 The Parallel Electrical Circuit
 Combination Circuits

3. INTRODUCTION
We will be discussing:
 – The kinds of electricity encountered in
corrosion control work
 – Explanations of various applicable
electrical
terms
 – How the electrical units represented by
these
terms interact
 – How the electrical units apply to various
types of electrical circuits

4. INTRODUCTION
It’s important that the principles set
forth here be thoroughly understood.
 This is the foundation of corrosion.
 If your building a house, you
always start with the foundation.
 What decides the integrity of the
house depends on the integrity of
the foundation.

5. ELECTRICAL
FUNDAMENTALS
Physical Matter
 Electricity is directly involved with
the make up of physical matter.

6. PHYSICAL MATTER
What is matter?
 Matter is that which makes up the
substance of everything. It occupies
space and will have mass.
◦ Solid
◦ Liquid
◦ Gas
 Whatever the form may be, however, it will
be made up of atoms or of atoms and/or
molecules.
 Atoms are the building blocks from which
elements are comprised.

7. PHYSICAL MATTER
What is an Element?
 Element being that form of matter
which cannot be changed by
chemical means.
 Example
◦ Copper (chemical symbol Cu)
◦ Oxygen (chemical symbol O)

8. PHYSICAL MATTER
What are molecules?
 Molecules are combinations of
atoms that comprised the smallest
part of a substance that retains the
physical characteristics of that
substance.

9. PHYSICAL MATTER
(CuSO4)
copper (Cu), sulphur (S), oxygen (O)
An example of a molecule would be
the smallest part of copper sulfate
(CuSO₄)

10. PHYSICAL MATTER

11. PHYSICAL MATTER

12. PHYSICAL MATTER
 Each elemental atom has its own
characteristic combination of
nucleus and electrons.
◦ Electrons are the key word insofar as
the relationship with electricity is
concerned.
◦ The flow of electrical current involves
the transfer of electrons through an
electrical circuit.

13. ELECTRICITY
Two types of electricity
 Alternating Current (AC)
 Direct Current (DC)

14. ELECTRICITY, AC
 Current flows first in one direction
then in the opposite direction in an
established pattern.
◦ Hertz – is a single cycle of the
produced wave form
◦ Alternating current, in the US, has a
frequency of 60 cycles per second,
referred to as 60 Hertz (or 60 Hz)

15. ELECTRICITY, AC

16. ELECTRICITY, AC
AC is a newly identified cause of
corrosion in very special cases
◦ AC is used for a power source for
Cathodic Protection such as rectifiers
(which converts AC power to DC
power)

17. ELECTRICITY, AC

18. ELECTRICITY, DC

19. ELECTRICITY, DC
Flows in one direction
 Examples
◦ flash light battery
◦ car battery

20. ELECTRICITY, DC

21. ELECTRICITY, DC
DC electricity is of prime importance
in the consideration of the corrosion
process.
 Involved in various types of
corrosion cells
 Involved in corrosion control by the
use of various types of cathodic
protection

22. CURRENT FLOW
Imagine, if you will, you want a cup of
water and went to a nearby faucet.
 You physically turn the knob “on” at
the faucet and water begins to pour
out.
 You just simulated an electrical circuit
and current flow. How??
 The water is your current (electrons).
The knob is your switch. The water
line, in which the water is traveling
threw, is your circuit.

23. CURRENT FLOW
When you turn on the knob, you
switch the current on (water flows),
the current will flow because of the
potential difference.
 On one side of the valve, at the
faucet, there is no water pressure
and at the opposite side of the
valve, water pressure exist, this
simulates the Voltage potential or
EMF (electro motive force).

24. CURRENT FLOW
The water pressure (PSI) simulates
the measurement of the volts or the
voltage in a circuit.

25. CURRENT FLOW
The resistance in the circuit is the
sum of the size of water lines, of the
opening on the valve and the opening
on the end of the faucet.
 The resistance restricts the amount
of water flowing into the cup in a
given time.
◦ In a electrical circuit, this is measured
in units of Ohms.

26. CURRENT FLOW
How much water will flow (gallons
per hour) in a given time will depend
on the resistance.
 The smaller the opening the more
resistance to the water flow.
◦ In a electrical circuit, the smaller the
wire, more the resistance on current
flow; the wire and components will
decide how much current will flow in a
given time better known as amperes.

27. BASIC TERMS
The following discussion defines and
explains the various electrical units
and terms which are involved in DC
electrical circuits.

28. VOLTS
Basic unit of electrical pressure which
forces an electrical current (electrons)
to flow through an electrical circuit
◦ (remember previous example of water
flow)
The comparable term in a water system
would be water pressure expressed as
pounds per square inch (psi)
◦ Another term to use for electrical
pressure is electromotive force or EMF

29. VOLTS

30. VOLTS

31. VOLTS
In corrosion testing work, DC
voltages used as a source of test
current may be batteries – such as
common dry cell batteries or storage
batteries.
 Flashlight battery – voltage – 1.5
volts
 Automobile type – voltage – 12
volts
 Larger currents needed, DC

32. VOLTS
In corrosion prevention work,
sources of DC voltage used to
provide cathodic protection current
include:
 Galvanic anodes
◦ Zinc
◦ Aluminum
◦ Magnesium
 Driving voltage of anodes may be
measured in tenths of a volt or in
millivolts.

33. VOLTS
In corrosion prevention, sources of
DC voltage used to provide cathodic
protection current include:
(continued)
 Higher capacity sources such as
AC to DC rectifiers or DC
generators of various types
◦ Normally available in a wide range or
voltages to match specific
requirements

34. AMPERES
Ampere (often abbreviated to amp) is
the basic unit to electrical current flow.
◦ (remember) What pushes the current
through the circuit, electrical pressure or
voltage
◦ (remember previous example of water
flow)
Ampere would be comparable term in a
water system to express the rate of
water flow, for an example, gallons per
hour.

35. AMPERES

36. AMPERES

37. GOLDEN RULE OF
CORROSION
When current discharges off of the
surface of metallic structure, there is
metal loss.
◦ Depending on the amount of current
discharge will decide the amount of
metal loss
◦ The greater the amount of current
discharge, the greater the amount of
metal loss.

38. COULOMBS
The term coulomb is seldom, if ever,
encountered in practical corrosion
work on underground structures.
◦ It is however, related to current.

39. COULOMBS
Earlier, in the section on “Physical
Matter”, it was indicated that electric
current flow involves the transfer (or
flow) of electrons through an electrical
circuit.
 The coulomb is a representation of
this electron flow and is the basis for
current, the Ampere
 One (1) DC Ampere = Is the flow of
one Coulomb per second = 6.241 x
10^18 electrons or 6.241 million
million million electrons per second

40. COULOMBS
The development of the coulomb is
based on the fact that when normal
(or neutral) atoms of a material either
gain or lose electrons, they will
develop positive or negative charges.
◦ When this condition exists, like
charges repel each other and opposite
charges attract each other.
◦ The attraction or repulsion force is the
source of electrical pressure, or
voltage, as discussed earlier.

41. COULOMBS
When a suitable current path is
established, there will be a transfer
between dissimilarly charged
materials to satisfy a natural
tendency to reestablish neutrality.

42. CURRENT FLOW

43. OHMS
The ohm is the basic unit for
resistance to the flow of electrical
current.
◦ (remember the water flow example)
what was the resistance? what decided
how much water flows in a given time?
◦ Resistance or Ohms opposes the flow
of current

44. OHMS
With a fixed driving voltage applied to
an electrical circuit,
 The amount of current flowing
through the circuit decreases as the
circuit resistance increases.
 The amount of current flowing
through the circuit increases as the
circuit resistance decreases.
◦ The usual symbol for resistance in
formulas is the letter R or the Greek letter
omega “Ω”.

45. OHMS
Resistance may also be measured in
terms such as milliohms (0.001
ohms) or in kiliohms (1000 ohms) or
in megaohms (1,000,000 ohms)

46. OHMS

47. RESISTIVITY
Resistivity a measure of the resisting
power of a specified material to the
flow of an electric current.
◦ It can be applied to both metallic and
non-metallic materials.
◦ Commonly expressed as ohm-
centimeters (ohm-cm)

48. RESISTIVITY
Unit ohm-cm is the resistance
between opposite faces of a cube of
material which is 1 centimeter x 1
centimeter x 1 centimeter in size.
◦ In practice, a cubic centimeter of a
material is never isolated for the
purpose of measuring its resistivity.
◦ Rather, the resistance across a body of
the material of known dimensions is
measured and, by calculation,
determining its unit resistivity.

49. WENNER 4 PIN METHOD

50. RESISTIVITY
Tools used for measuring resistivity
 – Collins rod
 – Four pin method
 – Soil box
Common areas of measurement taken
 – Soil
 – Water
◦ The symbol used for resistivity is “ρ”
(Greek letter rho)

51. RESISTIVITY – SOIL BOX

52. RESISTIVITY – COLLINS
ROD

53. WENNER 4 PIN TEST KIT

54. RESISTIVITY
The greatest need for measuring
resistivity is in connection with soils
and waters as needed for the design
of corrosion control systems. (Such
as)
 Current design
◦ Anode calculation
◦ Ground bed design for rectified system

55. POLARITY
The term polarity is important in
determining the direction of
conventional current flow in practical
usage.
◦ The direction of conventional current flow
is “+” positive to “-” negative.
◦ The flow of electrons is from “-” negative
to “+” positive which is in the opposite
direction from conventional current flow.
◦ When current flows one way, the
electrons are flowing the other way.

56. POLARITY
Polarity is used to identify which
direction the conventional current is
flowing.

57. CURRENT FLOW

58. CORROSION GOLDEN RULE
#2
Current always returns to it’s original
power source through a return path.

59. CURRENT FLOW

60. CONDUCTANCE
Opposite to resistivity
◦ Measured in Mho’s
◦ The term conductor is used to
designate a member of an electrical
circuit that readily carries an electrical
current.

61. CONDUCTANCE
Examples of conductors used in the
corrosion control –
 – Wire or Cable
 – Pipes or other metallic structures
Different metallic materials have different
capability for carrying electric current. This
is related to the characteristic resistivity of
the material as has been discussed earlier.
Can water or soil be considered a
conductor?
 Answer – yes, water or soil can carry an
electrical current but in most cases very
little conductance value due to resistance
in the material

62. CONDUCTANCE
Different materials have different levels
of conductance, for an example –
 Copper - 100%
 Aluminum - 60
 Magnesium - 36.8
 Zinc - 27.6
 Brass - 24.6
 Steel - 9.6
 Lead - 8.0

63. CONDUCTANCE
Although copper is the obviously the
best conductor material, a steel pipe
(even though a relatively poor
conductor material) can be a very
good practical conductor because,
particularly in larger sizes, the
amount of steel in the pipe is so
much greater than the amount of
copper in the usual copper wire or
cable.

64. CONDUCTANCE
For an example,
 – The resistance of 1000 feet of 4/0
American wire gage (AWG) copper
cable (which is approximately 0.54
inches in diameter) will have a
resistance in the order of 0.051 ohms
whereas the resistance of 1000 feet of
12-inch steel pipe with 0.375-inch wall
thickness will have a resistance of
only about 0.0058 ohms or roughly
one tenth that of the heavy copper
cable.

65. CONDUCTANCE
Remember the water example, the
bigger the pipe the more that water
will flow in a given time.
The bigger the conducting area, this
gives more room for the electrons to
flow; therefore it becomes a better
conductor.

66. INSULATOR
Insulator or insulating material will
have a very high resistance to the
flow of electrical current and is used
to confine or control the flow of
current in electrical circuits.

67. INSULATOR
Examples –
 Wire or cable jackets of rubber,
 Neoprene
 Plastics
 Fiberglass
 Coatings

68. OHM’S LAW
The worker in the field of
underground corrosion control must
have a thorough understanding of
Ohm’s law as it applies to DC circuits.
The law states that one volt of
electrical pressure will force one
ampere of current through a circuit
having a resistance of one ohm.

69. OHM’S LAW

70. OHM’S LAW

71. OHM’S LAW

72. OHM’S LAW
Example
If current flowing through a circuit is
3.6 amps
And the resistance of the circuit is 1.7
ohms
What does the voltage = ?
 E = I x R
◦ Volts = 3.6A x 1.7Ω = 6.12 V
◦ If any two components are known, the
other can be calculated.

73. OHM’S LAW
I is unknown, while the voltage (E)
and the resistance (R) are both
known.
Example
If E = 12.0 volts and R = 3.5 ohms, I =
?
◦ E = I x R, Ohm’s Law
◦ I = E/R, Ohm’s Law, solving for I
I = 12V/3.5Ω = 3.42 A

74. OHM’S LAW
R is unknown, while the voltage (E)
and the current (I) are both known.
Example
If E = 6.0 volts and I = 1.5 amps, R = ?
◦ E = I x R, Ohm’s Law
◦ R = E/I, Ohm’s Law, solving for R
R = 6.0V/1.5A = 4Ω

75. OHM’S LAW
Values entered in the formula must be
in the proper units.
Example
If voltage in the circuit is 2.0 volts and
the current measured in the circuit is
1.0 milliamp.
Do not set up the calculation as follows,
R = 2volts/1milliamps = 2 ohms
This is Wrong!

76. OHM’S LAW
What must be done is to convert the
1.0 milliamp to amps (1.0 mA =
0.001amps)
The correct calculation is:
R = 2volts/0.001amps = 2000 Ω

77. ELECTRICAL CIRCUIT

78. ELECTRICAL CIRCUIT
Figure 1-3 shows all of the circuit
resistance confined to a simple resistor.
Now assume that a DC power source
providing current to a cathodic
protection system for corrosion has
instruments indicating that the supply
voltage (E) is 20 volts, current flow (I) is
5 amps, the total circuit resistance (R)
can be calculated with Ohm’s law, R=
E/I.
◦ Circuit Resistance = 20V/5A = 4 Ohms

79. ELECTRICAL CIRCUIT
If a 20 volt DC power source were
connected across a known resistance
of 4 ohms,
 Current Flow ???
 Can be calculated by using I = E/R
 Current Flow (I) = 20V/4ohms = 5
amps

80. ELECTRICAL CIRCUIT,
SERIES

81. ELECTRICAL CIRCUIT,
SERIES
Figure 1-4 represents an electrical
circuit where the total circuit
resistance comprises two of more
load resistances where are
connected in series.
By “series”, it is meant that the
several load resistances are
connected end-to-end (in a row) and
that the entire circuit current has to
pass through each of the resistances.

82. ELECTRICAL CIRCUIT,
SERIES
The voltage will be distributed across
the several resistances in the circuit.
Assume the following values are known
with respect to the circuit of figure 1-4
 Power supply voltage (Es) = 10 volts
 Circuit current flow (I) = 2.0 amps
 Load resistance No. 1 (R1) = 3.0 ohms
 Load resistance No. 2 (R2) = 1.87
ohms
 Load resistance No. 3 (R3) = (20 ft of
No. 8
 wire) = 0.13 ohms

83. ELECTRICAL CIRCUIT,
SERIES
Calculate the voltage drops across the three
resistances by Ohm’s Law, E = I x R.
 V drop across R1 (V1) = 2A x 3.00 ohms =
6.0V
 V drop across R2 (V2) = 2A x 1.87 ohms =
3.74V
 V drop across R3 (V3) = 2A x 0.13 ohms =
0.26V
The sum of these voltages is:
 6 + 3.74 + 0.26 = 10 volts
The sum of the resistance in the example used
is :
 3.0 + 1.87 + 0.13 = 5 ohms
Perform self check: E = I x R

84. ELECTRICAL CIRCUIT,
SERIES
Assume we don’t know the
resistance values of the R1, R2, and
R3.
All we know is the power source
voltage (Es = 10 volts) and power
current (I = 2 amps)
 Determine the resistance of each
resistor.

85. ELECTRICAL CIRCUIT,
SERIES
Measure the voltage drop across
resistances R1 and R2 using a
suitable DC voltmeter, we get :
 V drop across resistance No. 1 = 6
volts
 V drop across resistance No. 2 =
3.74 volts
This gives all the needed information.

86. ELECTRICAL CIRCUIT,
SERIES
Using Ohm’s law,
 Resistance No. 1 = 6 volts/2 Amps =
3 ohms
 Resistance No. 2 = 3.74 volts/2
Amps = 1.87 ohms

87. ELECTRICAL CIRCUIT,
SERIES
Power source is known at 10V and
the current 2A
By Ohm’s law,
 Total Circuit Resistance = 10 volts/2
Amps = 5 ohms
Therefore, the value of resistance No.
3 has to be 5 ohms minus the sum of
Resistance R1 and R2.
 Resistance No. 3 = 5 – (3 + 1.87) =
0.13 ohm

88. ELECTRICAL CIRCUIT,
SERIES
Things to remember,
◦ The total current from the power source
flows through each resistance element in
the circuit.
◦ The sum of the voltage drops across the
several resistance elements in the circuit
must equal the voltage of the power
source.
◦ The sum of the ohms value of the several
resistance elements in the circuit must
equal the total circuit resistance:
RT = R1 + R2 + R3 ……

89. ELECTRICAL CIRCUIT,
PARALLEL
Two or more load resistances are
connected so that the plus (current
input) ends of all resistances are
connected together instead of being
connected end to end (in a row) as is
the case with a series circuit.
 Illustrated by figure 1-5

90. ELECTRICAL CIRCUIT,
PARALLEL

91. ELECTRICAL CIRCUIT,
PARALLEL
By looking at the Figure 1-5, we can see
the power source voltage will be
impressed on each resistance element
rather than being distributed as is the
case with a series circuit.
 Current will be divided among the
several resistance branches.

92. ELECTRICAL CIRCUIT,
PARALLEL
Power supply voltage (ES) = 20 volts
 Power supply current (IS) = 16.67
amps
 Load Resistance No. 1 (R1) = 3 ohms
 Load Resistance No. 2 (R2) = 2 ohms

93. ELECTRICAL CIRCUIT,
PARALLEL
First calculate the current flow (I1 and I2)
 The voltage drop across each branch
equals the power supply voltage (ES) or
20 volts.
 Current flow through R1
 I1 = 20 volts/ 3 Ohms = 6.67 amps
 Current flow through R2
 I2 = 20 volts/ 2 Ohms = 10 amps
 The sum of these two should equal the
power source output current of 16.67

94. ELECTRICAL CIRCUIT,
PARALLEL
Second, calculate the parallel resistance
of R1 and R2.
 Formula used, R1 x R2/R1 + R2 =
Parallel Resistance
 Parallel Resistance = 3 ohms x 2
ohms/3+2 = 6/5 = 1.2 ohms

95. ELECTRICAL CIRCUIT,
PARALLEL
Note that the parallel resistance of any
two resistors is always less than the
small resistance value.
 Perform check, RT = ES / IS

96. ELECTRICAL CIRCUIT,
PARALLEL

97. ELECTRICAL CIRCUIT,
PARALLEL
Remember the following important
points
 The full power supply voltage is
impressed across each parallel
branch.
 The sum of the currents through the
individual parallel branches must
equal the total current output of the
power source.
 The parallel resistance of two or more

98. ELECTRICAL CIRCUIT,
COMBINATION
Circuits can be a combination of series
elements and parallel elements.
 Figure 1-6

99. ELECTRICAL CIRCUIT,
COMBINATION

100. ELECTRICAL CIRCUIT,
COMBINATION
The power supply output current divides
between the two parallel branches R1
and R2 and then combines again after
passing through these resistance.
The full power supply current then
passes through the resistance element
R3

101. ELECTRICAL CIRCUIT,
COMBINATION
The voltage drop across parallel
branches R1 and R2 will be equal to
each other.
 However, the amount of the voltage
drop will be the power supply voltage
less the voltage drop across
resistance R3

102. ELECTRICAL CIRCUIT,
COMBINATION
The effective circuit resistance will be
the calculated parallel resistance of
branches R1 and R2 plus the resistance
of series resistance R3.

103. THANK YOU!
 Any questions or concerns?