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Electrical engineering background concepts
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Instrument Systems
Electrical Engineering Fundamentals
An instrument system is the correct combination of
sensor, controller, and final control element that will
allow a process to operate in an automatic mode.
The vary first instrument systems were hydraulic and for
the most part major examples of public works. The
Roman aquifers in addition to supplying drinking water
had a system of weirs and wheels designed to do specific
jobs.
Instrument Systems
Electrical Engineering Fundamentals
you need to figure out how to implant (embed) as rich
vibrant images the following terms and/or concepts.
Force
Newton
Energy
Joule
Power
Watt
Charge
Coulomb
1 electron’s charge
1 proton’s charge
Instrument Systems
Electrical Engineering Fundamentals
Force is the “push or pull” that alters the energy of an
entity (system).
1 Newton of force will be needed to just stop a 1 kg mass object that is
accelerating at 1 meter/ second squared.
1 Newton =
1 kg
.
1 meter / sec
2
Instrument Systems
Electrical Engineering Fundamentals
Energy is the characteristic of an entity (system) that is
associated with its motion, its potential motion and/or
its lack of motion.
Objects (systems) that are in an identical environment
are grouped together and said to be in the same Energy
State.
( The unit of energy is Joules which are defined as:
The energy used when 1 Newton of force
managed to move the object 1 meter. )
1 Joule = 1 Newton
. meter
Instrument Systems
Electrical Engineering Fundamentals
Power is a rate concept. Power is the rate energy is used.
Your guess is as good as mine. BUT here are some of its
properties!!!
1 Watt is the rate when 1 Joule is used in 1 second.
-19
1 electron
1.6 xsecond
10
Coulombs.
1 watt carries
= 1 Joule/
-19
1 proton carries 1.6 x 10
Coulombs.
Charge
is associated with objects.
What is it?
is packaged in electrons and protons.
is measured in Coulombs.
made Coulomb’s law famous!
Instrument Systems
Electrical Engineering Fundamentals
Coulombs Law sort of says that two charge particles in the
same elevator (or any where else for that matter) will not
stay still.
Coulomb determined that the force needed to keep them
from moving was equal to:
F (in Newtons) =
k ( Q ) (Q )/ r 2
1
2
Practice Problem
Show that two 1 Coulomb each charged particles in free space 1
meter apart need to have 1 million tons of force ( give or take a
few tons) applied in the correct direction to keep them from
moving.
Instrument Systems
Electrical Engineering Fundamentals
Potential
Note:
Difference
The second charged particle is not involved in
If one charged particle moves toward or away from a
the calculation but is the reason the first
second charge particle, the first charge particle is now in a
charged moved in the first place.
different energy state.
1 volt = 1 Joule/ 1 Coulomb
The second charged particle is sometimes
known as the reference or test charge.
The difference between these two energy states (the one
the first particle is in now and the one the first particle use
to be in) is the potential difference of the first particle.
Voltage is the potential difference per unit charge.
V =
Energy
difference
Amount of charge
on the first particle
Instrument Systems
Electrical Engineering Fundamentals
Electric Field
If one charged particle moves because of the presence of a
test charge, the first charge is said to move in (through) an
electric force field.
The charged particle can move toward or away from the
test charge but the field is still the region where the
influence of the test charge is felt by the first charge.
The electric field lines define the electric field with respect
to the test charge.
The direction the first charge travels across the electric
field lines determines if the change in energy can be used
by us to do work.
Instrument Systems
Electrical Engineering Fundamentals
Electric Field
Once a charge particle is in an electric field, the only thing
of interest is the voltage changes because the charged
particle moved to different places in the electric field
( Once you have purchased that stock, you only care about
the changes in the price, not what the stock costs.)
Knowledge of the voltage difference will let you compute
the energy difference between the two points in the electric
field if the charge on the particle is know.
Energy
= V (Amount of charge on the particle)
difference
Use a volt meter
for this value.
Many times the charge
particle is an electron and
the charge is know.
Instrument Systems
Electrical Engineering Fundamentals
Current
Current is the rate of movement of electric charge.
Current is not the rate of movement of the charged particle.
Current is measured in amperes.
current = (amount of charge) / ( change in time)
1 ampere =
1 Coulomb / second
Decision time !
We have to pick the sign of this 1 Coulomb charge.
Positive? Or negative?
Yep, Ben Franklin wins, the reference charge is the positive
charge, even though the electron is the most common
carrier of charge in systems we will deal with.
Instrument Systems
Electrical Engineering Fundamentals
Concepts that might seem confusing.
A charge moving from point “b” to point “a” (lower
voltage to higher voltage) requires energy from the
outside world and the charge is moving up the field
A charge moving from point “b” to point “a” produces an
electrical current that itself produces a net force. This force
produced by an electric current is referred to as magnetic
force, and it possess many of the properties of the magnetic
force associated with an ordinary bar magnet.
A charged particle has a static electric field around it.
A moving charged particle creates a current and has a
magnetic field around the path, usually a wire, it is
traveling on.
Instrument Systems
Electrical Engineering Fundamentals
Conventions that might seem confusing.
+
v
v
-
ab
is just the voltage between two points in a static electric field
ab
But by drawing convention;
When a charge moves from point
“b” to point “a” (lower voltage to
higher voltage) energy is required
from the outside world and the
charge is moving across the field
lines as it goes up the field
Terminal “a” is assigned to be
point at higher voltage
+
-
v
v
b
Energy is absorbed when v
ab
is positive
a
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
The passive sign convention for labeling the voltage and current
of a two-terminal electric circuit element.
When the element current and voltage are labeled with the assumed
current to enter the terminal of assumed higher voltage the element
is labeled using the “passive sign convention”
If the actual current and
voltage agree with their
passive assignment the
element is absorbing energy
Terminal “a” is assigned to be
point at higher voltage
+ b
i(t)
v(t)
a
-
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
If either (but not both) the current or voltage differ from the
assumed passive assignment, then that negative component is
opposite in direction of its assumed orientation and the energy
is flowing in the opposite direction. (The two port element is
delivering power.)
If the actual current and
voltage agree with there
passive assignment the
element is absorbing energy
Terminal “a” is assigned to be
point at higher voltage
+ b
i(t)
v(t)
a
-
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Practice Problem-- Assume the “passive labeling”
system is being practiced.
+ b
i(t)
If the actual current and voltages are in
the orientations as shown, which twoterminal element device is delivering
power?
-
b
i(t)
v(t)
v(t)
The one on the left
a
-
Terminal “a” is assigned to be point at higher voltage
True; terminal a is
high energy point
a
False; terminal a not
high energy point
+
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Practice Problem-- Assume the “passive labeling” system is being practiced.
If the actual current and voltages are in the orientations as shown,
what is the power and direction of power for each of the following?
1)
+ b
2A
p(t) = v(t) i(t)
2)
-
b
4A
p(t) = (-3V)(4A) =-12 Watts
p(t) = (3V)(2A) = 6 Watts
3V
a
-
p(t) = [-v(t)] i(t)
3V
6W of power
absorbed by
this element
a
+
12W of power
delivered by
this element
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Practice Problem-- Assume the “passive labeling” system is being practiced.
If the actual current and voltages are in the orientations as shown,
what is the power and direction of power for each of the following?
3)
+ b
2A
p(t) = v(t) [-i(t)]
4)
-
b
-3A
p(t) = (-3V)(-3A) =9 Watts
p(t) = (5V)(-2A) =-10 Watts
5V
a
-
p(t) = [-v(t)] i(t)
3V
10W of power
delivered by
this element
a
+
9W of power
absorbed by
this element
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Is there a voltage gain or drop across the following resistors?
This two terminal electric circuit element is the resistor. The resistance
of material used to make a resistor is always a positive number.
1)
2)
b
v(t) = [-i(t)]R
+
-
b
v(t)
v(t)
+
a
v(t) is negative
indicating a
voltage gain
v(t) = i(t)R
-a
v(t) is positive
indicating a
voltage drop
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Is there a charge gain or drop in each of the following capacitor?
This two terminal electric circuit element is the capacitor. The
capacitance of a capacitor is always a positive number
1)
2)
-
b
i(t) = [- dv/dt)]C
i(t)
+ a
Energy being stored
Energy being released
i(t) is negative indicating
the charge in the
capacitor is dropping.
+
b
i(t) = [dv/dt)]C
i(t)
- a
i(t) is positive
indicating a gain in the
charge of the capacitor.
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Are the following inductors absorbing or releasing energy?
This two terminal electric circuit element is the inductor. The
inductance of practical inductors is always a positive number
1)
2)
-
b
v(t) = [ - di/dt)]L
v(t)
+ a
Energy being stored
Energy being released
v(t) is negative
indicating the inductor
is releasing energy.
+
b
v(t) = [di/dt)]L
v(t)
-
a
v(t) is positive
indicating the inductor
is absorbing energy.
Instrument Systems
Electrical Engineering Fundamentals
Other concepts that might seem confusing.
The perfect world.
The ideal voltage supply v(t)
The ideal current source i(t)
Is a car battery an ideal voltage supply?
The voltage developed in an ideal voltage supply is a dictated value that
does not change with age, use (misuse), application, snow, sleet, rain, or
etc, it just keeps providing the voltage it was advertised to provide.
Is a lighting bolt, (like the one Ben Franklin felt when he was
flying his kite) an ideal current supply?
An ideal current source will provide the stated amount of current (a specific
number of moving charges/unit time) no matter what the demand for that
flow of charge becomes or how long that demand lasts.
Instrument Systems
Electrical Engineering Fundamentals
Other concepts that might seem confusing.
Is the energizer bunnies battery an ideal voltage supply?
Hmmmmmm! Life does seems to be full of difficult questions!
Circuit configurations
Voltage as function of time
that is available for use.
b
v (t) +
s
-
v(t)
a
1)
2)
3)
v (t) = v(t)
s
Voltage value profile as a function of time remains the same.
Voltage value profile can be fixed value, dc, shaped like a sine wave, ac, or any other
desired shape. The main point is that the specific voltage value of the ideal supply at
a specific instant in time is always that exact value desired no matter what the
circumstances are.
Instrument Systems
Electrical Engineering Fundamentals
Other concepts that might seem confusing.
Is the energizer bunnies battery an ideal voltage supply?
Hmmmmmm! Life does seems to be full of difficult questions!
Circuit configurations
b
b
v
v (t) +
s
-
v(t)
v
s1
s2
(t)
(t)
+
-
+
-
v(t)
a
a
Ideal voltage supplies
connected in series.
v(t) = v s1(t)
+ v s2(t)
Instrument Systems
Electrical Engineering Fundamentals
Other concepts that might seem confusing.
Is the energizer bunnies battery an ideal voltage supply?
Hmmmmmm! Life does seems to be full of difficult questions!
Circuit configurations
b
b
b
v
(t)
vv s1
(t)
s
a
+
+
-
-
v
+
(t)v(t)
s2
-
v(t)
s2
(t)
(t)
+
-
+
-
v(t)
a
a
Ideal voltage supplies
connected in parallel.
v(t) = v s1(t)
v
s1
= v s2(t)
Ideal voltage supplies
connected in series.
v(t) = v s1(t)
+ v s2(t)
Instrument Systems
Electrical Engineering Fundamentals
Other concepts that might seem confusing.
Ideal current source
Circuit configurations
i(t)
b
i
(t)
s1
i
(t)
s2
a
Ideal current sources
connected in parallel.
i(t) = i
(t) + i s2(t)
s1
b
i
(t)
s2
i
(t)
s1
i(t)
a
Ideal current sources
connected in series.
i(t) = i
(t) = i s2(t)
s1
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Things to remember about transformer passive diagram symbolism.
This four terminal electric circuit
element is a coupled inductor
(transformer).
1) A current i1(t) entering a dotted terminal in one inductor produces a voltage,v1 (t),
across the terminals of that inductor.
2) A current i1(t) entering a dotted terminal in one inductor produces an
induced voltage which is positive at the other dotted terminal. M
(Mutual inductance)
3) If a second current i 2 (t) enters the other
inductor at its dotted terminal there is a
voltage drop produced across that inductor.
4) If a second current i 2 (t) enters the other
inductor at its dotted terminal there is also
a voltage induced in the first inductor.
(Mutual induction)
i1 b
+
v (t) + v (t)
1
1
a
d
i2
+
c
v (t) + v2 (t)
2
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Things to remember about transformer passive diagram symbolism.
b
d
5)
+ i1
i2 +
This four terminal
electric circuit
v (t) + v (t)
v (t) + v (t) = v (t)
1
1
1
a
c
v (t) element
=
2
inductor
2
2 is a coupled
(transformer).
1) A current i1(t) entering a dotted terminal in one inductor produces a voltage,v1 (t),
across the terminals of that inductor.
2) A current i1(t) entering a dotted terminal in one inductor produces an
induced voltage which is positive at the other dotted terminal. M
(Mutual inductance)
3) If a second current i 2 (t) enters the other
inductor at its dotted terminal there is a
voltage drop produced across that inductor.
4) If a second current i 2 (t) enters the other
inductor at its dotted terminal there is also
a voltage induced in the first inductor.
(Mutual induction)
i1 b
+
v (t) + v (t)
1
1
a
d
i2
+
c
v (t) + v2 (t)
2
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Things to remember about transformer passive diagram symbolism.
M = MM
12 = M 21
i1 b
d
i2
+
+
v (t) = v (t) + v (t)
1
1
1
a
c
v (t) = L1 (d i1 /dt )
1
v (t) = v2 (t) + v (t)
2
2
v (t) = L 2(d i2 /dt)
2
v (t) = M (d i2 /dt )
1
M
12
=
v (t) = M (d i1 /dt )
2
M
21
The voltage difference v1(t) between point b and point a is the sum of the voltage, v
in current i1 and the mutually induced voltage v
1
1
induced by the change
produced by the change in the flow of current i2.
The voltage difference v2(t) between point d and point c is the sum of the voltage, v 2 induced by the change
in current i2 and the mutually induced voltage v
2
produced by the change in the flow of current i1.
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Passive diagram example
i 1 (t)
M
12
+
+
A) how many equations are
needed to describe this three
coupled inductor system?
-
M
13
-
three
v (t) = v (t) + v (t) + v (t)
2
2
2
2
v (t) = v (t) + v (t) + v (t)
3
3
3
3
M
23
An example of a three coupled
coil system that is complaint
with the passive diagram
symbolism.
+
i 3 (t)
v (t) = v (t) + v (t) + v (t)
1
1
1
1
i 2 (t)
But it is more convenient
to arrange them this way!
v (t) =
1
v (t)
+ v (t)
1
+ v (t)
1
v (t) = v (t)
2
2
+ v (t)
2
+ v (t)
2
v (t) = v (t)
3
3
+ v (t)
3
+
1
v (t)
3
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Passive diagram example
i1 /dt) +
v (t) = L1v(d(t)
1
+
v
(t)
i /dt) +
v (t) + L2v(d(t)
2 2
2
v
(t)
1
v (t) =
2
v (t) =
3
v
3
(t) +
v (t)
i 1 (t)
1
1
2
i /dt)
L3v(d(t)
3 3
v (t) +
3
M
12
+
+
+
i 3 (t)
After a bit more manipulation and substitution we get.
v (t) =
1
L1 (d i1 /dt)
+
M
12
(d i2 /dt )
An example of a three
coupled coil system that is
complaint with the passive
diagram symbolism.
-
M
13
i 2 (t)
M
23
v (t) = M (d i2 /dt )
1
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Passive diagram example
i1 /dt) +
v (t) = L1v(d(t)
1
1
v (t) =
2
v (t) =
3
v
2
v (t)
i 1 (t)
v
+
1
1
i /dt) +
(t) + L2v(d(t)
2 2
v (t)
+
3
v
3
(t)
M
12
(t)
v (t)
2
i /dt)
L3v(d(t)
3 3
+
+
+
i 2 (t)
-
-
M
13
+
i 3 (t)
L1 (d i1 /dt)
v (t) =
2
M
v (t) =
3
M
21
+
M
(d i1 /dt ) +
(d i1 /dt )
31
+ M
(d i2 /dt )
12
+
L2 (d i2 /dt)
+
(d i2 /dt )
32
+
M
23
v (t) = M (d i2 /dt )
1
Note:
After a bit more manipulation and substitution we get.
v (t) =
1
An example of a three
coupled coil system that is
complaint with the passive
diagram symbolism.
M
M
M
(d i3 /dt )
13
23
(d i3 /dt )
L3 (d i3 /dt)
M
M
31
23
21
=
M
=
M
=
M
13
32
12
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Passive diagram example
M
12
i 1 (t)
The equations below model the three
inductor circuit to the right.
If the circuit you have to deal with is
not wired this way, you can manipulate
the equations by multiplying specific
components by (-1) to get the correct
model.
+
+
i 2 (t)
-
-
M
13
+
i 3 (t)
Try the next practice problem to see how this works.
v (t) =
1
L1 (d i1 /dt)
v (t) =
2
M
v (t) =
3
M
21
+
M
(d i1 /dt ) +
(d i1 /dt )
31
+ M
(d i2 /dt )
+
M
L2 (d i2 /dt)
+
M
(d i2 /dt )
32
+
12
An example of a three
coupled coil system that is
complaint with the passive
diagram symbolism.
13
(d i3 /dt )
23
(d i3 /dt )
L3 (d i3 /dt)
M
23
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Practice Problem-- Assume the “passive labeling” system is being practiced.
If the actual current and voltages in the following diagram are in the
orientations as shown, what are the three equations that determine
the voltages across each of the coils, respectively?
i 1 (t)
M12
+
Note that either:
i 2 (t)
+
A) the defined voltages and corresponding currents do
not all conform to the passive sign convention.
or
B) the positive voltage terminal is not at the dot.
-
-
M
M 13
+
-
1) Reverse direction of i1
2) Reverse direction of i2
3) Reverse direction of v2
4) Reverse direction of v3
23
i 3 (t)
To get to the correct set of equations,
use negative signs in the model
equation set everywhere such a sign
corresponds to a reversal of an item in
the diagram to return the diagram to its
passive diagram form. 4 reversals must
be accomplished.
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Practice Problem-- Assume the “passive labeling” system is being practiced.
If the actual current and voltages in the following diagram are in the
orientations as shown, what are the three equations that determine
the voltages across each of the coils, respectively?
i 1 (t)
+
-
+
+
-
-
M
M 13
-
+
v (t) =
1
-
L1 (d i1 /dt)
-
v (t) =
2
-
M
-
v (t) =
3
-
M
i 2 (t)
M12
+
-
23
i 3 (t)
21
-
+
M
12
(d i2 /dt )
-
L2 (d i2 /dt)
-
M
(d i1 /dt ) +
(d i1 /dt ) +
31
+ M
2) Reverse direction of i2
3) Reverse direction of v2
4) Reverse direction of v3
(d i3 /dt )
+ M 23 (d i3 /dt )
(d i2 /dt ) +
32
1) Reverse direction of i1
13
L3 (d i3 /dt)
Instrument Systems
Electrical Engineering Fundamentals
Other conventions that might seem confusing.
Practice Problem-- Assume the “passive labeling” system is being practiced.
If the actual current and voltages in the following diagram are in the
orientations as shown, what are the three equations that determine
the voltages across each of the coils, respectively?
i i11(t)
(t)
M12
++
-
+
-
-
-
M
M 13
-
+
v (t) =
1
i (t)
i 22(t)
+
+
-
+
-
-
L1 (d i1 /dt)
-
-
M
+
12
(d i2 /dt )
+ M
13
(d i3 /dt )
-
i1 /dt ) + forL2
(d voltages
(d i2three
/dt) coupled
v (t) =
+ M 23 (d i3 /dt )
ThusM 21
the
the
2
coils are modeled with the following three
equations.
v (t) =
- M 31 (d i1 /dt ) +- M 32 (d i2 /dt ) + L3 (d i3 /dt)
3
23
v (t) =
1
i 3 (t)
-+v2(t)
-
-
L1 (d i1 /dt)
-
= -+ M 21 (d i1 /dt ) -+
-
-
+ v (t) = + M (d i1 /dt ) +
31
3
M
12
(d i2 /dt )
L2 (d i2 /dt)
M
+ M
13
(d i3 /dt )
-
+ M 23 (d i3 /dt )
-
(d i2 /dt ) +
32
L3 (d i3 /dt)