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Chapter 1 - Basic Concepts
• 1.1 System of Units
• 1.2 Basic Quantities
• 1.3 Circuit Elements
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1
Reading Assignment:
Chapter 1, Pages 1-11
Homework
End-of-Chapter Problems: P1.13, P1.18, P1.21
(PLUS WebCT Problems)
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“Today we live in a predominately electrical world.”
Electrotechnology pervades our day-to-day lives:
electrical appliances
 Refrigerators, stoves, stereos, hairdryers, ...
communication devices
 telephones, faxes, computer and data networks
computers
 main frames, personal (PCs), workstations, embedded systems,
PLCs
instrumentation
 meters, measurement systems, controls
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System of Units
The international system of units (SI) is composed of
the following basic units of measure:
Unit of Measure
meter
kilogram
second
ampere
degree Kelvin
candela
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Symbol
m
kg
s
A
K
cd
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Measures
distance
mass (weight)
time
electric current
temperature
luminous intensity
4
Standard SI Prefixes
Standard prefixes are employed with SI units to represent
powers of ten. These prefixes, sometimes referred to as
engineering notation, are used extensively in representing the
values of various circuit parameters.
Prefix
pico (p)
nano (n)
micro ()
milli (m)
Power
10-12
10-9
10-6
10-3
10-12 10-9 10-6 10-3
pico nano micro milli
(p) (n) ( or u) (m)
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Prefix
kilo (k)
mega (M)
giga (G)
tera (T)
1
103
106
Power
10+3
10+6
10+9
10+12
109
1012
kilo mega giga tera
(k) (M) (G) (T)
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TI-85 MODE Function
To Select Engineering Notation
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Example - Using Engineering Notation
If the local power company charges $0.12 per kWh
(kilowatt-hour) of energy used, how much would it
cost to illuminate a 100-Watt lamp for an entire year?
Solution:
Energy  Power  Time
W  P  t (kW  hours or kWh )
24hours
W  100W  365days 
day
W  876kWh
Cost  876kWh  $0.12 per kWh
Cost  $105.12
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TI-85 Screen
(Wh)
(kWh)
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Basic Quantities
• What is Electricity? - A natural force (like wind or water
flow) which can be harnessed for some beneficial purpose.
• What causes electricity? Just as wind is due to the
movement of air molecules, electricity is caused by the
movement of charged particles (i.e., electrons).
• The study of electricity has been going on since about 600
B.C. when it was noticed that amber, when rubbed,
acquired the ability to pick up light objects (we call this
static electricity).
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Definitions
Charge is an intrinsic property of matter (like mass and size).
Symbol1: q or Q
Unit: coulomb (C)
The elemental unit of charge is the charge of one electron, e:
e = -1.6022 x 10-19 C (1 proton = +1.6022 x 10-19 C)
Current is a measure of the amount of charge flowing past a
point in a given amount of time.
Symbol1: i or I
Unit: ampere (A)
1A  1C/s (coulomb per second)
1Lowercase
is used to indicate a variable (time-dependent) quantity while
upper case indicates a constant value.
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Current has magnitude and direction.
- -
2A
Conventional
Current Flow
- -
Electrical Conductor Segment
Electron Flow
(The flow of electrons is analogous to the flow
of water molecules in a pipe or hose.)
Note:
2A = 2C per second = (2/1.6022E-19) electrons per second
2A  12,500,000,000,000,000,000 electrons per second
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Typical Current Magnitudes
106
Lightning Bolt
4
10
Large Industrial Motor Current
2
Current in Amperes (A)
10
Typical Household Appliance Current
0
10
Causes Ventricular Fibrillation in Humans
-2
10
10-4
10-6
IC Memory Cell Current
-8
10
10-10
10-12
Synaptic Current (Brain Cell)
10
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11
Direct Current (dc) Versus Alternating Current (ac)
DC signals are constant with respect to time:
Ix
2A
t
AC signals vary with respect to time:
i(t) = 2e-at
i(t)
2A
2A
-1A
t
t
periodic ac signal
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nonperiodic ac signal
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Charge - Current Relationships
Current is proportional to the
rate of change of charge.
i(t)  dq(t)
dt
dq(t)  i(t)dt
t
t
0
0
t dq()  t i()d
t

 t
t
q()   i()d
t


0
0
t
q(t)  q(t )   i()d
0 t0
t
q(t)  q(t )   i()d
0 t0
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Charge is increased or decreased
depending on the area under the
current function.
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Example 1
Given:
Find:
q(t) = 12t C
i(t)
Solution: Since current is proportional to
the rate of change of charge, we take the
derivative of q(t) to find i(t):
i(t) = q(t)
i(t) = 12 A
Since the slope (rate of change) of q(t) is
constant at 12 Coulombs per second, the
current is constant.
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Example 2
Given: The current entering an element is described by the
function in the graph shown below.
i(t)
3
i(t)  1 0  t  1
t t 1



2
1
1
2
3
t
(seconds)
Find: The amount of change in the charge of the element
over the interval 0 to 3 seconds.
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Example 2 - Solution
To find the change in the charge of the element, we need to find the area
under the function i(t) over the interval 0 to 3 s. This can be done two
ways for this example: by integration or by using area formulas for
rectangles and triangles.
i(t)
Q  Area 1A 3s  1  2s 2A  5C
2
3
 2
1
3
Q  1dt   tdt  t 1   t   (1 0)  (4.5  0.5)  5C
0 2

1
0
1
3
2
1
1
2
3
t
Note: 5C is the increase in the charge of the element. We must
know the initial charge to determine the total charge of the element.
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Definition - Voltage
As stated earlier, current is the flow of charge through a
conductor. Voltage is the force required to move the charge
through the conductor. Hence, voltage is also referred to as
electromotive force and potential difference.
i (current flow)
+
a
+
Vab
Battery
+
Lamp
b
-
Vab
-
i
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i (current flow)
+
+
Vab
Battery
Vab
Lamp
-
-
The current from the battery flows through a voltage rise since the
battery is supplying energy. There is an equal voltage drop across the
lamp since the lamp is absorbing the energy supplied by the battery. This
is similar to the counter force created by friction when pushing an object
up an incline.
a
Fab
Fab
b
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Electromotive force is measured in volts. If one joule of energy is
required to move one coulomb of charge through an element, then
there is one volt (V) of potential difference across the element.
1volt 1 joule
coulomb
Like current, the voltage across an element may be positive
or negative. The assignment of a sign to a voltage is
somewhat arbitrary, but usually represents whether potential
difference is above or below some reference potential.
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(6 V above ground)
i
+
+
+
6V
Battery
Lamp
6V
-
-
“0” volts (ground)
As we traverse the circuit in a clockwise direction, there is a
voltage rise across the battery; we could assign this a “positive”
sign. Since we experience a voltage drop across the lamp, this
would be assigned the opposite sign or a “negative” value. We
could just as correctly assign all voltage drops a positive sign
and, therefore, all voltage rises a negative sign.
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Power and Energy
Energy is the capacity to do work. For example, energy is
stored in a battery and, when connected to a “load” such as a
lamp, there is a continuous exchange of energy from the
battery to the load. (There is also a transformation from
energy stored in a chemical form to energy emitted in the
form of heat and light.) Typically, electrical energy is measure
in joules (J) or watt-hours (W-h). (1W = 1J/s)
i
+
Vab
+
Battery
Lamp
-
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Vab
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Power is a measure of how much work is done per unit time,
or, in other words, a measure of the rate at which energy is
transferred (supplied or absorbed). There is always a
conservation of energy (power). That is, the battery supplies
exactly the amount of power consumed by the load, no more
no less. Typically, electrical power is measured in watts (W).
power flows from the source to the load
p  dw
dt
+
V
i +
i
Battery
-
(Passive sign
convention)
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Lamp
V
-
p = -vi
p = +vi
Supplies Power
Absorbs Power
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Mathematical Relationships
In general:
power  force  velocity
In electrical terms: p  v  i watts
1watt  1volt  amp  1va 1 joule
sec ond
Using previous
definitions:
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v p
i
but p  dw and i  dq
dt
dt
so,


 dw



dt

  dw
v 

dq
 dq





dt


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p  dw
dt
dw  pdt
t
t
t0
t0
 dw   pdt
t
w(t)  w(t )   pdt
0
t0
The total energy absorbed or supplied is equal to the
area under the power function in the interval t0 to t.
If the area is positive, then energy is absorbed by an
element; if the area is negative, the element is
supplying power.
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Example 3 - The current in a typical lightening bolt is 20KA, and
the bolt lasts typically for 0.1 seconds. If the voltage between the
earth and clouds is 500MV, find (a) the total charge transmitted
by the bolt and (b) the total energy released by the bolt.
4
4
Solution: (a) Q   i(t)dt   2x10 dt [2x10 t]00.1  2000C
0.1
0.1
0
0
8
w

Q

V

(
2000
C
)

(
5
x
10
V)
(b)
11
Note,
v (t ) 
12
w 10x10 C  V 1x10 J 1TJ
dq
If V is constant,
A typical automobile battery is rated at 12V
and 100 amp-hours (Ah).
100Ah = (100 C/s)(3600s) = 3.6 105C 
12V3.6 105C = 4.32 106J (MJ)
The lightening strike has as much energy as
230,000 batteries!
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Circuits I
dw
Then,
V 
W
Q
W  V  Q
25
Example 4 - (a) Find the power absorbed by the element
shown and (b) the total energy absorbed over a 10-second
interval.
10A +
Solution:
(a)
4v
-
p = vi =4v x 10A = 40w
(b) w  40dt  400J
10

0
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Example 5 - Find the power absorbed or supplied by
the elements shown below
4A -
2v +
-2A + 2v -
(a)
Solution:
(b)
(a) p = -vi = -2v x 4A = -8w
(element-a is supplying 8w of power)
(b) p = -vi = -2v x (-2A) = +4w
(element-b is absorbing 4w of power)
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Example 6 - Determine the unknown voltage or
current.
5A - V=? +
I=?
(a)
(b)
P = -20w
Solution:
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- 5v +
P = 40w
(a)
-20w = -V x 5A
V = 4v
(b)
40w = -(5v x I)
I = -8A
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Circuit Elements
• Circuits (electrical systems) consist interconnected devices.
• Electrical devices can be categorized as either a source or a load.
• A source can be either independent or dependent, a voltage source
or a current source.
• Independent voltage sources provide power at a constant voltage
• Independent current sources provide power at a constant current.
• The voltage level of a voltage source is not affected by the
amount of power demanded from it.
• Similarly, the current level of a current source is not affected by
the amount of power demanded from it.
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Independent Voltage Sources:
I
i(t)
+
v(t)
-
+
-
LOAD
Time-varying
+
V
-
LOAD
Constant (battery)
v
Since p = vi, as the
demand for power
increases, the current,
i, increases because v
is constant.
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V
p
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30
Independent Current Sources:
+
v(t)
or
V
-
i(t)
or
I
LOAD
i
Since p = vi, as the
demand for power
increases, the
voltage,v, increases
because i is constant.
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I
p
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Dependent Source (Controlled) Sources)
An independent source provides power to a circuit at a constant
voltage or current, and the level of that voltage or current is
unaffected by the circuit to which it is attached. Generally,
independent sources are used to represent devices such as
batteries or generators.
A voltage or current associated with a dependent source is
controlled or affected by the circuit to which it is attached.
Dependent sources, also called controlled sources, are used to
model electronic devices, sensors and other devices which are
influenced by some physical or electrical parameter. For
example, a device called a thermocouple generates a small
voltage (in the millivolt range) which is proportional to
temperature. So, a thermocouple can be thought of as a
temperature-controlled voltage source. That is, v(t) = KT, where
TFall
is 2001
the temperature and k is aCircuits
constant,
I
32
Dependent Sources
There are four types of controlled sources:
voltage-controlled voltage sources
current-controlled voltage sources
voltage-controlled current sources
current-controlled current sources
VCVS
+
vo
-
+
-
CCVS
io
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+
Circuits I
(VCVS)
(CCVS)
(VCCS)
(CCCS)
+
v = vo
-
+
v = rio
33
Dependent Sources
i = gvo
VCCS
+
vo
-
i =vo
CCCS
io
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Example 7 - Given the two networks shown below
we want to determine the outputs.
(a)
+
Vs = 2V
-
(b)
Is = 1mA
+
-
Solution:
+
Vo= 20Vs
-
(a) Vo = 20  2V = 40V
Io
(b) Io = 50  1mA = 50mA
50Is
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Example 8 - Compute the power that is supplied or
absorbed by each element in the network below.
Ix = 4A
1Ix
+ 12V -
2A
1
+
36V
-
Solution:
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- +
+
+
24V
+
3
2
-
-
2A
P36V = -(36V)(4A) = -144W
P1 = +(12V)(4A) = +48W
P2 = +(24V)(2A) = +48W
PDS = -(4V)(2A) = -8W
P1 = +(28V)(2A) = +56W
Total = 0W
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28V
-
(supplies 144W)
(absorbs 48W)
(absorbs 48W)
(supplies 8W)
(absorbs 56W)
36
Example 9 - Determine the value of Io.
Ix = 2A
- 6V +
+ 6V -
2A
- 12V +
Io
1
2
11A
+
+
10V
3A
9A
3
-
8A
+
4V
+
-
8Ix
-
Solution:
P2A = -(6V)(2A) = -12W
P1 = +(6V)(Io) = 6IoW = ?
P2 = -(12V)(9A) = -108W
P3 = -(10V)(3A) = -30W
P4V = -(4V)(8A) = -32W
PDS = +(16V)(11A) = 176W
Since PTotal = 0W = -182 + 6Io + 176  6Io = 6W  Io = 1A
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