Electrical Circuit Analysis Ch 01 basic concepts

CIRCUITS 1
DEVELOP TOOLS FOR THE ANALYSIS AND DESIGN OF
BASIC LINEAR ELECTRIC CIRCUITS

A FEW WORDS ABOUT ANALYSIS
USING MATHEMATICAL MODELS

BASIC STRATEGY USED IN ANALYSIS

MATHEMATICAL ANALYSIS
DEVELOP A SET OF MATHEMATICAL
EQUATIONS THAT REPRESENT THE CIRCUIT
- A MATEMATICAL MODEL -
LEARN HOW TO SOLVE THE MODEL TO
DETERMINE HOW THE CIRCUIT WILL BEHAVE
IN A GIVEN SITUATION
THIS COURSE TEACHES THE BASIC TECHNIQUES
TO DEVELOP MATHEMATICAL MODELS FOR
ELECTRIC CIRCUITS
THE MATHEMATICS CLASSES - LINEAR ALGEBRA,
DIFFERENTIAL EQUATIONS- PROVIDE THE TOOLS
TO SOLVE THE MATHEMATICAL MODELS
FOR THE FIRST PART WE WILL BE EXPECTED
TO SOLVE SYSTEMS OF ALGEBRAIC EQUATIONS
20642
0164
84912
321
321
321
=+−−
=++−
=−−
VVV
VVV
VVV
LATER THE MODELS WILL BE DIFFERENTIAL
EQUATIONS OF THE FORM
f
dt
df
y
dt
dy
dt
yd
fy
dt
dy
4384
3
2
2
+=++
=+
THE MODELS THAT WILL BE DEVELOPED HAVE
NICE MATHEMATICAL PROPERTIES.
IN PARTICULAR THEY WILL BE LINEAR WHICH
MEANS THAT THEY SATISFY THE PRINCIPLE OF
SUPERPOSITION
1 1 2 2 1 1 2 2
Model
Principle of Superposition
( ) ( ) ( )
y Tu
T u u T u T uα α α α
=
+ = +

a b
2 T E R M I N A L S C O M P O N E N T
c h a r a c t e r i z e d b y t h e
c u r r e n t t h r o u g h i t a n d
t h e v o l t a g e d i f f e r e n c e
b e t w e e n t e r m i n a l s
N O D E
N O D E
ELECTRIC CIRCUIT IS AN INTERCONNECTION OF ELECTRICAL COMPONENTS
LOW DISTORTION POWER AMPLIFIER
+
-
L
C
1R
2R
Sv −
+
Ov
TYPICAL LINEAR
CIRCUIT
The concept of node is extremely
important.
We must learn to identify a node
in any shape or form

BASIC CONCEPTS
LEARNING GOALS
•System of Units: The SI standard system; prefixes
•Basic Quantities: Charge, current, voltage, power and energy
•Circuit Elements: Active and Passive

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Electrical Circuit Analysis Ch 01 basic concepts

Information at the foundation of
modern science and technology
from thePhysics Laboratoryof NIST
Detailed contents
Values of the constantsand related information
Searchable bibliographyon the constants
In-depth information on the SI, the modern
metric system
Guidelinesfor the expression
of uncertainty in measurement
About this reference. Feedback.
Privacy Statement / Security Notice - NIST Disclaimer

SI DERIVED BASIC ELECTRICAL UNITS

ONE AMPERE OF CURRENT CARRIES ONE COULOMB OF CHARGE EVERY SECOND.
ELECTRONONEOFCHARGETHEIS(e)
(e)106.28COULOMB1 18
×=
sCA ×=
VOLT IS A MEASURE OF ENERGY PER CHARGE.
TWO POINTS HAVE A VOLTAGE DIFFERENCE OF ONE VOLT IF ONE COULOMB OF CHARGE
GAINS ONE JOULE OF CHARGE WHEN IT IS MOVED FROM ONE POINT TO THE OTHER.
C
J
V =
OHM IS A MEASURE OF THE RESISTANCE TO THE FLOW OF CHARGE.
THERE IS ONE OHM OF RESISTENCE IF IT IS REQUIRED ONE VOLT OF ELECTROMOTIVE FORCE
TO DRIVE THROUGH ONE AMPERE OF CURRENT
A
V
=Ω
IT IS REQUIRED ONE WATT OF POWER TO DRIVE ONE AMPER OF CURRENT AGAINST AN
ELECTROMOTIVE DIFFERENCE OF ONE VOLTS
AVW ×=

Strictly speaking current is a basic quantity and charge is derived. However,
physically the electric current is created by a movement of charged particles.
+
+
+
)(tq
+
What is the meaning of a negative value for q(t)?
PROBLEM SOLVING TIPPROBLEM SOLVING TIP
IF THE CHARGE IS GIVEN DETERMINE THE CURRENT BY
DIFFERENTIATION
IF THE CURRENT IS KNOWN DETERMINE THE CHARGE BY
INTEGRATION
A PHYSICAL ANALOGY THAT HELPS VISUALIZE ELECTRIC
CURRENTS IS THAT OF WATER FLOW.
CHARGES ARE VISUALIZED AS WATER PARTICLES

+
+
+
)(tq
+
EXAMPLE
])[120sin(104)( 3
Cttq π−
×=
=)(ti )120cos(120104 3
tππ×× − ][A
][)120cos(480.0)( mAtti ππ=
EXAMPLE



≥
<
= −
0
00
)( 2
tmAe
t
ti t
FIND THE CHARGE THAT PASSES
DURING IN THE INTERVAL 0<t<1
== ∫
−
1
0
2
dxeq x
)
2
1
(
2
1
2
1 02
1
0
2
eee x
−−−=− −−
FIND THE CHARGE AS A FUNCTION OF TIME
∫ ∫
∞− ∞−
−
==
t t
x
dxedxxitq 2
)()(
0)(0 =⇒≤ tqt
∫
−−
−==⇒>
t
tx
edxetqt
0
22
)1(
2
1
)(0
And the units for the charge?...
)1(
2
1 2−
−= eq Units?

1 2 3 4 5 610−
10
20
30
Charge(pC)
Time(ms)
Here we are given the
charge flow as function
of time.
)/(1010
0102
10101010 9
3
1212
sC
s
C
m −
−
−−
×−=
−×
×−×−
=
1 2 3 4 5 610−
10
20
30
Time(ms)
)Current(nA
40
20−
DETERMINE THE
CURRENT
To determine current we
must take derivatives.
PAY ATTENTION TO
UNITS

CONVENTION FOR CURRENTS
IT IS ABSOLUTELY NECESSARY TO INDICATE
THE DIRECTION OF MOVEMENT OF CHARGED
PARTICLES.
THE UNIVERSALLY ACCEPTED CONVENTION IN
ELECTRICAL ENGINEERING IS THAT CURRENT IS
FLOW OF POSITIVE CHARGES.
AND WE INDICATE THE DIRECTION OF FLOW
FOR POSITIVE CHARGES
-THE REFERENCE DIRECTION-
a b
a
a
ab
b
b
A3
A3− A3
A3−
THE DOUBLE INDEX NOTATION
IF THE INITIAL AND TERMINAL NODE ARE
LABELED ONE CAN INDICATE THEM AS
SUBINDICES FOR THE CURRENT NAME
a bA5 AIab 5=
AIab 3=
AIba 3−=
AIab 3−=
AIba 3=
POSITIVE CHARGES
FLOW LEFT-RIGHT
POSITIVE CHARGES
FLOW RIGHT-LEFT
baab II −=
A POSITIVE VALUE FOR
THE CURRENT INDICATES
FLOW IN THE DIRECTION
OF THE ARROW (THE
REFERENCE DIRECTION)
A NEGATIVE VALUE FOR
THE CURRENT INDICATES
FLOW IN THE OPPOSITE
DIRECTION THAN THE
REFERENCE DIRECTION

This example illustrates the various ways
in which the current notation can be used
b
a
I
A3
=
=
−=
ab
cb
I
AI
AI
4
2
A2
c

CONVENTIONS FOR VOLTAGES
ONE DEFINITION FOR VOLT
TWO POINTS HAVE A VOLTAGE DIFFERENTIAL OF
ONE VOLT IF ONE COULOMB OF CHARGE GAINS
(OR LOSES) ONE JOULE OF ENERGY WHEN IT
MOVES FROM ONE POINT TO THE OTHER
+ a
b
C1
IF THE CHARGE GAINS
ENERGY MOVING FROM
a TO b THEN b HAS HIGHER
VOLTAGE THAN a.
IF IT LOSES ENERGY THEN
b HAS LOWER VOLTAGE
THAN a
DIMENSIONALLY VOLT IS A DERIVED UNIT
sA
mN
•
•
==
COULOMB
JOULE
VOLT
VOLTAGE IS ALWAYS MEASURED IN A RELATIVE FORM AS THE VOLTAGE DIFFERENCE
BETWEEN TWO POINTS
IT IS ESSENTIAL THAT OUR NOTATION ALLOWS US TO DETERMINE WHICH POINT
HAS THE HIGHER VOLTAGE

THE + AND - SIGNS
DEFINE THE REFERENCE
POLARITY
V IF THE NUMBER V IS POSITIVE POINT A HAS V
VOLTS MORE THAN POINT B.
IF THE NUMBER V IS NEGATIVE POINT A HAS
|V| LESS THAN POINT B
POINT A HAS 2V MORE
THAN POINT B
POINT A HAS 5V LESS
THAN POINT B

THE TWO-INDEX NOTATION FOR VOLTAGES
INSTEAD OF SHOWING THE REFERENCE POLARITY
WE AGREE THAT THE FIRST SUBINDEX DENOTES
THE POINT WITH POSITIVE REFERENCE POLARITY
VVAB 2=
VVAB 5−= VVBA 5=
BAAB VV −=

ENERGY
VOLTAGE IS A MEASURE OF ENERGY PER UNIT CHARGE…
CHARGES MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB OR
RELEASE ENERGY – THEY MAY TRANSFER ENERGY FROM ONE POINT TO ANOTHER
BASIC FLASHLIGHT Converts energy stored in battery
to thermal energy in lamp filament
which turns incandescent and glows
The battery supplies energy to charges.
Lamp absorbs energy from charges.
The net effect is an energy transfer
EQUIVALENT CIRCUIT
Charges gain
energy here
Charges supply
Energy here

WHAT ENERGY IS REQUIRED TO MOVE 120[C] FROM
POINT B TO POINT A IN THE CIRCUIT?
VVAB 2=
JVQW
Q
W
V 240==⇒=
THE CHARGES MOVE TO A POINT WITH HIGHER
VOLTAGE -THEY GAINED (OR ABSORBED) ENERGY
THE CIRCUIT SUPPLIED ENERGY TO THE CHARGES
ENERGY
VOLTAGE IS A MEASURE OF ENERGY PER UNIT CHARGE…
CHARGES MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB OR
RELEASE ENERGY

VVAB 5=
−
+
V5
WHICH POINT
HAS THE HIGHER
VOLTAGE?
THE VOLTAGE
DIFFERENCE
IS 5V

EXAMPLE
A CAMCODER BATTERY PLATE CLAIMS THAT
THE UNIT STORES 2700mAHr AT 7.2V.
WHAT IS THE TOTAL CHARGE AND ENERGY
STORED?
CHARGE
THE NOTATION 2700mAHr INDICATES THAT
THE UNIT CAN DELIVER 2700mA FOR ONE
FULL HOUR
][1072.9
13600102700
3
3
C
Hr
Hr
s
S
C
Q
×=
××



×= −
TOTAL ENERGY STORED
THE CHARGES ARE MOVED THROUGH A 7.2V
VOLTAGE DIFFERENTIAL
][10998.6
][2.71072.9][
4
3
J
J
C
J
VCQW
×=
××=



×=
ENERGY AND POWER
2[C/s] PASS
THROUGH
THE ELEMENT
EACH COULOMB OF CHARGE LOSES 3[J]
OR SUPPLIES 3[J] OF ENERGY TO THE
ELEMENT
THE ELEMENT RECEIVES ENERGY AT A
RATE OF 6[J/s]
THE ELECTRIC POWER RECEIVED BY THE
ELEMENT IS 6[W]
HOW DO WE RECOGNIZE IF AN ELEMENT
SUPPLIES OR RECEIVES POWER?
VIP =
IN GENERAL
∫=
2
1
)(),( 12
t
t
dxxpttw

PASSIVE SIGN CONVENTION
POWER RECEIVED IS POSITIVE WHILE POWER
SUPPLIED IS CONSIDERED NEGATIVE
A CONSEQUENCE OF THIS CONVENTION IS THAT
THE REFERENCE DIRECTIONS FOR CURRENT AND
VOLTAGE ARE NOT INDEPENDENT -- IF WE
ASSUME PASSIVE ELEMENTS
a b
−+ abV
abI
ababIVP =
IF VOLTAGE AND CURRENT
ARE BOTH POSITIVE THE
CHARGES MOVE FROM
HIGH TO LOW VOLTAGE
AND THE COMPONENT
RECEIVES ENERGY --IT IS
A PASSIVE ELEMENT
a b
−+ abV
GIVEN THE REFERENCE POLARITY
a b
abI
IF THE REFERENCE DIRECTION FOR CURRENT
IS GIVEN
−+
THIS IS THE REFERENCE FOR POLARITY
REFERENCE DIRECTION FOR CURRENT
a b
−+ abV
VVab 10−=
EXAMPLE
THE ELEMENT RECEIVES 20W OF POWER.
WHAT IS THE CURRENT?
abI
SELECT REFERENCE DIRECTION BASED ON
PASSIVE SIGN CONVENTION
ababab IVIVW )10(][20 −==
][2 AIab −=
A2

Voltage(V) Current A - A' S1 S2
positive positive supplies receives
positive negative receives supplies
negative positive receives supplies
negative negative supplies receives
S2S1
We must examine the voltage across the component
and the current through it
0,0 <> ABAB IV
1SON
0,0 '''' >> BABA IV
2SON
A A’
B B’
''''2
1
BABAS
ABABS
IVP
IVP
=
=
UNDERSTANDING PASSIVE SIGN CONVENTION
0,0 '''' >< BABA IV
S2ON
−
+
V
I

CHARGES RECEIVE ENERGY.
THIS BATTERY SUPPLIES ENERGY
CHARGES LOSE ENERGY.
THIS BATTERY RECEIVES THE ENERGY
WHAT WOULD HAPPEN IF THE CONNECTIONS ARE REVERSED
IN ONE OF THE BATTERIES?

DETERMINE WHETHER THE ELEMENTS ARE SUPPLYING OR RECEIVING POWER
AND HOW MUCH
a
b
a
b
WHEN IN DOUBT LABEL THE
TERMINALS
OF THE COMPONENT
AIab 4=
VVab 2−=
WP 8−= SUPPLIES POWER
VVab 2=
2A−
2abI A= −
4P W= − ABSORBS POWER

1
2
1
2
AIVV 4,12 1212 −== AIVV 2,4 1212 ==
WHEN IN DOUBT LABEL THE
TERMINALS
OF THE COMPONENT

SELECT VOLTAGE REFERENCE POLARITY
BASED ON CURRENT REFERENCE DIRECTION
−
+
)5(][20 AVW AB ×=−
][4 VVAB −= −
+
IVW ×−= )5(][40
][8 AI −=

SELECT HERE THE CURRENT REFERENCE DIRECTION
BASED ON VOLTAGE REFERENCE POLARITY
A2−
)2(][40 1 AVW −×=
][201 VV −=
IVW ×=− ])[10(][50
][5 AI −=
WHICH TERMINAL HAS HIGHER VOLTAGE AND WHICH IS THE CURRENT FLOW DIRECTION

+
-
−
+
V24
−+ V6
−
+
V18
A2
A2
1
23
P1 = 12W
P2 = 36W
P3 = -48W
)2)(6(1 AVP =
)2)(18(2 AVP =
)2)(24()2)(24(3 AVAVP −=−=
IMPORTANT: NOTICE THE POWER BALANCE IN THE CIRCUIT
COMPUTE POWER ABDORBED OR SUPPLIED BY EACH ELEMENT

CIRCUIT ELEMENTS
PASSIVE ELEMENTS
INDEPENDENT SOURCES
VOLTAGE
DEPENDENT
SOURCES
CURRENT
DEPENDENT
SOURCES
?,,, βµ rgFORUNITS

EXERCISES WITH DEPENDENT SOURCES
OVFIND ][40 VVO = OIFIND mAIO 50=

DETERMINE THE POWER SUPPLIED BY THE DEPENDENT SOURCES
][40 V
][80])[2])([40( WAVP −=−=
TAKE VOLTAGE POLARITY REFERENCE TAKE CURRENT REFERENCE DIRECTION
][160])[44])([10( WAVP −=×−=

POWER ABSORBED OR SUPPLIED BY EACH
ELEMENT
][48)4)(12(1 WAVP ==
][48)2)(24(2 WAVP ==
][56)2)(28(3 WAVP ==
][8)2)(4()2)(1( WAVAIP xDS −=−=−=
][144)4)(36(36 WAVP V −=−=
NOTICE THE POWER BALANCE

USE POWER BALANCE TO COMPUTE Io
W12−
))(6( OI )9)(12( −
)3)(10( −
)8)(4( − )11)(28( ×
][1 AIO =
POWER BALANCE

Electrical Circuit Analysis Ch 01 basic concepts

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Electrical Circuit Analysis Ch 01 basic concepts