Download Lecture material (Chap. 1)

ELECTRIC CIRCUIT IS AN INTERCONNECTION OF ELECTRICAL COMPONENTS
The concept of node is extremely
important.
We must learn to identify a node
in any shape or form
2 TERMINALS COMPONENT
a
b
NODE
characterized by the
current through it and
the voltage difference
between terminals
NODE
L
R1
R2
vS

TYPICAL LINEAR
CIRCUIT
vO

+
-
C
LOW DISTORTION POWER AMPLIFIER
CONVENTION FOR CURRENTS
THE DOUBLE INDEX NOTATION
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 DIRECTIONA 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
IF THE INITIAL AND TERMINAL NODE ARE
LABELED ONE CAN INDICATE THEM AS
SUBINDICES FOR THE CURRENT NAME
a
5A
b
I ab  5 A
a 3A b a  3A b
I ab  3 A
I ab  3 A
a  3A b a 3A b
Iba  3 A
POSITIVE CHARGES
FLOW LEFT-RIGHT
I ba  3 A
POSITIVE CHARGES
FLOW RIGHT-LEFT
Iab   Iba
I  2 A
a
2A
I
b
I cb  4 A
I ab 
c
3A
This example illustrates the various ways
in which the current notation can be used
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
VAB  2V
VAB  5V
VBA  5V
VAB  VBA
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 glow
EQUIVALENT CIRCUIT The battery supplies energy to charges.
Lamp absorbs energy from charges.
The net effect is an energy transfer
Charges gain
energy here
Charges supply
Energy here
ENERGY
VOLTAGE IS A MEASURE OF ENERGY PER UNIT CHARGE…
CHARGES MOVING BETWEEN POINTS WITH DIFFERENT VOLTAGE ABSORB OR
RELEASE ENERGY
WHAT ENERGY IS REQUIRED TO MOVE 120[C] FROM
POINT B TO POINT A IN THE CIRCUIT?
THE CHARGES MOVE TO A POINT WITH HIGHER
VOLTAGE -THEY GAINED (OR ABSORBED) ENERGY
THE CIRCUIT SUPPLIED ENERGY TO THE CHARGES
VAB  2V
V
W
 W  VQ  240J
Q
THE VOLTAGE
DIFFERENCE
IS 5V

5V

WHICH POINT
HAS THE HIGHER
VOLTAGE?
VAB  5V
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]
IN GENERAL
P  VI
t2
w (t 2 , t1 )   p( x )dx
t1
HOW DO WE RECOGNIZE IF AN ELEMENT
SUPPLIES OR RECEIVES POWER?
PASSIVE SIGN CONVENTION
POWER RECEIVED IS POSITIVE WHILE POWER
SUPPLIED IS CONSIDERED NEGATIVE
 Vab 
a
b
I ab
P  Vab I ab
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 CONSEQUENCE OF THIS CONVENTION IS THAT
THE REFERENCE DIRECTIONS FOR CURRENT AND
VOLTAGE ARE NOT INDEPENDENT -- IF WE
ASSUME PASSIVE ELEMENTS
GIVEN THE REFERENCE POLARITY
 Vab 
a
b
REFERENCE DIRECTION FOR CURRENT
THIS IS THE REFERENCE FOR POLARITY


a
b
I ab
IF THE REFERENCE DIRECTION FOR CURRENT
IS GIVEN
EXAMPLE
 Vab 
2A
a
b
I ab
Vab  10V
THE ELEMENT RECEIVES 20W OF POWER.
WHAT IS THE CURRENT?
SELECT REFERENCE DIRECTION BASED ON
PASSIVE SIGN CONVENTION
20[W ]  Vab Iab  (10V ) Iab
I ab  2[ A]
UNDERSTANDING PASSIVE SIGN CONVENTION
We must examine the voltage across the component
and the current through it
I
A
S1
B
Current A - A'
positive
positive
positive negative
negative positive
negative negative
Voltage(V)
A’

V

S1
supplies
receives
receives
supplies
PS1  VAB I AB
S2
PS 2  VA'B ' I A'B '
B’
S2
ON S1
ON S2
receives VAB  0, I AB  0 VA B  0, I A B  0
supplies
ON S2
supplies
V A'B '  0, I A'B '  0
receives
'
'
'
'
DETERMINE WHETHER THE ELEMENTS ARE SUPPLYING OR RECEIVING POWER
AND HOW MUCH
a
a
Vab  2V
I ab  4 A
2A
I ab  2 A
Vab  2V
P  8W
SUPPLIES POWER
b
P  4W ABSORBS POWER
WHEN IN DOUBT LABEL THE TERMINALS
OF THE COMPONENT
b
WHEN IN DOUBT LABEL THE TERMINALS
OF THE COMPONENT
1
1
2
2
V12  12V , I12  4 A
V12  4V , I12  2 A
I  8[ A]
VAB  4[V ]
 20[W ]  VAB  (5 A)




SELECT VOLTAGE REFERENCE POLARITY
BASED ON CURRENT REFERENCE DIRECTION
40[W ]  (5V )  I
V1  20[V ]
40[W ]  V1  (2 A)
 2A
I  5[ A]
SELECT HERE THE CURRENT REFERENCE DIRECTION
 50[W ]  (10[V ])  I
BASED ON VOLTAGE REFERENCE POLARITY
WHICH TERMINAL HAS HIGHER VOLTAGE AND WHICH IS THE CURRENT FLOW DIRECTION
COMPUTE POWER ABDORBED OR SUPPLIED BY EACH ELEMENT
P1  (6V )(2 A)
2 A  6V 

24V


1
+
-
3
2
2A
18V

P1 = 12W
P2 = 36W
P3 = -48W
P2  (18V )(2 A)
P3  (24V )(2 A)  (24V )(2 A)
IMPORTANT: NOTICE THE POWER BALANCE IN THE CIRCUIT
CIRCUIT ELEMENTS
PASSIVE ELEMENTS
VOLTAGE
DEPENDENT
SOURCES
UNITS FOR  , g , r ,  ?
INDEPENDENT SOURCES
CURRENT
DEPENDENT
SOURCES
EXERCISES WITH DEPENDENT SOURCES
FIND VO
VO  40[V ]
FIND IO
IO  50mA
DETERMINE THE POWER SUPPLIED BY THE DEPENDENT SOURCES
40[V ]
P  (40[V ])(2[ A])  80[W ]
TAKE VOLTAGE POLARITY REFERENCE
P  (10[V ])(4  4[ A])  160[W ]
TAKE CURRENT REFERENCE DIRECTION
POWER ABSORBED OR SUPPLIED BY EACH
ELEMENT
P1  (12V )(4 A)  48[W ]
P2  (24V )(2 A)  48[W ]
P3  (28V )(2 A)  56[W ]
PDS  (1I x )(2 A)  (4V )(2 A)  8[W ]
P36V  (36V )(4 A)  144[W ]
NOTICE THE POWER BALANCE
USE POWER BALANCE TO COMPUTE Io
 12W
(12)(9)
(6)( IO )
(10)(3)
(4)(8)
POWER BALANCE
IO  1[ A]
(8  2)(11)