Transcript Lecture material (Chap. 2)

RESISTIVE CIRCUITS
Here we introduce the basic concepts and laws that are
fundamental to circuit analysis
LEARNING GOALS
• OHM’S LAW - DEFINES THE SIMPLEST PASSIVE ELEMENT: THE
RESISTOR
• KIRCHHOFF’S LAWS - THE FUNDAMENTAL CIRCUIT CONSERVATION
LAWS- KIRCHHOFF CURRENT (KCL) AND KIRCHHOFF VOLTAGE (KVL)
• LEARN TO ANALYZE THE SIMPLEST CIRCUITS
• SINGLE LOOP - THE VOLTAGE DIVIDER
• SINGLE NODE-PAIR - THE CURRENT DIVIDER
• SERIES/PARALLEL RESISTOR COMBINATIONS - A TECHNIQUE
TO REDUCE THE COMPLEXITY OF SOME CIRCUITS
• WYE - DELTA TRANSFORMATION - A TECHNIQUE TO REDUCE
COMMON RESISTOR CONNECTIONS THAT ARE NEITHER SERIES NOR
PARALLEL
• CIRCUITS WITH DEPENDENT SOURCES - (NOTHING VERY
SPECIAL)
RESISTORS
 v(t ) 
i(t )
A resistor is a passive element
characterized by an algebraic
relation between the voltage across
its terminals and the current
through it
v(t )  F (i (t )) General Model for a Resistor
A linear resistor obeys OHM’s Law
v(t )  Ri(t )
The constant, R, is called the
resistance of the component and
is measured in units of Ohm ()
From a dimensional point of view
Ohms is a derived unit of Volt/Amp
Since the equation is algebraic
the time dependence can be omitted
Standard Multiples of Ohm
M
Mega Ohm(106 )
k
Kilo Ohm(103 )
A common occurrence is
Volt
mA
resulting in resistance in k
Conductance
If instead
a function
current in
law can be
of expressing voltage as
of current one expresses
terms of voltage, OHM’s
written
i
1
v
R
1
as Conductanc e
R
of the component and write
i  Gv
We define G 
The unit of conductance is
Siemens
Some practical resistors
Symbol
i
Notice passive sign
convention


v0

Short
R
v
Two special resistor values

Circuit Represent ation
Circuit
R0
G
i
“A touch of
reality”
i0
Open
Circuit
R
G0
Linear approximation
v
Linear range
Actual v-I relationship
Ohm’s Law is an approximation valid
while voltages and currents remain
in the Linear Range



GIVEN VOLTAGE AND CONDUCTANCE
REFERENCE DIRECTIONS SATISFY
PASSIVE SIGN CONVENTION
i (t )  Gv (t ) OHM’S LAW
UNITS?
CONDUCTANCE IN SIEMENS, VOLTAGE
IN VOLTS. HENCE CURRENT IN AMPERES
i (t )  8[ A]
OHM’S LAW
v (t )  Ri (t )
UNITS?
 4[V ]  (2)i (t )  i (t )  2[ A]


4V


v (t )   Ri (t )
OHM’S LAW
THE EXAMPLE COULD BE
GIVEN LIKE THIS
DETERMINE CURRENT AND POWER ABSORBED
BY RESISTOR


 6mA
V2
P  VI  I R 
R
2
P  (12[V ])(6[mA ])  72[mW ]
0.6[mA ]
V 6[V ]
I 
R 10k
VS2
P
R
VS  6[V ]
VS2  (10 103 )(3.6 103W )
KIRCHHOFF CURRENT LAW
ONE OF THE FUNDAMENTAL CONSERVATION PRINCIPLES
IN ELECTRICAL ENGINEERING
“CHARGE CANNOT BE CREATED NOR DESTROYED”
NODES, BRANCHES, LOOPS
A NODE CONNECTS SEVERAL COMPONENTS.
BUT IT DOES NOT HOLD ANY CHARGE.
TOTAL CURRENT FLOWING INTO THE NODE
MUST BE EQUAL TO TOTAL CURRENT OUT
OF THE NODE
(A CONSERVATION OF CHARGE PRINCIPLE)
NODE: point where two, or more, elements
are joined (e.g., big node 1)
LOOP: A closed path that never goes
twice over a node (e.g., the blue line)
The red path is NOT a loop
BRANCH: Component connected between two
nodes (e.g., component R4)
NODE
KIRCHHOFF CURRENT LAW (KCL)
SUM OF CURRENTS FLOWING INTO A NODE IS
EQUAL TO SUM OF CURRENTS FLOWING OUT OF
THE NODE
5A

 5A
A current flowing into a node
is equivalent to the negative
flowing out of the node
ALGEBRAIC SUM OF CURRENT (FLOWING) OUT OF
A NODE IS ZERO
ALGEBRAIC SUM OF CURRENTS FLOWING INTO A
NODE IS ZERO
A node is a point of connection of two or more circuit elements.
It may be stretched out or compressed for visual purposes…
But it is still a node
A GENERALIZED NODE IS ANY PART OF A
CIRCUIT WHERE THERE IS NO ACCUMULATION
OF CHARGE
... OR WE CAN MAKE SUPERNODES BY
AGGREGATING NODES
Leaving 2 : i1  i6  i4  0
Leaving 3 :  i2  i4  i5  i7  0
Adding 2 & 3 : i1  i2  i5  i6  i7  0
INTERPRETATION: SUM OF CURRENTS LEAVING
NODES 2&3 IS ZERO
VISUALIZATION: WE CAN ENCLOSE NODES 2&3
INSIDE A SURFACE THAT IS VIEWED AS A
GENERALIZED NODE (OR SUPERNODE)
WRITE ALL KCL EQUATIONS
 i1 ( t )  i2 ( t )  i3 ( t )  0
i1 ( t )  i4 ( t )  i6 (t )  0
 i3 ( t )  i5 ( t )  i8 ( t )  0
THE FIFTH EQUATION IS THE SUM OF THE
FIRST FOUR... IT IS REDUNDANT!!!
FIND MISSING CURRENTS
KCL DEPENDS ONLY ON THE INTERCONNECTION.
THE TYPE OF COMPONENT IS IRRELEVANT
KCL DEPENDS ONLY ON THE TOPOLOGY OF THE CIRCUIT
WRITE KCL EQUATIONS FOR THIS CIRCUIT
•THE LAST EQUATION IS AGAIN LINEARLY
DEPENDENT OF THE PREVIOUS THREE
•THE PRESENCE OF A DEPENDENT SOURCE
DOES NOT AFFECT APPLICATION OF KCL
KCL DEPENDS ONLY ON THE TOPOLOGY
Here we illustrate the use
of a more general idea of
node. The shaded surface
encloses a section of the
circuit and can be considered
as a BIG node
SUM OF CURRENTS LEAVING BIG NODE  0
I 4  40mA  30mA  20mA  60mA  0
I 4  70mA
THE CURRENT I5 BECOMES INTERNAL TO THE
NODE AND IT IS NOT NEEDED!!!
FIND I x
Ix
 3mA
I X  I1  2 I X  0
I1  4mA  1mA  0
I1  3mA
VERIFICATION
I b  1mA  I X  2mA
1mA
2 I X  4mA  I b
Ib
2I x
4mA
KIRCHHOFF VOLTAGE LAW
ONE OF THE FUNDAMENTAL CONSERVATION LAWS
IN ELECTRICAL ENGINERING
THIS IS A CONSERVATION OF ENERGY PRINCIPLE
“ENERGY CANNOT BE CREATE NOR DESTROYED”

V 
A
B
 (V ) 
A
A VOLTAGE RISE IS
A NEGATIVE DROP
B
KIRCHHOFF VOLTAGE LAW (KVL)
KVL IS A CONSERVATION OF ENERGY PRINCIPLE
KVL: THE ALGEBRAIC SUM OF VOLTAGE
DROPS AROUND ANY LOOP MUST BE ZERO

V 
A
B
 (V ) 
A
A VOLTAGE RISE IS
A NEGATIVE DROP
B
PROBLEM SOLVING TIP: KVL IS USEFUL
TO DETERMINE A VOLTAGE - FIND A LOOP
INCLUDING THE UNKNOWN VOLTAGE
THE LOOP DOES NOT HAVE TO BE PHYSICAL

Vbe

 VS  VR  VR  VR  0
1
2
3
VR  12V
2
VR  18V
1
EXAMPLE : VR1 , VR3 ARE KNOWN
DETERMINE THE VOLTAGE Vbe
VR  Vbe  VR  30[V ]  0
1
LOOP abcdefa
3
BACKGROUND: WHEN DISCUSSING KCL WE SAW
THAT NOT ALL POSSIBLE KCL EQUATIONS
ARE INDEPENDENT. WE SHALL SEE THAT THE
SAME SITUATION ARISES WHEN USING KVL
A SNEAK PREVIEW ON THE NUMBER OF
LINEARLY INDEPENDENT EQUATIONS
IN THE CIRCUIT DEFINE
N
NUMBER OF NODES
B
NUMBER OF BRANCHES
N 1
LINEARLY INDEPENDEN T
KCL EQUATIONS
B  ( N  1) LINEARLY INDEPENDEN T
KVL EQUATIONS
EXAMPLE: FOR THE CIRCUIT SHOWN WE HAVE
N = 6, B = 7.
HENCE THERE ARE ONLY TWO INDEPENDENT
KVL EQUATIONS
THE THIRD EQUATION IS THE SUM OF THE
OTHER TWO!!
FIND THE VOLTAGES Vae ,Vec
GIVEN THE CHOICE USE THE SIMPLEST LOOP
DEPENDENT SOURCES ARE HANDLED WITH THE
SAME EASE
5k
10k

25V

 Vx 
+
-
V1
There are no loops with only
one unknown!!!
Vx/2 +
+
-

The current through the 5k and 10k
resistors is the same. Hence the
voltage drop across the 5k is one half
of the drop across the 10k!!!
Vx
4
VX VX
V1 

0
4
2
VX VX
 25[V ]  VX 

0
V
2
4
V1   X  5[V ]
4
VX  20[V ]
SINGLE LOOP CIRCUITS
BACKGROUND: USING KVL AND KCL WE CAN
WRITE ENOUGH EQUATIONS TO ANALYZE ANY
LINEAR CIRCUIT. WE NOW START THE STUDY
OF SYSTEMATIC, AND EFFICIENT, WAYS OF
USING THE FUNDAMENTAL CIRCUIT LAWS
a
1
2
b
6 branches
6 nodes
1 loop
3
c
4
WRITE 5 KCL EQS
OR DETERMINE THE
ONLY CURRENT
FLOWING
f
6
e
5
d
ALL ELEMENTS IN SERIES
ONLY ONE CURRENT
THE PLAN
• BEGIN WITH THE SIMPLEST ONE LOOP CIRCUIT
• EXTEND RESULTS TO MULTIPLE SOURCE
• AND MULTIPLE RESISTORS CIRCUITS
IMPORTANT VOLTAGE
DIVIDER EQUATIONS
VOLTAGE DIVISION: THE SIMPLEST CASE
KVL ON
THIS
LOOP
SUMMARY OF BASIC VOLTAGE DIVIDER
v R1 
R1
v (t )
R1  R2
EXAMPLE : VS  9V , R1  90k, R2  30k
VOLUME
CONTROL?
R1  15k 
A “PRACTICAL” POWER APPLICATION
HOW CAN ONE REDUCE THE LOSSES?
THE CONCEPT OF EQUIVALENT CIRCUIT
THIS CONCEPT WILL OFTEN BE USED TO SIMPLFY
THE ANALYSIS OF CIRCUITS. WE INTRODUCE IT
HERE WITH A VERY SIMPLE VOLTAGE DIVIDER
i
vS
R1
i
vS
+
-
R2
i
+
-
R1  R2
vS
R1  R2
AS FAR AS THE CURRENT IS CONCERNED BOTH
CIRCUITS ARE EQUIVALENT. THE ONE ON THE
RIGHT HAS ONLY ONE RESISTOR
SERIES COMBINATION OF RESISTORS
R1
R2

R1  R2
IN ALL CASES THE RESISTORS ARE
CONNECTED IN SERIES
FIRST GENERALIZATION: MULTIPLE SOURCES
 v2 
 v R1 
Voltage sources in series can be
algebraically added to form an
 equivalent source.
v3
+ +
-

v5
i(t)
R2
+
-

R1
+
-

v1


We select the reference direction to
 move along the path.
v Voltage drops are subtracted from rises
R2

+ -
KVL
 v4 
vR1  v2  v3  vR 2  v4  v5  v1  0
R1
Collect all sources on one side
v1  v2  v3  v4  v5   vR1  vR 2
v   v
eq
R1
 vR 2
veq
+
-
R2
SECOND GENERALIZATION: MULTIPLE RESISTORS
FIND I ,Vbd , P (30k )
APPLY KVL
TO THIS LOOP
APPLY KVL
TO THIS LOOP
LOOP FOR Vbd
Vbd  12  20[k ] I  0 (KVL)  Vbd  10V
POWER ON 30k  RESISTOR
P  I 2 R  (104 A) 2 (30 *103 )  30mW
v R  Ri i 
i
VOLTAGE DIVISION FOR MULTIPLE RESISTORS
SINGLE NODE-PAIR CIRCUITS
THESE CIRCUITS ARE CHARACTERIZED BY ALL
THE ELMENTS HAVING THE SAME VOLTAGE
ACROSS THEM - THEY ARE IN PARALLEL

V

EXAMPLE OF SINGLE NODE-PAIR

V

THIS ELEMENT IS INACTVE (SHORT-CIRCUITED)
BASIC CURRENT DIVIDER
Rp
THE CURRENT DIVISION
APPLY KCL
THE CURRENT i(t) ENTERS THE NODE AND
SPLITS - IT IS DIVIDED BETWEEN THE
CURRENTS i1(t) AND i2(t)
USE OHM’S LAW TO REPLACE
CURRENTS
DEFINE “PARALLEL RESISTANCE COMBINATION”
i (t ) 
1
v (t )
Rp
v (t ) 
R1 R2
i (t )
R1  R2
4
1
I

I

I

(5)
I1 
(5)  1mA 2
1
1 5
1 4
FIND I1 , I2 , VO
WHEN IN DOUBT… REDRAW THE CIRCUIT TO
HIGHLIGHT ELECTRICAL CONNECTIONS!!
IS EASIER
TO SEE THE
DIVIDER
 80k * I 2
24V
FIRST GENERALIZATION: MULTIPLE SOURCES
APPLY KCL TO THIS NODE
EQUIVALENT SOURCE
DEFINE “PARALLEL RESISTANCE COMBINATION”
iO ( t ) 
1
v (t )
Rp
v (t ) 
R1 R2
iO ( t )
R1  R2
FIND VO AND THE POWER
SUPPLIED BY THE SOURCES
6k
10mA
3k
Rp
5mA

VO

15mA

VO  10V
VO
P15mA  VO (15mA )

6k * 3k
Rp 
 2 k
6k  3k
P6 mA
 150mW
 VO (10mA )
 100mW
SECOND GENERALIZATION: MULTIPLE RESISTORS
APPLY KCL TO THIS NODE
Ohm’s Law at every resistor
v ( t )  RP i O ( t ) 
R
v (t )   i K (t )  p iO (t )
ik (t ) 
Rk
Rk 
General current divider
SERIES PARALLEL RESISTOR COMBINATIONS
UP TO NOW WE HAVE STUDIED CIRCUITS THAT
CAN BE ANALYZED WITH ONE APPLICATION OF
KVL(SINGLE LOOP) OR KCL(SINGLE NODE-PAIR)
WE HAVE ALSO SEEN THAT IN SOME SITUATIONS
IT IS ADVANTAGEOUS TO COMBINE RESISTORS
TO SIMPLIFY THE ANALYSIS OF A CIRCUIT
NOW WE EXAMINE SOME MORE COMPLEX CIRCUITS
WHERE WE CAN SIMPLIFY THE ANALYSIS USING
THE TECHNIQUE OF COMBINING RESISTORS…
… PLUS THE USE OF OHM’S LAW
SERIES COMBINATIONS
PARALLEL COMBINATION
G p  G1  G2  ...  GN
FIRST WE PRACTICE COMBINING RESISTORS
3k
SERIES
6k||3k
(10K,2K)SERIES
6k || 12k  4k
5k
3k
12k
4k || 12k 12k
FIRST REDUCE IT TO A SINGLE LOOP CIRCUIT
SECOND: “BACKTRACK” USING KVL, KCL OHM’S
6k
I3
V
OHM' S : I 2  a
6k
KCL : I1  I 2  I 3  0
OHM' S : Vb  3k * I 3
…OTHER OPTIONS...
6k || 6k
KCL : I 5  I 4  I 3  0
OHM' S : VC  3k * I 5
I1 
12V
12k
Va 
3
(12)
39
12
I3
4  12
Vb  4k * I 4
I4 
2k || 2k  1k
VOLTAGE DIVIDER : VO 
LEARNING
BY DOING
1k
(3V )  1V
1k  2k
1k  1k  2k
CURRENT DIVIDER : I O 
1k
(3 A)  1A
1k  2k
AN EXAMPLE OF “BACKTRACKING”
 1.5mA
I1  3mA
V xz  6V
3V
 1.5mA
1mA
VO  36V
3V
 0.5mA
A STRATEGY. ALWAYS ASK: “WHAT ELSE CAN I
COMPUTE?”
Vb  6k * I 4
I3 
Vb
3k
I2  I3  I4
Va  2k * I 2
V xz  Va  Vb
V
I 5  xz
4k
I1  I 2  I 5
VO  6k * I1  V xz  4k * I1
FIND VO
60k
V1 6V
FIND VS
2V
30k || 60k  20k
STRATEGY : FIND V1
USE VOLTAGE DIVIDER
9V
12V
VOLTAGE DIVIDER
20k
VO 
V1
20k  40k
0.05mA
I1 
VS  20k * 0.15mA  6V

20k
0.15mA 
6V

THIS IS AN INVERSE PROBLEM
WHAT CAN BE COMPUTED?
20k
+
-
V1  60k * 0.1mA
V1

20k
(12)  6V
20k  20k

SERIES
PARALLEL
6V
120k
http://www.wiley.com/college/irwin/0470128690/animations/swf/D2Y.swf
Y   TRANSFORMATIONS
THIS CIRCUIT HAS NO RESISTOR IN
SERIES OR PARALLEL
IF INSTEAD
OF THIS
WE COULD
HAVE THIS
THEN THE CIRCUIT WOULD
BECOME LIKE THIS AND
BE AMENABLE TO SERIES
PARALLEL TRANSFORMATIONS
Rab  R2 || ( R1  R3 )
 Y
Rab  Ra  Rb
Y 
Ra R1
Rb R1
R2 ( R1  R3 )
R
R
1
2


R

Ra  Rb 
3
Ra 
R
R
Ra
b
3
R1  R2  R3
R1  R2  R3
Rb R2
RR

 R2  b 1
Rc R1
Rc
REPLACE IN THE THIRD AND SOLVE FOR R1
R2 R3
R ( R  R2 ) Rb 
Ra Rb  Rb Rc  Rc Ra
Rb  Rc  3 1
R1  R2  R3
R

1
R1  R2  R3
Rb
R3 R1
Rc 
R R  Rb Rc  Rc Ra
R1  R2  R3
R2  a b
R1 ( R2  R3 )
Rc
Rc  Ra 
 Y
R1  R2  R3
R R  Rb Rc  Rc Ra
R3  a b
Ra
SUBTRACT THE FIRST TWO THEN ADD
TO THE THIRD TO GET Ra
Y 
LEARNING EXAMPLE: APPLICATION OF WYE-DELTA TRANSFORMATION
c
R1

R3
R2
12k  6k
12k  6k  18k
R1 R2
Ra 
R1  R2  R3
Rb 
R2 R3
R1  R2  R3
Rc 
R3 R1
R1  R2  R3
 Y
a
b
a
c
DELTA CONNECTION
b
COMPUTE IS
REQ  6k  3k  9k  || (2k  6k )  10k
IS 
12V
 1.2mA
12k
ONE COULD ALSO USE A
WYE - DELTA TRANSFORMATION ...
CIRCUITS WITH DEPENDENT SOURCES
A CONVENTION ABOUT DEPENDENT SOURCES.
UNLESS OTHERWISE SPECIFIED THE CURRENT
AND VOLTAGE VARIABLES ARE ASSUMED IN SI
UNITS OF Amps AND Volts
DEPENDENT
VARIABLE


GENERAL STRATEGY
TREAT DEPENDENT SOURCES AS REGULAR
SOURCES AND ADD ONE MORE EQUATION FOR
THE CONTROLLING VARIABLE
 VA 
FIND VO
VD   I X
CONTROLLING
VARIABLE
FOR THIS EXAMPLE THE MULTIPLIER MUST HAVE
UNITS OF OHM
OTHER DEPENDENT SOURCES
VD  VX (  scalar)
I D  VX
( Siemens)
I D  I X
(  scalar)
A PLAN:
SINGLE LOOP CIRCUIT.
USE KVL TO DETERMINE CURRENT
KVL :  12  3k * I1  VA  5k * I1  0
ONE EQUATION, TWO UNKNOWNS. CONTROLLING
VARIABLE PROVIDES EXTRA EQUATION
AN ALTERNATIVE DESCRIPTIO N
 V  UNITS
VD  I X ,   2 
 mA 
ASSUMES CURRENT IN mA
KVL
ARE EXPLICIT
VA  2k * I1
REPLACE AND SOLVE FOR THE CURRENT
I1  2mA
USE OHM’S LAW
VO  5k * I1  10V
KCL TO THIS NODE. THE
DEPENDENT SOURCE IS JUST
ANOTHER SOURCE
FIND VO
A PLAN:
IF V_s IS KNOWN V_0 CAN BE DETERMINED USING VOLTAGE DIVIDER.
TO FIND V_s WE HAVE A SINGLE NODE-PAIR CIRCUIT
THE EQUATION FOR THE CONTROLLING
VARIABLE PROVIDES THE ADDITIONAL EQUATION
ALGEBRAICALLY, THERE ARE TWO UNKNOWNS
AND JUST ONE EQUATION
SUBSTITUTION OF I_0 YIELDS
VOLTAGE DIVIDER
* / 6k  5VS  60 VO 
NOTICE THE CLEVER WAY OF WRITING mA TO
HAVE VOLTS IN ALL NUMERATORS AND THE
SAME UNITS IN DENOMINATOR
4k
2
VS  (12)V
4k  2k
3
FIND VO
KVL TO
THIS LOOP
A PLAN:
ONE LOOP PROBLEM.
FIND THE CURRENT
THEN USE OHM’S LAW.
THE DEPENDENT SOURCE IS ONE MORE VOLTAGE
SOURCE
THE EQUATION FOR THE CONTROLLING VARIABLE
PROVIDES THE ADDITIONAL EQUATION
REPLACE AND SOLVE FOR CURRENT I
… AND FINALLY
FIND G 
vO ( t )
vi (t )
KCL
A PLAN:
ONE LOOP ON THE LEFT - KVL
ONE NODE-PAIR ON RIGHT - KCL
KVL
KVL
KCL
gm v g ( t ) 
ALSO A VOLTAGE DIVIDER
vO ( t )
0
RL