Download ch2.3_singleloop

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
THE DIFFERENCE BETWEEN ELECTRIC
CONNECTION AND PHYSICAL LAYOUT
THIS CONCEPT WILL OFTEN BE USED TO SIMPLFY
THE ANALYSIS OF CIRCUITS. WE INTRODUCE IT SOMETIMES, FOR PRACTICAL CONSTRUCTION
HERE WITH A VERY SIMPLE VOLTAGE DIVIDER
REASONS, COMPONENTS THAT ARE ELECTRICALLY
CONNECTED MAY BE PHYSICALLY QUITE APART
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
CONNECTOR SIDE
ILLUSTRATING THE DIFFERENCE
BETWEEN PHYSICAL LAYOUT AND
ELECTRICAL CONNECTIONS
PHYSICAL NODE
PHYSICAL NODE
SECTION OF 14.4 KB VOICE/DATA MODEM
CORRESPONDING POINTS
COMPONENT SIDE
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
THE “INVERSE” VOLTAGE DIVIDER
R1

VS
+
-
R2
VO

VOLTAGE DIVIDER
VO 
R2
VS
R1  R2
"INVERSE" DIVIDER
VS 
R1  R2
VO
R2
COMPUTE VS
" INVERSE" DIVIDER
220  20
VS 
458.3  500k
220
Find I and Vbd
APPLY KVL
TO THIS LOOP
 6  80kI  12  40kI  0  I  0.05mA
Vbd  40kI  12V  0  Vbd  10V
If Vad = 3V, find VS

3V

INVERSE DIVIDER PROBLEM
25  15  20
VS 
3  9V
20
Notice use of
passive sign
convention
 80k  * i ( t ) 
i(t )
KVL :  6V  80k  * i ( t )  12V  40k  * i ( t )  0

40k  * i ( t )

i(t )  
6V
 0.05mA
120k 
Knowing the current one can compute ALL
the remaining voltages and powers
EXAMPLE
A
9V
20k
B
+ -
C
I
+
-
KVL FOR
E
30k
DETERMINE I
VDAUSING KVL
D
10k
V DA 
VCD 
30k * I  1.5V
I DE 
0.05mA
KVL : - 12  20k * I  9  30k * I  10k * I  0
3V
I
 0.05mA
60k
KVL : VDA  12  10k * I  0
VDA  11.5V

Vab

KVL HERE
EXAMPLE
 4V 
b
+-
4k
VS
Sometimes you may want
to vary a bit
3Vx
+
-
a
I
APPLY KVL
TO THIS LOOP
VS  12V
 Vx
V X Vab


 P(3Vx ) 
OR KVL HERE
P(3Vx ) is the power absorbed or
supplied by the dependent source
KVL :  12  4  3VX  VX  0  V X  2V
KVL : Vab  4  3V X  0  Vab  10V
KVL : Vab  VS  V X  0
P(3V )  3VX I (PASSIVE SIGN CONVENTION)
X
4V
 1mA
4k
 2[V ] *1[mA]  2mW
OHMS' LAW : I 
P(3V
X
)