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SINGLE NODE-PAIR CIRCUITS
THESE CIRCUITS ARE CHARACTERIZED BY ALL
THE ELMENTS HAVING THE SAME VOLTAGE
ACROSS THEM - THEY ARE IN PARALLEL
IN PRACTICE NODES MAY ASSUME STRANGE
FORMS

V

EXAMPLE OF SINGLE NODE-PAIR

V

THIS ELEMENT IS INACTVE (SHORT-CIRCUITED)
LOW DISTORTION POWER AMPLIFIER
LOW VOLTAGE POWER SUPPLY FOR CRT - PARTIAL VIEW
SAMPLE PHYSICAL NODES
COMPONENT SIDE
CONNECTION SIDE
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
CAR STEREO AND CIRCUIT MODEL
215mA
215mA
POWER PER SPEAKER
LEARNING EXTENSION - CURRENT DIVIDER
I1 
120
(16) I1  12mA
120  40
THERE IS MORE THAN ONE
OPTION TO COMPUTE I2
USING CURRENT DIVIDER
KCL : I 2  16  I1  0
40
I2  
(16)  4mA
120  40
POWER : I 2 R
RESISTANCE IN k, P  144 * 40mW  5.76W
CURRENT IN mA YIELD POWER IN mW
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
 150mW
P6 mA  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
Notice use of passive
sign convention

v(t )

KCL :6mA 
v
4 k
v
6 k
v
v
v

 4mA 
0
4k  6k 
12k 
72V  3v  2v  48V  v  0
24V  6v  0
v  4V
Once v(t) is known
all other variables can
be determined; e.g.,
v
12k
12k 
P6 k
v2
16V


 2.667 mW
6k  6k 
FIND i1 AND THE POWER
SUPPLIED BY THE SOURCE
20k||5k
i1
4k
20k
5k
8mA
1
1
1
1 5 1 4 1
 
 
  Rp  2k
R p 4k 20k 5k
20k
2k
2k
i1 
(8)  4mA
4k
v  4k * i1  16V
AN ALTERNATIVE
APPROACH
i1
P  v (8mA )  128mW
v ( t )  RP i O ( t ) 
R
v (t )   i K (t )  p iO (t )
ik (t ) 
Rk
Rk 
General current divider
4k
8mA
4k
FIND THE CURRENT
IL
COMBINE RESISTORS
COMBINE THE SOURCES
1mA
STRATEGY: CONVERT THE PROBLEM INTO A
BASIC CURRENT DIVIDER BY COMBINING
SOURCES AND RESISTORS.
THE NEXT SECTION EXPLORES IN MORE
DETAIL THE IDEA OF COMBINING RESISTORS
NOTICE THE MINUS SIGN
I1
6k
6k I2
C
B
3
I1  9[mA]  3mA
9
I 2   I1
3k
3k
9mA
A
I1
B
6k
C
6k
3k
I2
6k
B
C
9mA
A
I1
3k
9mA
3k
6k
A I2
DIFFERENT LOOKS FOR THE SAME
ELECTRIC CIRCUIT
3k
I1
6k
I2
6k
C
B
3k
3k
9mA
A
B
REDRAWING A CIRCUIT
MAY, SOMETIMES, HELP TO
VISUALIZE BETTER THE
ELECTRICAL CONNECTIONS
I1
9mA
A
6k
I2
6k
3k
C
3k
Determine power
delivered by source
+
2k
4k
V
_
3k
20mA
P  Rp * (20mA)
1
1
1
1 63 4




Rp 2k 4k 3k
12k
12
Rp  k
13
12
P  *103  * (20 *103 ) 2 [ A]
13
4.800
P
W
13
2