14
RESISTIVE CIRCUITS •SINGLE NODE-PAIR CIRCUIT ANALYSIS
RESISTIVE CIRCUITS
•SINGLE NODE-PAIR CIRCUIT ANALYSIS
SINGLE NODE-PAIR CIRCUITS
THESE CIRCUITS ARE CHARACTERIZED BY ALLTHE ELMENTS HAVING THE SAME VOLTAGEACROSS THEM - THEY ARE IN PARALLEL
−
+V
−
+V
EXAMPLE OF SINGLE NODE-PAIR
THIS ELEMENT IS INACTVE (SHORT-CIRCUITED)
IN PRACTICE NODES MAY ASSUME STRANGEFORMS
LOW DISTORTION POWER AMPLIFIER
COMPONENT SIDE CONNECTION SIDE
LOW VOLTAGE POWER SUPPLY FOR CRT - PARTIAL VIEW
SAMPLE PHYSICAL NODES
BASIC CURRENT DIVIDER
THE CURRENT i(t) ENTERS THE NODE ANDSPLITS - IT IS DIVIDED BETWEEN THECURRENTS i1(t) AND i2(t)
APPLY KCL
USE OHM’S LAW TO REPLACE CURRENTS
DEFINE “PARALLEL RESISTANCE COMBINATION”
THE CURRENT DIVISION
mAI 1)5(41
11 =
+= )5(
414
12 +=−= III
)()(
)(1)(
21
21 tiRR
RRtv
tvR
tip
+=
=
pR
O21 VI I FIND ,,
WHEN IN DOUBT… REDRAW THE CIRCUIT TOHIGHLIGHT ELECTRICAL CONNECTIONS!!
2*80 Ik=V24
IS EASIERTO SEE THEDIVIDER
LEARNING EXTENSION - CURRENT DIVIDER
CAR STEREO AND CIRCUIT MODEL
mA215 mA215
POWER PER SPEAKER
)16(40120
1201 +=I
016 12 =−+ II :KCL
mAI 121 =mW INPOWERYIELDmA IN CURRENT
,k IN RESISTANCE :POWER
ΩRI 2
WmWP 76.540*144 ==
mAI 4)16(40120
402 −=
+−=
DIVIDER CURRENT USING
THERE IS MORE THAN ONE OPTION TO COMPUTE I2
FIRST GENERALIZATION: MULTIPLE SOURCESAPPLY KCL TO THIS NODE
)()(
)(1)(
21
21 tiRR
RRtv
tvR
ti
O
pO
+=
=
DEFINE “PARALLEL RESISTANCE COMBINATION”
SOURCE EQUIVALENT
mA10 mA15Ωk3
Ωk6
−
+
OV
SOURCES THEBY SUPPLIEDPOWER THE AND VFIND O
Ω=+
= kkkkkRp 2
363*6
−
+
OVpRmA5
mWmAVP
mWmAVP
VV
OmA
OmA
O
100)10(
150)15(
10
10
15
=−=
−==−=
APPLY KCL TO THIS NODE
SECOND GENERALIZATION: MULTIPLE RESISTORS
)()()()(
)()(ti
RR
tiR
tvti
tiRtvO
k
pK
kk
OP
=⇒⎪⎭
⎪⎬⎫
=
=
General current divider
Ohm’s Law at every resistor
mA8k4 k20 k5
1iSOURCE THEBY SUPPLIED
POWER THE AND FIND 1i
⇒=++
=++=kkkkkRp 21
20415
51
201
411
kRp 2=
mA8k4 k4
1i
AN ALTERNATIVEAPPROACH
mWmAvPVikv
mAkki
128)8(16*4
4)8(42
1
1
−=−===
==
20k||5k
)()()()(
)()(ti
RR
tiR
tvti
tiRtvO
k
pK
kk
OP
=⇒⎪⎭
⎪⎬⎫
=
=
General current divider
FIND THE CURRENTLI
COMBINE THE SOURCESCOMBINE RESISTORS
STRATEGY: CONVERT THE PROBLEM INTO ABASIC CURRENT DIVIDER BY COMBININGSOURCES AND RESISTORS.THE NEXT SECTION EXPLORES IN MOREDETAIL THE IDEA OF COMBINING RESISTORS
NOTICE THE MINUS SIGN
mA1
k3
k3
k6
k6A
B
C
mA9A
B
C
k6
k6
k3
k3mA9
k6
k3
k6
k3
A
CB
9mA
I1 I2
I mA mA
I I
1
2 1
39
9 3= =
= −
[ ]
I1
I2
I1
I2
DIFFERENT LOOKS FOR THE SAMEELECTRIC CIRCUIT
k3k3 k6k6
A
B
C
mA9
k6
k3
k6
k3
A
BC
mA9
I1 I2
I1
I2
REDRAWING A CIRCUITMAY, SOMETIMES, HELP TOVISUALIZE BETTER THE ELECTRICAL CONNECTIONS
k2 k4 k3mA20
Determine powerdelivered by source
kRp
kkkkRp
1312
12436
31
41
211
=
++=++=
2)20(* mARpP =
WP
AP
13800.4
][)10*20(*10*1312 233
=
Ω= −
+
V_
