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Engineering 43
Chp 2.4
Single Node-Pair
Circuits
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
1
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
Single Loop Ckts - Background
 Using KVL And KCL We Can Write
Enough Equations To Analyze Any
Linear Circuit
 Begin The Study Of Systematic and
Efficient Ways Of Using These
Fundamental Circuit Laws
• This Time → Single NODE-PAIR Circuits
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
Single Node-Pair (SNP) Circuits
OF SINGLE
 EXAMPLE
SNP Example
 SNP Circuits Are
Characterized By
ALL the Elements
Having The SAME
VOLTAGE Across
Them → They Are In
PARALLEL

NODE-PAI

V

This Element is INACTIVE
V • The Inactive Element

Engineering-43: Engineering Circuit Analysis
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Has NO Potential Across
it → SHORT Circuited
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
Untangling Reminder
 Nodes Can Take STRANGE Shapes
NODE →
A region of
Constant
Electrical
Potential
Low
Distortion
Power
Amplifier
Engineering-43: Engineering Circuit Analysis
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e.g.; a group
of connected
WIRES is
ONE Node
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
LOW VOLTAGE POWER SUPPLY FOR CRT - PARTIAL VIEW
SOME PHYSICAL NODES
Engineering-43: Engineering Circuit Analysis
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COMPONENT SIDE
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
CONNECTION SIDE
The Current Divider
 Basic Circuit
APPLY KCL
 The Current i(t)
Enters The Top
Node And Splits, or
DIVIDES, into the
the Currents
i1(t) and i2(t)
Engineering-43: Engineering Circuit Analysis
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 Apply KCL at Top
Node
 Use Ohm’s Law to
Replace Currents
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
The Current Divider cont.
 Basic Circuit
Rp
1
it  
vt 
Rp
vt  
 By KCL & Ohm
R1 R2
i t 
R1  R2
 The Current Division
 Define PARALLEL
Resistance
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
Current Divider Example
 For This Ckt Find:
I1, I2, Vo
 When in doubt…
REDRAW the circuit
to Better Visualize
the Connections
 By I-Divider
Vo  I 2  80kΩ
 24V
2-Legged
Divider is
more
Evident
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
Real World Example
 Car Stereo and Circuit Model
215mA
 Use I-Divider to Find
Current thru the 4Ω
Speakers
215mA
 Thus the Speaker
Power
 Power Per Speaker
by Joule
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
Current & Power Example
KCL
 For This Ckt Find:
• I 1, I 2,
• P40k  Power
ABSORBED by
40 kΩ Resistor
 By I-Divider
120
16mA
I1 
120  40
 12mA
 Find I2 by I-Divider
OR KCL
• Choose KCL
Engineering-43: Engineering Circuit Analysis
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I 2  16mA  I1  0
 I 2  4mA
 The 40k Power by RI2
P40k  12mA 40k
2
 milli   k   milli
P40k  5760mW  5.76W
2
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
Generalization: Multi I-Sources
 Given Single Node-Pair Ckt w/ Multiple Srcs
KCL
 KCL on Top Node:
 Combine Src Terms
To Form Equivalent
Source
Engineering-43: Engineering Circuit Analysis
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 The Equivalent Ckt
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
I-Source Example
 For This Ckt Find
Vo, and the
Power Supplied
by the I-Srcs 10mA
6k 
3k 
15mA

VO

 Combine Srcs to
 Vo by Ohm’s Law
Yield Equivalent Ckt V  R I  2k 5mA   10V
o
p Src


Use
PASSIVE
SIGN
R
V
Convention for Power
p

5mA
Rp 
6k * 3k
 2 k
6k  3k
Engineering-43: Engineering Circuit Analysis
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O
P10m  10V 10mA   100mW
P15m  10V  15mA 
 150mW
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
Generalization: Multi Resistors
 Given Single Node-Pair Ckt w/ Multiple R’s
vt   R p io t  

iK t   vt  RK 
iK t  
Rp
RK
io t 
 KCL on Top Node:
 The Equivalent
Resistance & v(t)
N
1
1

R p j 1 R j
vt   io t   R p
Engineering-43: Engineering Circuit Analysis
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Ohm’s Law at
Each Resistor
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
KCL
Multi-R Example
 For This Ckt Find
i1, and the
Power Supplied
by the I-Source 8mA
 Find Rp
1
1
1
1



R p 4kΩ 20kΩ 5kΩ
5 1 4
1


20kΩ
2kΩ
 R p  2kΩ
Engineering-43: Engineering Circuit Analysis
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i1
4k
20k
5k
 Recall the General
Current Divider Eqn
vt   R p io t  

iK t   vt  RK 
iK t  
Rp
RK
io t 
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
Multi-R Example cont.
 Find i1, by Divider
• Take Care with
Passive Sign Conv
2kΩ
8mA   4mA
i1 
4kΩ
 Find v for SingleNode-Pair by Ohm
v  isrc R p
 8mA 2kΩ   16V
 Find Psrc by v•i
Engineering-43: Engineering Circuit Analysis
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i1
4k
20k
5k
8mA
Psrc  visrc

v

  16V 8mA   128mW
• Note: this time For
Passive Sign
Convention CURRENT
Direction assigned
as POSITIVE
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
Multi-R: Alternative Approach
 Start by
Combining R’s
NOT associated
8mA
with i1
 The Ckt After the
R-Combination
i1
4k
4k
8mA
Engineering-43: Engineering Circuit Analysis
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20k||5k
i1
4k
20k
5k
 Now Have 1:1 Current
Divider so
4kΩ
i1 
isrc
4kΩ  4kΩ
1
i1  isrc  4mA
2
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
Example: Multi-R, Multi-Isrc SNP
 Given Single Node-Pair Ckt: Find IL
 Soln Game Plan:
Convert The
Problem Into A
Basic CURRENT
DIVIDER By
Combining Sources
And Resistors
Engineering-43: Engineering Circuit Analysis
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 Combine Sources
• Assume DOWN =
POSITIVE
I S ,eq
I S ,eq  1mA  4mA  2mA
 1mA
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
Multi-R, Multi-Isrc SNP cont.
 Given Single Node-Pair Ckt: Find IL
 Next Combine
Parallel Resistors
 IL by I-Divider
 Then the Equivalent
Circuit →
Engineering-43: Engineering Circuit Analysis
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Note MINUS Sign
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
I1
6k
6k
I2
C
B
3k
3k
9mA
The SAME Ckt
Can Look
Quite Different
I1  3 99mA   3mA
A
I 2   I1  3mA
I1
B
6k
C
6k
3k
I2
6k
9mA
A
3k
I1
B
C
3k
9mA
3k
6k
A I2
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
UnTangling Utility
 Redrawing A Circuit May, Sometimes, Help To
Better Visualize The Electrical Connections
B
I1
6k
6k
I1
C
B
3k
I2
9mA
3k
9mA
A
I2
A
6k
6k
3k
C
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
3k
Another Example
 For This Ckt
Find the I-Src
Power, P20
+
2k
 Alternatives for P
20mA
1
1
1
1



R p 2kΩ 4kΩ 3kΩ
P20  VI  V 20mA 
• By Joule and Energy
Balance
P20   PR j  RP 20mA 
2
21
3k
 Use ||-Resistance
• By vi
Engineering-43: Engineering Circuit Analysis
4k V
_
63 4
12

 R p  kΩ
12kΩ
13
2
P20  12 13 k 20mA 
4800
P20 
mW supplied
13
Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
All Done for Today
A Different
Type of
NODE
BRANCHES Connect to NODES
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt
WhiteBoard Work
 Let’s Work This
Problem
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-44_Lec-02-4_Single_Node-Pair_Ckts.ppt