NETWORKS 1: ECE 09'201'01

1
CHAPTER 3

• NETWORKS 1 ECE 09.201.01
• 13 SEPTEMBER 2006 Lecture 4
• ROWAN UNIVERSITY
• College of Engineering
• Dr Peter Mark Jansson, PP PE
• DEPARTMENT OF ELECTRICAL COMPUTER ENGINEERING
• Autumn Semester 2006 - Quarter One

2
Networks I

• Todays Learning Objectives
• Use KVL and KCL
• What is voltage and current division?
• Parallel and series sources

3
chapter 3 - overview

• electric circuit applications - done
• define node, closed path, loop - done
• Kirchoffs Current Law - done
• Kirchoffs Voltage Law- done
• a voltage divider circuit in progress
• parallel resistors and current division
• series V-sources / parallel I-sources
• resistive circuit analysis

4
Kirchhoffs laws

• Kirchhoffs Current Law (KCL)
• The algebraic sum of the currents into a node at
  any instant is zero.
• Kirchhoffs Voltage Law (KVL)
• The algebraic sum of the voltages around any
  closed path in a circuit is zero for all time.

5
KVL
Use KVL and Ohms Law VOLTAGE DIVIDER
i V/(R1 R2) vR1 iR1 VR1 /(R1 R2) vR2
iR2 VR2/(R1 R2)
6
SERIES RESISTORS
NOTE
i V/(R1 R2) vR1 iR1 VR1 /(R1 R2) vR2
iR2 VR2/(R1 R2) VOLTAGE DIVIDER
7

KCL
Node 1
Node 2
v2
v3
Node 3
Node 1 I - i1 0 Node 2 i1 - i2 - i3
0 Node 3 i2 i3 - I 0 i2 v2/R2 i3
v3/R3
Use KCL and Ohms Law CURRENT DIVIDER
8
PARALLEL RESISTORS

• resistors attached in parallel can be simplified
  by adding their conductances (G) together to get
  an equivalent resistance (R1/G).

Geq Gr1 Gr2 etc.. When you only have
two Req (R1R2)/(R1R2)
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Equivalent parallel resistors

• Example 3 parallel resistors 6?, 9?, 18? ? what
  is the equivalent resistance? Geq Gr1 Gr2
  etc..
• 1/6 1/9 1/18 6/18 1/3
• If Geq 1/3 then R ?

10
Learning check 1

• What is effective resistance value of three
  parallel resistors with values of 4?, 5?, 20??
• Hint calculate Geq , then R

11
parallel current sources

• when connected in parallel, a group of current
  sources can be treated as one current source
  whose equivalent current
• ? all source currents
• unequal current sources are not to be connected
  in series

12
Learning check 2

• 2a. What is effective value of i for the example
  of parallel current sources on the board (5?,
  10?, 7?, 4?)?
• 2b. What is the power dissipated in the
  resistor of 6??

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PROBLEM SOLVING METHOD
va
vb
_
_
node3
node1
node2


Rb
Ra
ib
ia

ivs

vc

ic
vis
vs
Rc
is
_
_
_
node4
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steps taken

• Apply P.S.C. to passive elements.
• Show current direction at voltages sources.
• Show voltage direction at current sources.
• Name nodes and loops.
• Name elements and sources.
• Name currents and voltages.

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WRITE THE KCL EQUATIONS
node1
node3
node2
node4
16
WRITE THE KVL EQUATIONS
loop1
loop2
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WRITE SUPPLEMENTARYEQUATIONS
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sample problems

• Problem 3.3-7
• example of series V sources, KVL and p.s.c.
• Finish for LC 3 answer
• Problem 3.3-10
• example of V source, KCL, KVL and Ohms Law
• Finish for LC 4 answer

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CIRCUIT REDUCTION
20

• Begin with loop on far right.
• Combine the three resistors that are in series.
• Req 4550100 195?

21

• Again using the loop on the far right.
• The 90 ? and 195 ? resistors are in parallel.
• Req (90)(195)/(90195) 61.58 ?

22

• Still working with the loop on the far right.
• The 30 ? and the 61.58 ? resistors are in series.
• Req 30 61.58 91.58 ?

23

• Again, the far right loop.
• The 15 ? and 91.58 ? resistors are in parallel.
• Req(15)(91.58)/(1591.58)12.9?

24

• Now there is only one loop.
• All the resistors are in series.
• Req 1012.95 27.9 ?

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a
b

• Use Ohms Law to determine iT.
• iT 5/27.9 0.179A
• iT flows in all three resistors, the 12.9 ?
  resistor is the equivalent resistance of the
  entire circuit beyond points a and b.

26
a
ix

• iT divides at a to flow through the 15 ? and the
  91.58 ? resistors (the 91.58 ? is an equivalent
  resistance for the rest of the circuit).
• Use current divider ix (0.179)(15)/(1591.58)
  0.0252A.

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a
0.0252A
b

• No calculations are required at this step because
  the 0.0252A is flowing through both resistors in
  the right loop.
• This circuit must be drawn however, because the
  61.58 ? resistor is an equivalent for the circuit
  to the right of a and b.

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a
0.0252A
b

• Use the current divider equation again to
  determine i1.
• i1 (0.0252)(195)/(90195) 0.01724A 17.24mA.
• The current through the 195 ? resistor is 0.0252
  - 0.01724 Write answer as LC 5

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More sample problems

• P 3.4-5 write answer as LC 6
• P 3.4-8 write answer as LC 7
• on page 91

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Test next Wednesday

• Chapters 1-3, part of 4

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Assignment 2

• Due Tuesday, 19 September, 925 AM  
• Dorf Svoboda, pp. 46-52
• Problems, 2.4-7, 2.4-10, 2.5-1, 2.5-2, 2.6-1,
  2.7-2, 2.8-2, 2.9-1, 2.9-2, 2.10-2Design
  Problems 2-1, 2-3
• Dorf Svoboda, pp. 86-100Problems 3.2-3, 3.2-4,
  3.2-5, 3.2-9, 3.2-10, 3.3-1, 3.3-2, 3.3-4, 3.4-1,
  3.4-4,   3.6-3, 3.6-6, 3.6-11, 3.6-14, 3.8-6
• Design Problem 3-1