Resistive Circuits

1
Resistive Circuits

• NETWORKS 1 ECE 09.201.02
• 09/11/07 Lecture 3
• ROWAN UNIVERSITY
• College of Engineering
• Prof. John Colton
• DEPARTMENT OF ELECTRICAL COMPUTER ENGINEERING
• Fall 2007 - Quarter One

2
Some Administrative items

• 09/18 Test 1 08/00 915
• Ch 1-3
• Problem Sets 1 and 2
• Questions so far?

3
Resistive Circuits Topics

• Define node, closed path, loop
• Kirchoffs Current Law (KCL)
• Kirchoffs Voltage Law (KVL)
• Equivalent circuits
• Series resistors and voltage division
• Parallel resistors and current division
• Series voltage sources
• Parallel current 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.

5
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

6
Equivalent Circuits
Appearance of Circuit B to A
7
Equivalent Circuits
Appearance of Circuit B to A
Same appearance of Circuit B to A
Replacing circuit B by an equivalent circuit
Beq does not change the current or voltage of
any Circuit element of Circuit A
8
Series resistor equivalent voltage division

• The series connection of n resistors having
  resistances R1, R2, , Rn is externally
  equivalent to a single resistor having resistance
  Req

n Req R1 R2 Rn ? Rk

k 1
9
Series resistor equivalent voltage division

• The series connection of n resistors having
  resistances R1, R2, , Rn is externally
  equivalent to a single resistor having resistance
  Req

n Req R1 R2 Rn ? Rk

k 1
The voltage appearing across one of n resistors
connected in series with a voltage source, is the
ratio of its resistance to the total resistance
times the voltage source value




n vk vs Rk / (R1 R2 Rn ) vs Rk /?
Rk k1
10
Parallel Resistance Equivalent

• Conductances attached in parallel can be added
  together to get an equivalent conductance.

Req
Req
11
Parallel Resistance Equivalent

• Conductances attached in parallel can be added
  together to get an equivalent conductance.

Req
Req
is i1 i2 v/R1 v/R2 G1v G2v (G1G2)v
Geq G1G2 1/R1 1/R2 1/Req
Req 1/(G1G2) R1 R2 /(R1 R2)
12
Parallel Current Division

• The current into one of a series of resistors
  connected in parallel with a current source, is
  the ratio of its conductance to the total
  conductance times the current source value.

e.g. i2 G2V G2 is/(G1G2) is G2/(G1G2)
is R1/(R1R2)

13
Parallel resistor equivalent current division

• The parallel connection of n resistors having
  conductances G1, G2, , Gn is externally
  equivalent to a single resistor having
  conductance Geq

n Geq G1 G2 Gn ? Gk

k 1
14
Parallel resistor equivalent current division

• The parallel connection of n resistors having
  conductances G1, G2, , Gn is externally
  equivalent to a single resistor having
  conductance Geq

n Geq G1 G2 Gn ? Gk

k 1
The current into one of n resistors connected in
parallel with a current source, is the ratio of
its conductance to the total conductance times
the current source value.




n ik is Gk / (G1 G2 Gn ) is Gk / ?
Gk k1
15
Example KCL
v1
_
Node 1
Node 2

R110?


i1
I5A
i3
i2
v2
v3
R2 20?
_
R3 5?
_
Node 3
16
Example KCL
v1
_
Node 1
Node 2

R110 ?


i1
I5A
i3
i2
v2
v3
R2 20 ?
_
R3 5 ?
_
Node 3
i3 i1 G3 /(G2 G3) 5(1/5)/1/201/5 4A
i2 i1 R3 /(R2 R3) 5(5)/205 1A
v v2 v3 i2 R2 i3 R3 Req i1 20V
17
Example Many parallel resistances

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

What is the equivalent resistance of the parallel
connection of N identical resistors of
resistance R?
Geq
18
Example Many parallel resistances

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

What is the equivalent resistance of the parallel
connection of N identical resistors of
resistance R?
Geq
19
Series voltage sources

• When connected in series, a group of voltage
  sources can be treated as one voltage source
  whose equivalent voltage value is the algebraic
  sum of all source voltages
• This is a direct result of KVL which sums
  voltages around the loop that includes the series
  voltage source connection
• Replacing the multiple series voltage sources
  with a single equivalent voltage source does not
  change the current in or voltage across any other
  circuit element
• Unequal voltage sources may not be connected in
  parallel violates the constraints imposed by
  the model

20
Example Series voltage source reduction
Determine the value of voltage v
21
Example Series voltage source reduction
Determine the value of voltage v
Combine series voltage sources and the 5, 15, and
20 ? resistors
6V
22
Example Series voltage source reduction
Determine the value of voltage v
Combine series voltage sources and the 5, 15, and
20 ? resistors
6V
Voltage divider v (6)10/50 1.2V
23
Parallel current sources

• When connected in parallel, a group of current
  sources can be treated as one current source
  whose equivalent current is the algebraic sum of
  all source currents
• This is a direct result of KCL which sums current
  at nodes at the nodes associated with the
  parallel connection
• Replacing the multiple parallel current sources
  with the single equivalent current source does
  not change the voltage across or current in any
  other circuit element
• Unequal current sources may not be connected in
  series violates the constraints imposed by the
  model

24
Example Parallel current source reduction
Combine parallel current sources and other
resistors
25
Example Parallel current source reduction
Combine parallel current sources and other
resistors
Use current division
i -3 (4/7) -12/7A
26
Source equivalents
Circuit Equivalent
Circuit Equivalent Circuit
Circuit
27
Example Source reduction
Find the current in each resistor
Convenient to combine the two current source and
combine the two voltage sources before solving
for this power dissipation. First, name the
currents and the nodes, then combine the sources.
28
Example Source reduction
Node 1
Node 2
3A and 1.75A current sources are in parallel for
nodes 1 and 2, opposing. 3V and 8V voltage
sources are in series for the outer loop,
opposing.
Node 1
Node 2
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Example Source reduction
Write equations
Apply KCL at the top node of the current source
to get i1 1.75 i2 Apply KVL to the outside
loop to get 52i2 2i1 0 Solve i1
-2.125A i2 -0.375A
Node 1
Node 2
30
Circuit Analysis General Methodology (so far)

• Simplify circuit configuration as much as
  possible
• Determine what you need to retain explicitly
• Create equivalent sources and resistance
  configurations around your retained elements

31
Circuit Analysis General Methodology (so far)

• Simplify circuit configuration as much as
  possible
• Determine what you need to retain explicitly
• Create equivalent sources and resistance
  configurations around your retained elements
• Name nodes and/or loops on simplified circuit
• Show current direction at voltage sources.
• Show voltage direction at current sources.
• Apply PSC to passive elements, assuming current
  and voltage directions

32
Circuit Analysis General Methodology (so far)

• Simplify circuit configuration as much as
  possible
• Determine what you need to retain explicitly
• Create equivalent sources and resistance
  configurations around your retained elements
• Name nodes and/or loops on simplified circuit
• Show current direction at voltage sources.
• Show voltage direction at current sources.
• Apply PSC to passive elements, assuming current
  and voltage directions
• Write equations to solve using developed
  techniques
• KVL, KCL, Ohms Law
• Combining resistors
• Current and voltage dividers

33
Example
Calculate v
34
Example
v
By series voltage division, v (4ia)5/8 By KVL
on the left hand loop, 24 8ia 4ia 12 ia
ia 2 v 5V
35
Example
What is the power supplied by the dependent
source?
36
Example
What is the power supplied by the dependent
source?
By series voltage division, va (120)10/60
20V ia 4A power from dependent
source 480W
37
Circuit Analysis Example
Find i1 and v1
38
Circuit Analysis Example
Find i1 and v1
6O
39
Circuit Analysis Example
40
Circuit Analysis Example
41
Test next Tuesday 09/18

• Chapters 1-3

42
Problem Set 2

• Due Monday, 09/17, 800 AM
• Dorf Svoboda, pp. 46-52
• 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,
• Design Problems 2-1, 2-3
• Dorf Svoboda, pp. 86-100
• Problems 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