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Transcript of Rowan Hall 238A [email protected] September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K....
![Page 1: Rowan Hall 238A beard@rowan.edu September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.](https://reader035.fdocuments.in/reader035/viewer/2022062804/5697bf741a28abf838c7f9f4/html5/thumbnails/1.jpg)
Rowan Hall 238A
http://rowan.jkbeard.com
September 18, 2006
Networks I for M.E.ECE 09.201 - 2
James K. Beard, Ph.D.
![Page 2: Rowan Hall 238A beard@rowan.edu September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.](https://reader035.fdocuments.in/reader035/viewer/2022062804/5697bf741a28abf838c7f9f4/html5/thumbnails/2.jpg)
Slide 2
Admin
1 – week from tomorrow Test 1 Cruise course website Questions thus far?
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Slide 3
Networks I Today’s Learning Objectives –
Analyze DC circuits with passive elements including: resistance -- DONE
Learn about switches -- DONE Introduce KVL and KCLWhat is voltage and current division?Parallel and series sourcesSolve some more difficult problems
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Slide 4
chapter 2 - overview engineering and linear models - doneengineering and linear models - done active and passive circuit elements - doneactive and passive circuit elements - done resistors – Ohm’s Law - doneresistors – Ohm’s Law - done dependent sources – donedone independent sources – donedone transducers – done switches – in progress
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Slide 5
chapter 3 - overview electric circuit applications define: node, closed path, loop Kirchoff’s Current Law Kirchoff’s Voltage Law a voltage divider circuit parallel resistors and current division series V-sources / parallel I-sources resistive circuit analysis
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Slide 6
Gustav Robert Kirchhoff
1824-1887 two profound scientific laws published
in 1847
how old was he? LC #1
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Slide 7
Kirchhoff’s laws
Kirchhoff’s Current Law (KCL): The algebraic sum of the currents into a
node at any instant is zero.
Kirchhoff’s Voltage Law (KVL):The algebraic sum of the voltages around
any closed path in a circuit is zero for all time.
![Page 8: Rowan Hall 238A beard@rowan.edu September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.](https://reader035.fdocuments.in/reader035/viewer/2022062804/5697bf741a28abf838c7f9f4/html5/thumbnails/8.jpg)
Slide 8
KCL
Assume passive sign convention
R2= 20
I=5A
R1=10
R3= 5+
_
+
_
+ _Node 1 Node 2
Node 3
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Slide 9
R2= 20
I=5A
R1=10
R3= 5
+
_
+
_
v1=50v+ _
i1
i2 i3
I
Node 1 +I - i1 = 0
Node 2 +i1 - i2 - i3 = 0
Node 3 +i2 + i3 - I = 0
i2 = v2/R2 i3 = v3/R3
Node 1 Node 2
Node 3
Use KCL andOhm’s Law
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Slide 10
R2= 20
I=5A
R1=10
R3= 5
+
_
+
_
v1=50v+ _
i1
i2 i3
I
Node 1 +I - i1 = 0
Node 2 +i1 - i2 - i3 = 0
Node 3 +i2 + i3 - I = 0
i2 = v2/R2 i3 = v3/R3
Node 1 Node 2
Node 3
Use KCL andOhm’s Law
CURRENT DIVIDER
v2 v3
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Slide 11
Learning check #1
what is relationship between v2 and v3 in previous example?
<, >, =
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Slide 12
KVL
+V - vR1 - vR2 = 0
iV = iR1 = iR2 = i
+V = iR1 + iR2
V = i(R1 + R2)
R2= 20
V= 5v
R1=10
+
_
+ _
LOOP 1+_
Start
i = V/(R1 + R2)
vR1 = iR1 = VR1 /(R1 + R2)
vR2 = iR2 = VR2/(R1 + R2)
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Slide 13
SERIES RESISTORS
+V - vR1 - vR2 = 0
iV = iR1 = iR2 = i
+V = iR1 + iR2
V = i(R1 + R2)
R2= 20
V= 5v
R1=10
+
_
+ _
LOOP 1+_
Start
i = V/(R1 + R2)
vR1 = iR1 = VR1 /(R1 + R2)
vR2 = iR2 = VR2/(R1 + R2)
VOLTAGE DIVIDER
NOTE
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Slide 14
SERIES RESISTORS
resistors attached in a “string” can be added together to get an equivalent resistance.
R = 2 R = 3
R = 4R = 9
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Slide 15
Learning check #2
what is value of Req in previous example when the three resistors are replaced with the following 4 new resistors?
1 k, 100, 10, and 1
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Slide 16
PARALLEL RESISTORS
resistors attached in parallel can be simplified by adding their conductances (G) together to get an equivalent resistance (R=1/G).
R = 9R = 1Geq = Gr1 + Gr2 + etc..
When you only have two:
Req = (R1*R2)/(R1+R2)
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Slide 17
Learning checks #3 & #4
4. what is value of Req in previous example?
5. what is the new value of Req when the two parallel resistors are replaced 2 new resistors shown below?
10 and 40
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Slide 18
series voltage sources
when connected in series, a group of voltage sources can be treated as one voltage source whose equivalent voltage = all source voltages
unequal voltage sources are not to be connected in parallel
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Slide 19
Learning check #5
6a. What is effective value of V for the series voltage sources in the example on board?
6b. What is the power dissipated in the resistor of 30?
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Slide 20
Networks I Today’s Learning Objectives –
Use KVL and KCLWhat is voltage and current division?Parallel and series sources
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Slide 21
chapter 3 - overview electric circuit applications - done define: node, closed path, loop - done Kirchoff’s Current Law - done Kirchoff’s Voltage Law- done a voltage divider circuit – in progress parallel resistors and current division series V-sources / parallel I-sources resistive circuit analysis
![Page 22: Rowan Hall 238A beard@rowan.edu September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.](https://reader035.fdocuments.in/reader035/viewer/2022062804/5697bf741a28abf838c7f9f4/html5/thumbnails/22.jpg)
Slide 22
Kirchhoff’s laws
Kirchhoff’s Current Law (KCL): The algebraic sum of the currents into a
node at any instant is zero.
Kirchhoff’s Voltage Law (KVL):The algebraic sum of the voltages around
any closed path in a circuit is zero for all time.
![Page 23: Rowan Hall 238A beard@rowan.edu September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.](https://reader035.fdocuments.in/reader035/viewer/2022062804/5697bf741a28abf838c7f9f4/html5/thumbnails/23.jpg)
Slide 23
KVL
+V - vR1 - vR2 = 0
iV = iR1 = iR2 = i
+V = iR1 + iR2
V = i(R1 + R2)
R2= 20
V= 5v
R1=10
+
_
+ _
LOOP 1+_
Start
i = V/(R1 + R2)
vR1 = iR1 = VR1 /(R1 + R2)
vR2 = iR2 = VR2/(R1 + R2)
Use KVL andOhm’s LawVOLTAGEDIVIDER
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Slide 24
SERIES RESISTORS
+V - vR1 - vR2 = 0
iV = iR1 = iR2 = i
+V = iR1 + iR2
V = i(R1 + R2)
R2= 20
V= 5v
R1=10
+
_
+ _
LOOP 1+_
Start
i = V/(R1 + R2)
vR1 = iR1 = VR1 /(R1 + R2)
vR2 = iR2 = VR2/(R1 + R2)
VOLTAGE DIVIDER
NOTE
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Slide 25
R2= 20
I=5A
R1=10
R3= 5
+
_
+
_
v1=50v+ _
i1
i2 i3
I
Node 1 +I - i1 = 0
Node 2 +i1 - i2 - i3 = 0
Node 3 +i2 + i3 - I = 0
i2 = v2/R2 i3 = v3/R3
Node 1 Node 2
Node 3
Use KCL andOhm’s Law
CURRENT DIVIDER
v2 v3
KCL
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Slide 26
PARALLEL RESISTORS
resistors attached in parallel can be simplified by adding their conductances (G) together to get an equivalent resistance (R=1/G).
R = 9R = 1Geq = Gr1 + Gr2 + etc..
When you only have two:
Req = (R1*R2)/(R1+R2)
![Page 27: Rowan Hall 238A beard@rowan.edu September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.](https://reader035.fdocuments.in/reader035/viewer/2022062804/5697bf741a28abf838c7f9f4/html5/thumbnails/27.jpg)
Slide 27
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 = ?
![Page 28: Rowan Hall 238A beard@rowan.edu September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.](https://reader035.fdocuments.in/reader035/viewer/2022062804/5697bf741a28abf838c7f9f4/html5/thumbnails/28.jpg)
Slide 28
Learning check #1
What is effective resistance value of three parallel resistors with values of 4, 5, 20?
Hint: calculate Geq , then R
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Slide 29
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
![Page 30: Rowan Hall 238A beard@rowan.edu September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.](https://reader035.fdocuments.in/reader035/viewer/2022062804/5697bf741a28abf838c7f9f4/html5/thumbnails/30.jpg)
Slide 30
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?
![Page 31: Rowan Hall 238A beard@rowan.edu September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.](https://reader035.fdocuments.in/reader035/viewer/2022062804/5697bf741a28abf838c7f9f4/html5/thumbnails/31.jpg)
Slide 31
PROBLEM SOLVING METHOD
+
_
+ _++
_
+
__
node1 node2 node3
node4
Ra Rb
Rcvs is
ia ib
ic
va vb
vc
ivs
visloop1 loop2
![Page 32: Rowan Hall 238A beard@rowan.edu September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.](https://reader035.fdocuments.in/reader035/viewer/2022062804/5697bf741a28abf838c7f9f4/html5/thumbnails/32.jpg)
Slide 32
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.
![Page 33: Rowan Hall 238A beard@rowan.edu September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.](https://reader035.fdocuments.in/reader035/viewer/2022062804/5697bf741a28abf838c7f9f4/html5/thumbnails/33.jpg)
Slide 33
WRITE THE KCL EQUATIONS
0 avs ii
0 cba iii
0 sb ii
0 vssc iii
node1:
node2: node4:
node3:
+
_
+ _+
+
_
+
__
node1 node2 node3
node4
Ra Rb
Rcvs is
ia ib
ic
va vb
vc
ivs
visloop1 loop2
![Page 34: Rowan Hall 238A beard@rowan.edu September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.](https://reader035.fdocuments.in/reader035/viewer/2022062804/5697bf741a28abf838c7f9f4/html5/thumbnails/34.jpg)
Slide 34
WRITE THE KVL EQUATIONS
+
_
+ _+
+
_
+
__
node1 node2 node3
node4
Ra Rb
Rcvs is
ia ib
ic
va vb
vc
ivs
visloop1 loop2
0 cas vvv 0 isbc vvv
loop1: loop2:
![Page 35: Rowan Hall 238A beard@rowan.edu September 18, 2006 Networks I for M.E. ECE 09.201 - 2 James K. Beard, Ph.D.](https://reader035.fdocuments.in/reader035/viewer/2022062804/5697bf741a28abf838c7f9f4/html5/thumbnails/35.jpg)
Slide 35
WRITE SUPPLEMENTARYEQUATIONS
+
_
+ _+
+
_
+
__
node1 node2 node3
node4
Ra Rb
Rcvs is
ia ib
ic
va vb
vc
ivs
visloop1 loop2
cccbbbaaa R/viR/viR/vi