Download Networks I for M.E. ECE 09.201 - 2

Networks I for M.E.
ECE 09.201 - 2
James K. Beard, Ph.D.
Rowan Hall 238A
[email protected]
http://rowan.jkbeard.com
September 18, 2006
Admin



1 – week from tomorrow Test 1
Cruise course website
Questions thus far?
September 18, 2006
Class Name
Slide 2
Networks I

Today’s Learning Objectives –
 Analyze
DC circuits with passive elements
including: resistance -- DONE
 Learn about switches -- DONE
 Introduce KVL and KCL
 What is voltage and current division?
 Parallel and series sources
 Solve some more difficult problems
September 18, 2006
Class Name
Slide 3
chapter 2 - overview
engineering and linear models - done
 active and passive circuit elements - done
 resistors – Ohm’s Law - done
 dependent sources – done
 independent sources – done

 transducers – done

switches – in progress
September 18, 2006
Class Name
Slide 4
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
September 18, 2006
Class Name
Slide 5
Gustav Robert Kirchhoff
1824-1887
 two profound scientific laws published
in 1847


how old was he? LC #1
September 18, 2006
Class Name
Slide 6
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.
September 18, 2006
Class Name
Slide 7
KCL
R1=10
Node 1
Node 2
+
_
+
+
_
_
I=5A
R3= 5
R2= 20
Node 3
Assume passive sign convention
September 18, 2006
Class Name
Slide 8
Node 1
I=5A
R1=10
i1
+ v1=50v _
Node 2
+
+
I
i2
R2= 20
Node 1 +I - i1 = 0
Node 2 +i1 - i2 - i3 = 0
Node 3 +i2 + i3 - I = 0
i2 = v2/R2
September 18, 2006
R3= 5
_
i3
_
Node 3
Use KCL and
Ohm’s Law
i3 = v3/R3
Class Name
Slide 9
Node 1
I=5A
R1=10
Node 2
i1
+ v1=50v _
+
I
v2
i2
R2= 20
+
R3= 5
_
v3
i3
_
Node 3
Node 1 +I - i1 = 0
Node 2 +i1 - i2 - i3 = 0
Use KCL and
Ohm’s Law
Node 3 +i2 + i3 - I = 0
i2 = v2/R2
September 18, 2006
i3 = v3/R3
Class Name
CURRENT DIVIDER
Slide 10
Learning check #1
what is relationship between v2 and v3 in
previous example?
 <, >, =

September 18, 2006
Class Name
Slide 11
KVL
R1=10
_
+
+
V= 5v
+
_
LOOP 1
R2= 20 _
Start
+V - vR1 - vR2 = 0
i = V/(R1 + R2)
iV = iR1 = iR2 = i
vR1 = iR1 = VR1 /(R1 + R2)
+V = iR1 + iR2
V = i(R1 + R2)
September 18, 2006
vR2 = iR2 = VR2/(R1 + R2)
Class Name
Slide 12
SERIES RESISTORS
R1=10
_
+
+
V= 5v
+
_
LOOP 1
R2= 20 _
Start
NOTE
+V - vR1 - vR2 = 0
i = V/(R1 + R2)
iV = iR1 = iR2 = i
vR1 = iR1 = VR1 /(R1 + R2)
+V = iR1 + iR2
vR2 = iR2 = VR2/(R1 + R2)
V = i(R1 + R2)
September 18, 2006
Class Name
VOLTAGE DIVIDER Slide 13
SERIES RESISTORS

resistors attached in a “string” can be
added together to get an equivalent
resistance.
R = 2
R = 3
R = 9
R = 4
September 18, 2006
Class Name
Slide 14
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

September 18, 2006
Class Name
Slide 15
PARALLEL RESISTORS

resistors attached in parallel can be
simplified by adding their conductances (G)
together to get an equivalent resistance
(R=1/G).
Geq = Gr1 + Gr2 + etc..
R = 1
R = 9
When you only have two:
Req = (R1*R2)/(R1+R2)
September 18, 2006
Class Name
Slide 16
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

September 18, 2006
Class Name
Slide 17
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

September 18, 2006
Class Name
Slide 18
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?

September 18, 2006
Class Name
Slide 19
Networks I

Today’s Learning Objectives –
 Use
KVL and KCL
 What is voltage and current division?
 Parallel and series sources
September 18, 2006
Class Name
Slide 20
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
September 18, 2006
Class Name
Slide 21
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.
September 18, 2006
Class Name
Slide 22
KVL
R1=10
Use KVL and
Ohm’s Law
VOLTAGE
DIVIDER
+
V= 5v
+
_
LOOP 1
R2= 20 _
Start
+V - vR1 - vR2 = 0
i = V/(R1 + R2)
iV = iR1 = iR2 = i
vR1 = iR1 = VR1 /(R1 + R2)
+V = iR1 + iR2
V = i(R1 + R2)
September 18, 2006
_
+
vR2 = iR2 = VR2/(R1 + R2)
Class Name
Slide 23
SERIES RESISTORS
R1=10
_
+
+
V= 5v
+
_
LOOP 1
R2= 20 _
Start
NOTE
+V - vR1 - vR2 = 0
i = V/(R1 + R2)
iV = iR1 = iR2 = i
vR1 = iR1 = VR1 /(R1 + R2)
+V = iR1 + iR2
vR2 = iR2 = VR2/(R1 + R2)
V = i(R1 + R2)
September 18, 2006
Class Name
VOLTAGE DIVIDER Slide 24
KCL
Node 1
I=5A
R1=10
Node 2
i1
+ v1=50v _
+
I
v2
i2
R2= 20
+
R3= 5
_
v3
i3
_
Node 3
Node 1 +I - i1 = 0
Node 2 +i1 - i2 - i3 = 0
Use KCL and
Ohm’s Law
Node 3 +i2 + i3 - I = 0
i2 = v2/R2
September 18, 2006
i3 = v3/R3
Class Name
CURRENT DIVIDER
Slide 25
PARALLEL RESISTORS

resistors attached in parallel can be
simplified by adding their conductances (G)
together to get an equivalent resistance
(R=1/G).
Geq = Gr1 + Gr2 + etc..
R = 1
R = 9
When you only have two:
Req = (R1*R2)/(R1+R2)
September 18, 2006
Class Name
Slide 26
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 = ?

September 18, 2006
Class Name
Slide 27
Learning check #1

What is effective resistance value of three
parallel resistors with values of 4, 5,
20?

Hint: calculate Geq , then R
September 18, 2006
Class Name
Slide 28
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

September 18, 2006
Class Name
Slide 29
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?

September 18, 2006
Class Name
Slide 30
PROBLEM SOLVING
METHOD
+
node1
va
Ra
ia
ivs
vs
_ node2 +
+
_
loop1
+
ic R
c
vb
node3
Rb
ib
+
vc
_
_
loop2
is
vis
_
node4
September 18, 2006
Class Name
Slide 31
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.

September 18, 2006
Class Name
Slide 32
WRITE THE KCL
EQUATIONS
v
_
+
node1
a
Ra
vs
node2 +
ia
ivs
+
_
loop1
vb
+
vc
_
node3
Rb
ib
+
ic R
c
_
loop2
is
vis
_
node4
node1:
ivs  ia  0
node3:
ib  i s  0
node2:
ia  ib  ic  0
node4:
ic  i s  ivs  0
September 18, 2006
Class Name
Slide 33
WRITE THE KVL EQUATIONS
+
node1
va
Ra
ia
ivs
vs
_ node2 +
+
_
loop1
+
ic R
c
vb
node3
Rb
ib
+
vc
loop2
_
_
is
vis
_
node4
loop1:
loop2:
 vc  vb  vis  0
 v s  va  vc  0
September 18, 2006
Class Name
Slide 34
WRITE SUPPLEMENTARY
EQUATIONS
+
node1
va
Ra
ia
ivs
vs
_ node2 +
+
_
loop1
vb
+
vc
_
node3
Rb
ib
+
ic R
c
_
loop2
is
vis
_
node4
ia  va / Ra
September 18, 2006
ib  vb / Rb
Class Name
ic  vc / Rc
Slide 35