Networks I for M.E. ECE 09.201 - 2

Rowan Hall 238A

beard@rowan.edu

http://rowan.jkbeard.com

September 18, 2006

Networks I for M.E.ECE 09.201 - 2

James K. Beard, Ph.D.

Slide 2

Admin

1 – week from tomorrow Test 1 Cruise course website Questions thus far?

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

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

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

Slide 6

Gustav Robert Kirchhoff

1824-1887 two profound scientific laws published

in 1847

how old was he? LC #1

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.

Slide 8

KCL

Assume passive sign convention

R2= 20

I=5A

R1=10

R3= 5+

_

+

_

+ _Node 1 Node 2

Node 3

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

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

Slide 11

Learning check #1

what is relationship between v2 and v3 in previous example?

<, >, =

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)

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

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

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

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)

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

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

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?

Slide 20

Networks I Today’s Learning Objectives –

Use KVL and KCLWhat is voltage and current division?Parallel and series sources

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

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.

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

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

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

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)

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 = ?

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

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

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?

Slide 31

PROBLEM SOLVING METHOD

+

_

+ _++

_

+

__

node1 node2 node3

node4

Ra Rb

Rcvs is

ia ib

ic

va vb

vc

ivs

visloop1 loop2

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.

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

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:

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