ece205a_w12_circuits

7/21/2019 ECE205A_w12_circuits

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Week 1, 21-23 Jan ECE 205A Spring 2014 1

ECE 205A Lecture 1.2

Ohms law, KCL and KVL, series and parallel resistors;Analysis Method(s)

Announcements

205AGHW1 due Tuesday 28 Jan, in class206Orientation and Lab 0 next week

Due

Nothing today

For next time, readKnI 2.7, 2.8


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What will we talk about today?

Ohms Law, Power

Nodes, Branches, Loops, Meshes

KCL, KVL

Node-Pair

Series, ParallelThe Sudoku Method


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R

+

-

Ohmss Law

R Rresistance (1

= 1V/A)

or

G Gconductance (1S = 1A/V)

R

+ -

a b

Vab

R

For an ideal resistor, the -curve

is a straight line, passing through 0.

viR

1


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Resistance is a property of an object.

Resistivity is a property of a material

R L

a is a bulk property

of the material

has units of m (ohm-meter)

R depends on both the material and the geometry.We will see this again with capacitors and inductors

a

Length

L


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Ohms Law and Power

and R so

and

A resistor is a passive element. It can not generate (or

store) energy. It can only dissipate. 0

If you compute P


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Reality is a little more complicated

R=0 is a superconductor

R= is a superinsulator

These are physics magic. You are unlikely to encounter

them in real life

You may have to settle for:

R=0.3(copper penny)

R=1010 (your ECE 206 multimeter)

(FETs can be 1012-1015)

Questions?


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Kirchoffs Laws- Definitions

Node:a point at which two

or more circuit elements

are connected. a, d, e, f, g, h

Note: a, b, and c

are really the same node.

However, if you want to say

that R=0 for wire, then Vab = iR = i*0 = 0 = Vbc = Vac

So it wont kill you to say a, b, c are nodes

R3R2R4

R6

R5

R1 V(t)

+

-

I(t)

a b c

hgf

ed


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Branch:a two-terminal circuit element,

connected between two nodes

R1, R2, R3, R4, R5, R6, V(t), I(t)

Loop:a closed path through

(some part of) the circuit in which

no node is crossed more than once

a-b-e-d-a

a-b-c-h-g-f-d-a etc. There are 7 loops here

R3R2

R4

R6

R5

R1 V(t)+

-

I(t)

a b c

hgf

ed


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Mesh:a loop that does not contain within it another loop

R3R2R4

R6

R5

R1 V(t)

+-

I(t)

a b c

hgf

ed

mesh

mesh

mesh


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Kirchoffs Current Law- KCL

the algebraic sum of the currents entering a node is zero

i1

i2 i3

i1 - i2 - i3= 0

An alternative form for KCL is current in = current out

i1 = i2 + i3Questions?

Our convention:

in+

out-


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Kirchoffs Voltage Law- KVL

the algebraic sum of the voltages around a loop is zero

R1

Vx+

- R2

Vy

+

-


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and

VY+

-

d

R1

a bfrom to

VX

+

-

a

c

Questions?


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Kirchoffs Voltage Law- KVL

the algebraic sum of the voltages around a loop is zero

R1

Vx+

- R2

Vy

+

-

a b

c d

0

Vx Vy 0

Ohms Law

R1*

R2*

-VxR1*iVyR2*i 0

VxVy

R 1 R 2


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The greatest danger is in loosing track of the signs:R1

Vx+

- R2

Vy+

-

a b

c d

Vx

+

-

a

c

Vy

+

-

b

d

+ + +

0Vx Vy 0


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There are several ways to write KVL;

this will be our convention:

Vx

+

-

a

c

Vy

+

-

b

d

+ + +

Questions?

0

Vx Vy 0


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Node-Pair Circuits

If your house was constructed after (e.g.) 1930

R1Ix Iy

R2 R3

1 2 3

Since this is a common topology for AC power circuits, it

is important to understand

Ix, Iy? Two sources? Is that a mistake?The next time you are on I-57. Count the wind turbines


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R1Ix Iy

R2 R3

1 2 3

Lets use KCL:

Ix 1 Iy 2 3 0

And Ohms Law

1

R1

2

R2

3

R3

Ix 1 Iy 2 3 0 Ix

R1 Iy

R2

R3


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0 Ix

R1 Iy

R2

R3

R1

R2

R3

Ix Iy

Ix Iy

1R1 1R2 1R3

Remember conductance? G=1/R =G

Think about this problem using G1, G2, G3Its so much easier!

Questions?


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Resistors, in Series and Parallel

Series:two resistors

which share

a common nodethat has

no other connections

R8+R10, R10+R11, R11+R9 are series-pairsWe say R8 is in series with R10

or R8 and R10 are in series

R2

R1 R5 R8

R10

R11R9

R7

R6

R3 R4

Vx

+

-


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Parallel:

two resistorswhich form a loop

with

no other elements

R3||R4R3 is parallel with R4

R3 and R4 are in parallel

Note ||parallel operator

R2

R1 R5 R8

R10

R11R9

R7

R6

R3 R4

Vx+

-


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Series

ReqV

V1V2

V1

V2

R1R2

In general, R R=

R1

R2

ReqV

+

-

V

+

-

-

-

+

+

V1

V2


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VReq

VR1R2

V1 R1* R1

R1R2V

V2 R2* R2

R1R2V

R1

R2

ReqV

+

-V

+

-

-

-

+

+

V1

V2

} VoltageDivision


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Parallel

1

Req

V

12

V

1

V

2

V

1

R1

1

R2

In general, R

R=

R1 R2

V+

-

21

ReqV+

-


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R1||R2

Req

R1

R2 Req

R1*R2

R1R2

Strangely, this form

is more common in books

But this form is easieron your calculator!

R1 +R2=

R1 R2

V+

-

21

ReqV+

-


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V Req 1||2

R1*R2

R1R2

1 V

R1

R2

R1R2

2 VR2

R1R1R2

}

Current

Division

R1 R2

V+

-

21

ReqV+

-


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Current Divider, 3 resistors

I2I

R1

I1

R2I

R3I2

I

R1I1

R2

I3

I

1

1

1

1

1

1

1


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2+

-

1

+

-

+

-

2

+

-

1

+

-

+

-

Sources, in Series and in Parallel

1 2 Bursts into flames!

Unless 1 2


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21

Sources, in Series and in Parallel

Also nota good idea! 1 2

Unless 1 2

Questions?

2

1


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The Sudoku Method

Write down what you know,

where you know it

8

5

2 4

2

5

3

14

1

4

4

12v+

-

Hintmaking copies of the diagram, helps


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Cluster

4+2+5+3=14

4||4=2

8

5

2 4

2

5

3

14

1

4

4

12v+

-


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Expand Clusters

14||14=7

8+2=10

8

5

2 4

2

5

3

14

1

4

4

12v+

-


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Keep Expanding

2+7+1=10

10||10=5

8

5

2 4

2

5

3

14

1

4

4

12v+

-


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Start applying Ohms Law

5+5=10

= 12V/10= 1.2A

8

5

2 4

2

5

3

14

1

4

4

12v+

-


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More Ohms Law

= 1.2A*5= 6V

8

5

2 4

2

5

3

14

1

4

4

12v+

-


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The Sudoku Method

12V-6V=6V 6V/10=0.6A

8

5

2 4

2

5

3

14

1

4

4

12v+

-


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The Sudoku Method

0.6A*8= 4.8V

6-4.8=1.2V 1.2V/4 = 0.3A

8

5

2 4

2

53

14

1

4

4

12v+

-


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Fast vsDangerous

The Sudoku Methodcan be very fast for simple circuits

-

even circuits that are large, but have simple topology

However, the method has two fundamental weaknesses

1.

If you make a mistake, the error propagates2.

It can be very difficult to identify wherethe chain

of errors begins

If the results matter, formal methods are better.


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Next Week

Formal Methods

Node Voltage

Loop/mesh Current

Superposition

Questions?


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= = = = = = = = =

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