EEE 33 Basic Circuit Analysis University Of The Philippines

7/29/2019 EEE 33 Basic Circuit Analysis University Of The Philippines

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EEE 31 Introduction to Electrical and Electronics Engineering 2nd Sem2004

Linear CircuitsPart 1

EEE 31

Introduction to Electrical and

Electronics Engineering

EEE 31 Introduction to Electrical and Electronics Engineering p 2

Lecture Outline

Lumped and Distributed Models

Network/Circuit

Conservation of Charge

KCL

Conservation of Energy

KVL


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Consider the system

- Meralco views the total consumption of an entirehousehold as load-The only information you get is the KWH youconsumed when you get an electric bill

Lumped and Distributed Modeling

NPC

Melchor Hall UP Chapel DEEE Bldng.

Meralco

Consider also anelectric fan- motorwires, stator, rotor resistance per line


Lumped and Distributed Modeling

Distributed Parameter Modeling modeling of an electrical circuit into an infinite number

of vanishingly small elements

Lumped Parameter Modeling modeling of an electrical circuit into a number of simple

elements

Wire (0.5m)

Wire (0.5m)

501.5V

Wire has resistanceof 1/m

Voltagesource hasresistance

of 2

Zero-resistance wires

531.5V


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Network / Circuit

Current is measured as the flow of chargethrough the cross section of a circuit element

Voltageis measuredacross the ends of acircuit element

Electrical circuit consist of various types ofcircuit elementsconnected byconductors Circuit elements: sources, resistors, inductors and

capacitors

Conductors are wires connecting circuit elements


Network / Circuit

Topology

Branch of geometry which is concerned with thoseproperties of a geometrical figure which areunchangedwhen the figure is twisted, bent, folded, stretched,squeezed, or tied in knots, with the provision thatNOparts of the figure are to be cut apart or to be joinedtogether

Figures of the same topology.


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Nodes- where two or more circuit elements areconnected

Network / Circuit

5 circuit elements4 nodes

A B C

D


Path a set of elements that may be traversed inorder without passing through the same node twice

Branch a single path, containing one simpleelement, which connects one node to any othernode

Network / Circuit

5 circuit elements

5 branches


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Network / Circuit

Path

A B C

D

Path 1: going through source, 1k and 2.7k resistors

Path 2: going through 2.2pF, 2.7k and 10mH


Network / Circuit

Loop a closed path

In an electrical circuit, a loop is a closed path starting at anode and proceeding through circuit elements, eventuallyreturning to the starting node.

L1 L2

L1 and L2 are loops.


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Mesh a loop that does not contain any otherloops within it

Network / Circuit

M1 M2

M1 and M2 are meshes.


Planar circuits -a circuit which may be drawn ona plane surface in such a way that no branchpasses over or under any other branch

Network / Circuit


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Non-planar circuits - any circuit that is not planar.

Network / Circuit


Series Connection

Consider the circuit

The circuit comprises a loop. Every node has only 2 elements connected to it. Thecurrent througheach element is thesame.

a

b

i i

i i

i i Z

Water flow analogy


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Parallel Connection

Consider the circuit

Same circuit as before except for an extraelement.

Thevoltage acrosstheelementsX & Y is thesame.

i i

i i

i+

V

-

ix iyX Y



The algebraic sum of all electric chargesin any closed system is constant.

Charge can be transferred from one body to

another, but it cannot be created ordestroyed.


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Consider the junctionin the circuit:


21 QQQ +=

Taking the derivative

QQQ 21

+

=

Butt

Qi

=

21 iii +=

Q Q1

Q2


Kirchhoffs Current Law

The algebraic sum of the currents enteringa node equals zero.

KCL is a re-statement of conservation of charge

Convention:

+ for currents entering a node- for currents leaving a node

Total currententering a node

Total currentleaving a node=


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Example: KCL

sw ii =(a)

a

= zyxs iiii zyxs iiii ++=(b)

b

or



Energy can not be created ordestroyed,

it is converted

from one form to another.


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Consider the circuit: Dissipated energy foreach element


- Voltage Source

Ps = -VsIs

- Elementx

Px= VxIx

- Elementy

Py= VyIy


Consider the circuit: At any instant,


Ps +Px+ Py = 0

0=P

-VsIs+ VxIx+ VyIy= 0

CancelingIs

-Vs+ Vx+ Vy= 0

But Is=Ix= Iy-VsIs+ VxIs+ VyIs= 0


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Kirchhoffs Voltage Law

KVL is a re-statement of conservation of energy

Convention: Use the first polarity markencountered at each voltage in the algebraic sum

The algebraic sum of the voltages equalszero for any closed path (loop) in an

electrical circuit.

Totalvoltage rise

Totalvoltage drop=


3

Loop 3: 9 +1 3 +12 4 +3 =0

Example: KVL

1

Loop 1: 9 +1 +5 +3 =0

2

Loop 2: 3 +VX 4 5 =0 or VX =12


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KVL

WaterfallBy M.C. Escherhttp://www.worldofescher.com/gallery/A63L.html


Resistors in Series

KVL: Vtot = VR1 + VR2

Ohms Law: IReq = IR1 + IR2

Equivalent Resistance: Req = R1 + R2


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Voltage Source in Series

Equivalent Voltage:

You cannot put two voltage sources in parallel

unless they have the same value.

Veq = VS1 + VS2


Resistors in Parallel

Ohms Law:21 R

V

R

V

R

V

eq

+=

KCL: Itot = IR1 + IR2 Two Resistors inParallel:

21

21eq

RR

RRR

+=

Equivalent Resistance:21

111

RRReq+=


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Current Sources in Parallel

You cannot put two current sources in seriesunless they have same value.

IS1

>IS2

: IS1

< IS2

:

Equivalent Current: Itot = IS1 - IS2


Example 1

Find thecurrent IS.

I1 I2 I3

KCL: IS = I1 + I2 + I3

KVL: mAVI 5010051 =

= mAVI 1005052 =

= AV

I 15

53 =

=

Plugging in the values,

IS =50mA +100mA +1A =1.15A


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Example 1 : Alternative Solution

DetermineReq:

+

+

=

5

1

50

1

100

11

eqR

Simplify the resistors.

23.01=

eqRor Req = 4.348

KVL:

==

348.4

55 V

R

VI

eqs

IS =1.15A

Resistors in parallel: If R1


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Example 4

Find the current IS.

From previous example, we found I1=40 mA andV3=10V

KCL at node a: Is = I1+ I2

Plugging in to the first equation,

Is=40mA +40 mA =80 mA

I1 IS I2

a

Ohms Law: mA40250

32 =

=

VI



Get the equivalent resistance

From previous example, we foundVS =10V

IS

a

R1in series withR

2: R

eq1=R

1+R

2Req1 =100 +150 = 250

Req1is in parallelwithR3

Equivalent circuit:


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Get the equivalent resistance

IS

a

Req1is in parallelwithR3 : 31

31 )(

RR

RRR

eq

eqeq

+=

=+

= 125250250

)250(250

Ohms Law: mAR

VI

eq

SS 80

125

10===

The resistance of twoequal resistors in

parallel is equivalentto half the original

resistance.


Example 5

Find theequivalentresistance

between nodesA and B.

Simplify the circuit from left to right.

Series resistance: Req1 = R1 + R2 + R3

=80 +30 +40

Req1 =150


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Example 5

Req1parallel R4 :

75150

)75(150)(

41

412

+=

+=

RR

RRR

eq

eqeq

Req2=50

Req2seriesR5 : Req3=50 +50 =100

Req3parallel R6 :

150100

)150(1003

+=eqR =60

Req3seriesR7 :

RAB =60 +20 =80

EEE 33 Basic Circuit Analysis University Of The Philippines

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