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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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EEE 31 Introduction to Electrical and Electronics Engineering p 3
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
EEE 31 Introduction to Electrical and Electronics Engineering p 4
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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EEE 31 Introduction to Electrical and Electronics Engineering p 5
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
EEE 31 Introduction to Electrical and Electronics Engineering p 6
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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EEE 31 Introduction to Electrical and Electronics Engineering p 7
Nodes- where two or more circuit elements areconnected
Network / Circuit
5 circuit elements4 nodes
A B C
D
EEE 31 Introduction to Electrical and Electronics Engineering p 8
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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EEE 31 Introduction to Electrical and Electronics Engineering p 9
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
EEE 31 Introduction to Electrical and Electronics Engineering p 10
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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EEE 31 Introduction to Electrical and Electronics Engineering p 11
Mesh a loop that does not contain any otherloops within it
Network / Circuit
M1 M2
M1 and M2 are meshes.
EEE 31 Introduction to Electrical and Electronics Engineering p 12
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
EEE 31 Introduction to Electrical and Electronics Engineering p 14
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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EEE 31 Introduction to Electrical and Electronics Engineering p 15
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
EEE 31 Introduction to Electrical and Electronics Engineering p 16
Conservation of Charge
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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EEE 31 Introduction to Electrical and Electronics Engineering p 17
Conservation of Charge
Consider the junctionin the circuit:
Conservation of Charge
21 QQQ +=
Taking the derivative
QQQ 21
+
=
Butt
Qi
=
21 iii +=
Q Q1
Q2
EEE 31 Introduction to Electrical and Electronics Engineering p 18
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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EEE 31 Introduction to Electrical and Electronics Engineering p 19
Example: KCL
sw ii =(a)
a
= zyxs iiii zyxs iiii ++=(b)
b
or
EEE 31 Introduction to Electrical and Electronics Engineering p 20
Conservation of Energy
Energy can not be created ordestroyed,
it is converted
from one form to another.
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EEE 31 Introduction to Electrical and Electronics Engineering p 21
Consider the circuit: Dissipated energy foreach element
Conservation of Energy
- Voltage Source
Ps = -VsIs
- Elementx
Px= VxIx
- Elementy
Py= VyIy
EEE 31 Introduction to Electrical and Electronics Engineering p 22
Consider the circuit: At any instant,
Conservation of Energy
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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EEE 31 Introduction to Electrical and Electronics Engineering p 23
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=
EEE 31 Introduction to Electrical and Electronics Engineering p 24
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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EEE 31 Introduction to Electrical and Electronics Engineering p 25
KVL
WaterfallBy M.C. Escherhttp://www.worldofescher.com/gallery/A63L.html
EEE 31 Introduction to Electrical and Electronics Engineering p 26
Resistors in Series
KVL: Vtot = VR1 + VR2
Ohms Law: IReq = IR1 + IR2
Equivalent Resistance: Req = R1 + R2
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EEE 31 Introduction to Electrical and Electronics Engineering p 27
Voltage Source in Series
Equivalent Voltage:
You cannot put two voltage sources in parallel
unless they have the same value.
Veq = VS1 + VS2
EEE 31 Introduction to Electrical and Electronics Engineering p 28
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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EEE 31 Introduction to Electrical and Electronics Engineering p 29
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
EEE 31 Introduction to Electrical and Electronics Engineering p 30
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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EEE 31 Introduction to Electrical and Electronics Engineering p 31
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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EEE 31 Introduction to Electrical and Electronics Engineering p 33
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
EEE 31 Introduction to Electrical and Electronics Engineering p 34
Example 4 : Alternative Solution
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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EEE 31 Introduction to Electrical and Electronics Engineering p 35
Example 4 : Alternative Solution
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.
EEE 31 Introduction to Electrical and Electronics Engineering p 36
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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EEE 31 Introduction to Electrical and Electronics Engineering p 37
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
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