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7/28/2019 Dc Ckt Analysis Ppt2
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& 7 Choi ( ( & 6 6 S U L Q J Lecture
CIRCUIT ELEMENTS AND CIRCUIT ANALYSIS
Lecture 2 review:
• Terminology: Nodes and branches
• Introduce the implicit reference (ground) node – definesnode voltages• Introduce fundamental circuit laws: Kirchhoff’s Current
and Voltage Laws
Today:• Basic circuit elements: Ideal voltage sources and current
sources; linear resistors• Resistor formulas• Resistors in series and the parallel• Formal nodal analysis
+ −
+ −
+
− +
−
& 7
Choi ( ( & 6 6 S U L Q J Lecture
ELECTRIC CIRCUITSCircuit: Interconnection of electrical elements; modeled with two (or more)terminals Terminology: Nodes and Branches
There are also 3-terminal elements:
% U D Q F K
1 R G H
9 (Branches)
1 R G H V
& 7 Choi ( ( & 6 6 S U L Q J Lecture
CIRCUIT TERMINOLOGY
Labels:Nodes: connection points of two or more circuit elements(together with “wires”)Branches: one two-terminal element and the nodes at either end
Idealizations:Wires are “perfect conductors” – voltage is the same at any pointon the wire (add capacitor circuit element to model thisphenomenon)Is this reasonable? Yes, in many cases.What are the limits?
“long” wires” ⇒ voltage propagates at nearly c = 3 × 10 8 m/s
& 7 Choi ( ( & 6 6 S U L Q J Lecture
Circuit Elements of DC Circuits
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& 7 Choi ( ( & 6 6 S U L Q J Lecture
Ideal Voltmeter and Ideal Ammeter
• Ideal voltmeter is a device which measures thevoltage across its terminals while drawing nocurrent. The current thru an ideal voltmeter is zero.
• Power = Voltage X Current = ?
• Ideal ammeter is a device which measures thecurrent going thru it while maintains the voltageacross its terminal to be zero.
• Power = Voltage X Current = ?
& 7 Choi ( ( & 6 6 S U L Q J Lecture
CURRENT-VOLTAGE CHARACTERISTICS OF VOLTAGE &CURRENT SOURCES
Describe a two-terminal circuit element by plotting current vs. voltage
U H O H D V L Q J S R Z H U
V
i
D E V R U E L Q J S R Z H U
U H O H D V L Q J S R Z H U
D E V R U E L Q J S R Z H U
F K D U J L Q J
V
i
: H D V V X P H
: 3 D V V R F L D W H G V L J Q V V R L
F R P H V R X W R I
W H U P L Q D O
Ideal voltage source
V
ι
+−
i
Ideal current source
: H D V V X P H
: 3 D V V R F L D W H G V L J Q V V R L
F R P H V R X W R I W H U P L Q D O
Y
& 7 Choi ( ( & 6 6 S U L Q J Lecture
Ideal/Independent Voltage Sources
(a) + terminal is higher than - terminal by V volt.(b) v s(t) is time varying(c) The voltage vary sinosoidally (sine wave) with amplitude of
160v.
& 7 Choi ( ( & 6 6 S U L Q J Lecture
How (not) to jump start a car?
Good battery-
+Dead battery
-
+
Good battery-
+Dead battery
-
+
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& 7 Choi ( ( & 6 6 S U L Q J Lecture
Example• Find the potential difference V AB.
Remember the “Chain-rule”?
VAB = VAC+VCD+VDB = (VA-VC) + (V C-VD) + (V D-VB)or = -6v + 4v + (-3v) = -5v.VAB = VAD+VDBIf CE is any circuit element, will it affect the result?
& 7 Choi ( ( & 6 6 S U L Q J Lecture
Ideal Current Sources• Ideal current source (similar to voltagesource)
27 milliampere is goingin the direction as shown
A sinusoidal current sourcewith a magnitude of 16milliampere is going in thedirection as shown
& 7 Choi ( ( & 6 6 S U L Q J Lecture
Current Source Example
• Find the current I.
ABy itself, the 10V voltagesource does not affectthe current.
Apply KCL at node A:Sum of all current into nodeA equal to zero.
I + 3A = 0I = -3A
& 7 Choi ( ( & 6 6 S U L Q J Lecture
Exercise
• Find the current I. A
By itself, the 10V voltage sourcedoes not affect the current.
At node A, apply KCL:Sum of all the current going intonode A:-3mA -5mA + I =0I = 8mA.
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& 7 Choi ( ( & 6 6 S U L Q J Lecture
Ideal Resistance
• Ohm’s law stated that:IA→B = (VA-VB)/ R
& 7 Choi ( ( & 6 6 S U L Q J Lecture
Examples• Terminal A is at potential -3v with respect to ground,
and terminal B is at -2v. The resistance is 10 Ω. What
is the current?• Apply Ohm’s law:
IA→B = (VA-VB)/ R = (-3 - (-2))/(10) = -0.1A.
• Suppose the current IA→B is known to be positive, doesthe voltage drop from A to B positive or negative?
If IA→B > 0, then V A - VB >0 => V A > VB
& 7 Choi ( ( & 6 6 S U L Q J Lecture
Examples (cont.)
• Suppose the “actual direction” of current is from B to Awith a magnitude of 10A, the resistant is 3 Ω. What is V AB?
Remember the direction of potential drop is also the directionof the current flow.
By Ohm’s law: I B→A = (VB-VA)/ R => (VB-VA)= VBA = R I B→A => VAB = -VBA = - R I B→A = - 3 (10) = -30v.
& 7 Choi ( ( & 6 6 S U L Q J Lecture
Example• Reference for the voltage V 1 and current I 1 have beenchosen as indicated. Measurements show that the valueof I1 is -200A. If R 1=3Ω, what is V 1?
I1= (VA-VB)/R 1Notice the direction of I 1 andVAB(=V1) is consistent .V1=I1R1 = (-200)(3) = -600v
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& 7 Choi ( ( & 6 6 S U L Q J Lecture
Example
• R 1=3000 Ω, R 2=4700 Ω, and R 3=3300 Ω. From measurement
VA=-2v, V B=3v, V C=-3v, V D=12.735V (w.r.t.). Find I 1, I2,and I 3.
I1= (VB-VA)/R 1=(3-(-2))/3000 = 1.666mA
I2= (VC-VB)/R 2=(-3-3)/4700= -1.28mA
I3= (VB-VD)/R 3
=(3-12.735)/3300=-2.95mA.
& 7 Choi ( ( & 6 6 S U L Q J Lecture
I-V Graph for Ideal Resistance
• For an ideal resistor,the slope (1/R) is constant.
& 7 Choi ( ( & 6 6 S U L Q J Lecture
6 O R S H 5
$ Q V Z H U 9 ⇒ ,
9 , 5 ⇒ , P $
Z K H Q 9 9
IV CHARACTERISTICSOF A RESISTOR
If we use associated current andvoltage (i.e., i is defined as into +terminal), then v = iR (Ohm’slaw)
−
+
v
i
R
Question: What is the I-V characteristic for a 10K Ω resistor? Draw on axesbelow.
V10 20 30
I (mA)
1
2
3
& 7 Choi ( ( & 6 6 S U L Q J Lecture
Resistors in Parallel and Series
In series:
v = i R1+i R2 = i (R1+R 2) = i (R), where R=R 1+R 2
Series Parallel
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& 7 Choi ( ( & 6 6 S U L Q J Lecture
continue
• In Parallel
Apply KCL at node Di = i 1 +i 2 = (v/R 1-v/R 2)
= v (1/R 1-1/R 2)= v (R 2 + R 1)/(R 1R2)= v /[(R 1R2)/(R 2 + R 1)]= v /R
R = [(R 1R2)/(R 2 + R 1)], where R 1 is in parallel with R 2
& 7 Choi ( ( & 6 6 S U L Q J Lecture
Resistors example• The resistors shown has 2 terminals A and B. It is desirable to replace it
with a single resistor connected between terminals A and B. What
should the value of this resistor be so that the resistance between theterminals is unchanged?By inspection, R 2 and R 3 are inseries. And this series combinationis in parallel with R 1.
Resistant between A and B:
R = (R 2+R 3) || R 1
= R 1(R2+R 3)/(R 1+(R 2+R 3))
remember product over sum.=(10,000)(100,000+47,000)/(10,000+100,000+47,000)= 9360 Ω
& 7 Choi ( ( & 6 6 S U L Q J Lecture
Power Dissipation in Ideal ResistorResistor can convert electrical energy into thermal energy(heat).In Chapter 1, we learned that in any circuit element:Power = Voltage across the circuit element X Current
flow thru the circuit elementPower = V I
But in the case of resistor, Ohm’s law holdV = I R (voltage difference between resistor terminals
equals current flow thru resistor X resistance)Power = ( I R) I =I 2 R
Or Power = V (V/R) = V2
/RIn practice, manufacturers state the max. power dissipationof a resistor in watts.
& 7 Choi ( ( & 6 6 S U L Q J Lecture
Example• The power dissipation of a 47000 Ω resistor is stated by the
manufacturer to be 1/4 Watt. What is the maximum dc voltage thatmay be applied? What is the largest dc current that can be madeto flow through the resistor without damaging it?
From the formula: P = V 2 /RP max = Vmax
2 /Rget V max
2 = P max RVmax = √[(1/4)(47000)]
= 108.4V
From the formula: P = I 2RP max = I max
2 R
get I max 2 = P max /R= √[(1/4)/(47000)]= 2.3mA
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& 7 Choi ( ( & 6 6 S U L Q J Lecture
CONDUCTIVITY, RESISTIVITY, AND RESISTANCE(optional-for those of you who are interested)
If we apply a voltage V across a block of conductingmaterial of length L and cross-sectional area A, thecurrent, , , is proportional to V and A, and inversely,proportional to L.
ResistorA t
/
:
The proportionality constant σ is called the conductivity and has units:
9
/
$
,
6 L H P H Q V 6 6 F P O H Q J W K R O W D J H & X U U H Q W Y
σ =Ω
===
More familiar, resistivity, ρ F P
$
9
F P O H Q J W K X U U H Q W Y R O W D J H F X Q L W V =Ω=×= σ ρ
Define resistance and Ohm’s Law:
: W
/
5
W : $ L I
$
/
$
/
,
9
5
ρ
ρ σ
=
×===≡
, ~ 9 $ /
& 7 Choi ( ( & 6 6 S U L Q J Lecture
FORMAL CIRCUIT ANALYSIS USING KCL:NODAL ANALYSIS
2 Define unknown node voltages (those not fixed byvoltage sources )
3 Write KCL at each unknown node, expressing currentin terms of the node voltages (using the constitutiverelationships of branch elements*)
* with inductors or floating voltages we will use a modified Step 3
1 Choose a Reference Node
4 Solve the set of equations (N equations for N unknownnode voltages)
& 7 Choi ( ( & 6 6 S U L Q J Lecture
NODAL ANALYSIS USING KCL – The Voltage Divider –
1 Choose reference node
2 Define unknown node voltages
9
3 Write KCL at unknown nodes
$ $
#
'
#
' ' −=−
4 Solve:
$ $
# #
#
' '
+⋅=
9
−
R1
V SS
− L
L
R2
& 7 Choi ( ( & 6 6 S U L Q J Lecture
EXAMPLE OF NODE ANALYSIS
Define the node voltages (except reference node and the one set bythe voltage source); write down set of equations for node voltages V aand V b
node voltage known Ì
V a V b
←reference nodeApply KCL:
=+−
+−
#
'
#
' '
#
' '
, - , ,
=−+− $
- , -
,
5
9
5
9 9
R 4V 1 R 2
+- IS
R 3R1
You can solve for V a ,Vb .
: K D W L I Z H X V H G G L I I H U H Q W U H I Q R G H "
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& 7 Choi ( ( & 6 6 S U L Q J Lecture
RESISTORS IN SERIES
Circuit with several resistors in series – Can we find an equivalent resistance?
. & / W H O O V X V V D P H F X U U H Q W
I O R Z V W K U R X J K H Y H U \ U H V L V W R U
. 9 / W H O O V X V
$ $
9 5 , 5 , 5 , 5 , =⋅+⋅+⋅+⋅
& O H D U O \
5 5 5 5 9 ,
$ $
+++=
7 K X V H T X L Y D O H Q W U H V L V W D Q F H R I U H V L V W R U V L Q V H U L H V L V W K H V L P S O H V X P
R 2
R 1
V SS
I ?
R 3
R 4
−+
(Here its more convenient to use KVL than node analysis)
& 7 Choi ( ( & 6 6 S U L Q J Lecture
GENERALIZED VOLTAGE DIVIDER
Circuit with several resistors in series
R 2
R 1
V SS
I
R 3
R 4
−+−
+
−+ V 1
V 3
• We know 5 5 5 5 9 ,
$ $
+++=
• Thus,
$ $
'
# # # #
#
' ⋅+++
=
and $ $
'
# # # #
#
' ⋅+++
=
0 9 . 0 9 .
& 7 Choi ( ( & 6 6 S U L Q J Lecture
WHEN IS VOLTAGE DIVIDER FORMULA CORRECT?
R 2
R 1
V SS
I
R 3
R 4
−+ −
+
6 6
9
5 5 5 5
5
9 j
+++=
Correct if nothing else
connected to nodes
3
V SS
i
R 2
R 1
R 3
R 4
−+ −
+
'
R 5
i X
$ $
9
5 5 5 5
5
9
j
+++≠
because R 5 removes condition ofresistors in series – i.e . , L
≠
What is V 2? Answer: $ $
9
5 5 5 5 5
5
j
+++
V2