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((&6 6SULQJ Lecture &7
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((&6 6SULQJ Lecture CIRCUIT ELEMENTS AND CIRCUIT ANALYSIS
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Lecture %UDQFK
9 (Branches)
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Today:
• Basic circuit elements: Ideal voltage sources and current
sources; linear resistors
• Resistor formulas
• Resistors in series and the parallel
• Formal nodal analysis
((&6 6SULQJ Circuit: Interconnection of electrical elements; modeled with two (or more)
terminals Terminology: Nodes and Branches
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−
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• Terminology: Nodes and branches
• Introduce the implicit reference (ground) node – defines
node voltages
• Introduce fundamental circuit laws: Kirchhoff’s Current
and Voltage Laws
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ELECTRIC CIRCUITS
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Lecture 2 review:
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There are also 3-terminal elements:
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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 point
on the wire (add capacitor circuit element to model this
phenomenon)
Is this reasonable? Yes, in many cases.
What are the limits?
“long” wires” ⇒ voltage propagates at nearly c = 3 × 108 m/s
((&6 6SULQJ Lecture Circuit Elements of DC Circuits
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((&6 6SULQJ Lecture &7
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((&6 6SULQJ Lecture ι
• Ideal voltmeter is a device which measures the
voltage across its terminals while drawing no
current. The current thru an ideal voltmeter is zero.
• Power = Voltage X Current = ?
V
+
−
Ideal current source
Ideal voltage source
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• Ideal ammeter is a device which measures the
current going thru it while maintains the voltage
across its terminal to be zero.
• Power = Voltage X Current = ?
Lecture Choi
CURRENT-VOLTAGE CHARACTERISTICS OF VOLTAGE &
CURRENT SOURCES
Describe a two-terminal circuit element by plotting current vs. voltage
Ideal Voltmeter and Ideal Ammeter
((&6 6SULQJ &7
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Ideal/Independent Voltage Sources
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V
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Lecture &7
How (not) to jump start a car?
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+
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-
+
+
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Good battery
(a) + terminal is higher than - terminal by V volt.
(b) vs(t) is time varying
(c) The voltage vary sinosoidally (sine wave) with amplitude of
160v.
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Dead battery
Good battery
Dead battery
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((&6 6SULQJ Lecture &7
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Example
Remember the “Chain-rule”?
VAB = VAC+VCD+VDB = (VA-VC) + (VC-VD) + (VD-VB)
or
= -6v
+ 4v
+ (-3v) = -5v.
VAB = VAD+VDB
If CE is any circuit element, will it affect the result?
Lecture Current Source Example
I + 3A = 0
I
= -3A
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((&6 6SULQJ A sinusoidal current source
with a magnitude of 16
milliampere is going in the
direction as shown
Lecture &7
Exercise
• Find the current I.
• Find the current I.
Apply KCL at node A:
Sum of all current into node
A equal to zero.
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27 milliampere is going
in the direction as shown
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By itself, the 10V voltage
source does not affect
the current.
Lecture Ideal Current Sources
• Ideal current source (similar to voltage
source)
• Find the potential difference VAB.
((&6 6SULQJ ((&6 6SULQJ A
By itself, the 10V voltage source
does not affect the current.
At node A, apply KCL:
Sum of all the current going into
node A:
-3mA -5mA + I =0
I = 8mA.
A
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((&6 6SULQJ Lecture &7
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((&6 6SULQJ Lecture &7
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Examples
Ideal Resistance
• Terminal A is at potential -3v with respect to ground,
and terminal B is at -2v. The resistance is 10Ω. What
is the current?
• Apply Ohm’s law:
• Ohm’s law stated that:
IA→B = (VA-VB)/R
IA→B = (VA-VB)/R = (-3 - (-2))/(10) = -0.1A.
• Suppose the current IA→B is known to be positive, does
the voltage drop from A to B positive or negative?
If IA→B > 0, then VA - VB >0 => VA > VB
((&6 6SULQJ Lecture &7
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Examples (cont.)
• Suppose the “actual direction” of current is from B to A
with a magnitude of 10A, the resistant is 3Ω. What is VAB?
Remember the direction of potential drop is also the direction
of the current flow.
By Ohm’s law: IB→A = (VB-VA)/R
=> (VB-VA)= VBA = R IB→A
=> VAB = -VBA = - R IB→A = - 3 (10) = -30v.
((&6 6SULQJ Lecture &7
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Example
• Reference for the voltage V1 and current I1 have been
chosen as indicated. Measurements show that the value
of I1 is -200A. If R1=3Ω, what is V1?
I1= (VA-VB)/R1
Notice the direction of I1 and
VAB(=V1) is consistent.
V1=I1R1 = (-200)(3) = -600v
((&6 6SULQJ Lecture &7
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((&6 6SULQJ Lecture &7
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I-V Graph for Ideal Resistance
Example
• R1=3000Ω, R2=4700Ω, and R3=3300Ω. From measurement
VA=-2v, VB=3v, VC=-3v, VD=12.735V (w.r.t.). Find I1, I2,and I3.
• For an ideal resistor,
the slope (1/R) is constant.
I1= (VB-VA)/R1=(3-(-2))/3000
= 1.666mA
I2= (VC-VB)/R2=(-3-3)/4700
= -1.28mA
I3= (VB-VD)/R3
=(3-12.735)/3300=-2.95mA.
((&6 6SULQJ Lecture &7
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IV CHARACTERISTICS
OF A RESISTOR
v
R
−
Lecture Resistors in Parallel and Series
i
+
((&6 6SULQJ If we use associated current and
voltage (i.e., i is defined as into +
terminal), then
v = iR (Ohm’s
law)
Question: What is the I-V characteristic for a 10KΩ resistor? Draw on axes
I (mA)
below.
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⇒ ,
3
)
2
1
6ORSH
V
5
9
,5
⇒
ZKHQ 9
,
Series
P$
Parallel
9
In series:
10 20 30
v = i R1+i R2 = i (R1+R2) = i (R), where R=R1+R2
((&6 6SULQJ Lecture &7
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((&6 6SULQJ Lecture continue
•
Apply KCL at node D
i = i1+i2 = (v/R1-v/R2)
= v (1/R1-1/R2)
= v (R2 + R1)/(R1R2)
= v /[(R1R2)/(R2 + R1)]
= v /R
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 the
terminals is unchanged?
By inspection, R2 and R3 are in
series. And this series combination
is in parallel with R1.
Resistant between A and B:
R = (R2+R3) || R1
= R1(R2+R3)/(R1+(R2+R3))
remember product over sum.
=(10,000)(100,000+47,000)/(10,000+100,000+47,000)
= 9360Ω
R = [(R1R2)/(R2 + R1)], where R1 is in parallel with R2
Lecture Choi
Resistors example
• In Parallel
((&6 6SULQJ &7
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Power Dissipation in Ideal Resistor
Resistor 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 element
Power = V I
But in the case of resistor, Ohm’s law hold
(voltage difference between resistor terminals
V =IR
equals current flow thru resistor X resistance)
Power = (I R) I =I2 R
Or Power = V (V/R) = V2/R
In practice, manufacturers state the max. power dissipation
of a resistor in watts.
((&6 6SULQJ Lecture &7
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Example
• The power dissipation of a 47000Ω resistor is stated by the
manufacturer to be 1/4 Watt. What is the maximum dc voltage that
may be applied? What is the largest dc current that can be made
to flow through the resistor without damaging it?
From the formula: P
P max
get Vmax2
Vmax
= V2/R
= Vmax2/R
= PmaxR
= √[(1/4)(47000)]
= 108.4V
From the formula: P
Pmax
get Imax2
= I2R
= Imax2 R
= Pmax/R
= √[(1/4)/(47000)]
= 2.3mA
((&6 6SULQJ Lecture &7
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((&6 6SULQJ CONDUCTIVITY, RESISTIVITY, AND RESISTANCE
(optional-for those of you who are interested)
A t
Resistor
More familiar, resistivity, ρ
ρ = σ
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Define resistance and Ohm’s Law:
((&6 6SULQJ 6 = 6LHPHQV =
, =σ $9
/
9
5
≡
5
=ρ
,
=
/
σ
$
=ρ
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1 Choose a Reference Node
:
The proportionality constant σ is called the conductivity and has units:
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FORMAL CIRCUIT ANALYSIS USING KCL:
NODAL ANALYSIS
/
If we apply a voltage V across a block of conducting
material of length L and cross-sectional area A, the
current, ,, is proportional to V and A, and inversely,
proportional to L.
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Lecture =
/
$
Ω
9
$
2 Define unknown node voltages (those not fixed by
voltage sources)
3 Write KCL at each unknown node, expressing current
in terms of the node voltages (using the constitutive
relationships of branch elements*)
FP
LI $
4 Solve the set of equations (N equations for N unknown
node voltages)
= :×W
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* with inductors or floating voltages we will use a modified Step 3
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Lecture &7
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((&6 6SULQJ Lecture NODAL ANALYSIS USING KCL
– The Voltage Divider –
L
1 Choose reference node
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EXAMPLE OF NODE ANALYSIS
R1
Define the node voltages (except reference node and the one set by
the voltage source); write down set of equations for node voltages Va
and Vb
9
2 Define unknown node voltages
3 Write KCL at unknown nodes
VSS −
R2
9
−
L
node voltage known
R1
Ì
+
-
966 − 9 9 − =
5
5
4 Solve:
9 = 966 ⋅
5
5 + 5 Va
V1
R2
Apply KCL:
9D − 9 9D − 9E 9D
+
+
=
5
9E
− 9D
5
+
5
5
9E
− ,6 = 5
R3
Vb
IS
R4
← reference node
You can solve for Va,
Vb.
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((&6 6SULQJ Lecture &7
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((&6 6SULQJ Lecture &7
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GENERALIZED VOLTAGE DIVIDER
RESISTORS IN SERIES
(Here its more convenient to use KVL than node analysis)
Circuit with several resistors in series – Can we find an equivalent resistance?
I
I ?
‡
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R2
+
−
VSS
R3
VSS
‡
.9/ WHOOV XV
, ⋅ 5
R4
‡
+ , ⋅ 5 + , ⋅ 5 + , ⋅ 5 = 966
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,
)
= 966 5 + 5 + 5 + 5 Lecture &7
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WHEN IS VOLTAGE DIVIDER FORMULA CORRECT?
I
i
R1
R1
+
R2
V SS
+
−
−
+
V2
VSS
R3
+
−
R4
9
5
=
5
j9
+5 +5 +5
9
− R2
X
i
R3
R5
3
R4
66
Correct if nothing else
connected to nodes
What is V2?
9
≠
5
5 + 5 + 5 + 5
j966
because R5 removes condition of
resistors in series – i.e. L ≠ ,
Answer:
+
−
R1
R2
R3
R4
+V
− 1
+
− V3
• We know
,
= 966 5 + 5 + 5 + 5 • Thus,
9 =
and
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((&6 6SULQJ Circuit with several resistors in series
5
5 + 5 + 5 5 + 5
j966
9 =
5
⋅9
5 + 5 + 5 + 5 66
5
⋅9
5 + 5 + 5 + 5 66
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