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Lecture 23. Impedance, Resonance in R-C-L Circuits
1
Preparation for the Final Exam
(a) Start earlier!(b) Review the concepts (lectures + textbook) and prepare your equation
sheet. Think how you can use every equation on your sheet, what types of
problems can be solved with these equations.
(c) Work on practice exams.
(d) Review all HW and Iclicker questions.
(e) Go over the end-of-chapter problems (you don’t need to solve them,
just check that you know how to approach them).
At the Exam(a) Make sure you understand the problem, read the problem formulation
carefully. Make a drawing!!! If you remain uncertain raise your hand and
ask the proctors.
(b) Get the units right. It is easy to eliminate the answers with wrong units.
This applies to formulas too.
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Reactance (recap)
2
= =
Resistor
Capacitor
Inductor
=
=1
=
=
=
AC (cos + ) driven circuits!
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Impedance
3
=
Impedance is a measure of how much the circuit
impedes the flow of current. The impedance is a complex
number (time-independent phasor), it relates time-
dependent phasors V(t) and I(t).
=
RLC
Reactances: = =1
=
Impedances: = =
1
=
= =
all terms are real
=
−
is the reference
phasor
=
−
≡
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C circuit
Can we plug a 1-µF capacitor into a wall outlet
( = 2 ∙ 60
, = 120) if the circuit breakers
can take 15A ?
= =1
=
1
2 ∙ 60 ∙ 1 ∙ 10−Ω = 2650Ω
=
=
120
2650Ω = 45
This current is sufficiently small. Theprimary concern is the voltage rating
of the capacitor, which should be
around 200V.
4
= =
Voltage = =
Current (reference phasor)
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L circuit
What happens when we plug a 1-H inductor into a wall
outlet?
= = = 2 ∙ 60 ∙ 1Ω = 377Ω
=
= 120377Ω
= 0.32
Again, the current won’t blow a circuit
breakers. The inductor must be designed tocarry 0.32A without overheating or
saturating the iron core.
5
= =
Voltage
Current (reference phasor)
= =
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Series R-C circuit
6
= = ()
=
+
=
+
1
= + =
= + =
=
=
−
=
1
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R-C circuits: Example
7
() = () =
−
= 1
tan =
1
= 1
1 ∙ 10 ∙ 100 ∙ 0.2 ∙ 10− = 50
1
Note that is negative (as it should be for the RC circuits).
=
=
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Low-Pass Filter
8
=
Goal: to suppress high-frequency ( > )
components in the spectrum of a signal.
=
=
+
=
1
+
1
=
1
+ 1=
1
+ 1
=
= +
=
=1
+ 1
≪1
1
≫1
1
Cutoff frequency: = 2 =1
We want to suppress the high-frequency ( > 10)
components in the output of an audio amplifier with the
output resistance 100 Ω. What capacitance do you need?
=1
2
=1
210
∙ 100
= 160
Output power:= two
times
=
= +
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Series R-L-C Circuits
9
= + = + 1
For R, C , and L in series:
() = () +
1
= ∙ ∗ = + 1
= = +
1
?1
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Series R-L-C circuits: Example 1
10
=
= = +
1
= 2.2 40 + 80 110 = 110
= 2.2 ∙ 80 = 176
=
= 2.2 ∙ 110 = 242
= 2.2 ∙ 40 = 88
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Series R-L-C circuits: Example 2
11
= ∙ cos = 151 ∙ 0.446 ∙ 0.72 = 48.6
An R-L-C series circuit with an inductance of 0.119H , a resistance of
244 Ω, and a capacitance of 7.27 µF carries an rms current of 0.446A
with a frequency of 391Hz .
= + = + 1
=
+ 1
= 339Ω
tan =
1
=
2455 ∙ 0.119 2455 ∙ 7.27 ∙ 10− −
244 ≈ 0.97
= 2455 /
arctan 0.97 ≈ 0.77
cos 0.77 = 0.72
3. What is the rms voltage of the source?
4. What average power is delivered by the source?
- average rate at which electrical energy is converted to thermal energy in the resistor
2. What is the phase angle?
- power factor for this circuit
1. What is the impedance of the circuit?
= ∙ = 0.446 ∙ 339Ω = 151
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Parallel R-L-C Circuit: Example
12
= ∙ cos
=
=
1
+
1
=10
2
1
0.25 + 6 3 =
50
2
1
=1
+ 1
1
=
1
+
1
=
1
1
+ 1
=
1
1
1
+ 1
tan =
1
1/ =
3
4 cos =
1
1 + =
4
5
= ∙ cos =10
2∙
50
2∙
4
5 = 200
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Series Resonance in the R-L-C circuits
13
= + = + 1
For R, C , and L in series:
= ∙ ∗
=
+
1
=
=
+ 1
=
= 1
- resonancefrequency
At = minimum (real) impedance, max current.
> < =
Note that at =, and can be greater than .
Resonancecondition:
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Parallel Resonance in the R-L-C circuits
16
1
=
1
+
1
+
=
1
1
+
1
=
1
+
1
1
=
1
+
1
=
=
1
+
1
min at =
=1
At the resonance frequency
R = 1Ω, C = 1F, L = 1H, and V = 1V
is at its minimum
→ 0 is a “short”
→ ∞ is a “short”
Note that at =, and can be
greater than .
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Transformer
17
Φ - the flux
per turn
ℇ = Φ
ℇ =
Φ
ℇ
ℇ=
For an ideal transformer
( = = 0):
=
Energy conservation: =
=
1
=
- as if the source had been
connected directly to a resistance
“impedance transformation”
=
Using mutual inductance = :
≡Φ
=Φ
= Φ
ℇ =
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Example
18
ℇ =
=
Sloppy formulation
=
=0.4 ∙ cos 377 12
377 ∙ 5 ∙ cos 377 = 2.55
ℇ =
= 6 = 10 ∙ 10− ∙ 6 ∙ 3
= 0.18
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19
Next time: Lecture 24. Electromagnetic Waves,
§§ 32.1 - 4
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