Fundamentals of Electrical Engineering
Fundamentals of Electrical Engineering 2
Grundlagen der Elektrotechnik 2
Chapter: Magnetically Coupled Circuits /
Magnetisch verkoppelte Kreise
Michael E. Auer
Source of figures:
Alexander/Sadiku: Fundamentals of Electric Circuits,
McGraw-Hill
Michael E.Auer
19.03.2010
GET25
Fundamentals of Electrical Engineering
Course Content
Sinusoids and Phasors / Harmonische Funktionen und Zeiger
Sinusoidal Steady State Analysis /
Netzwerkanalyse bei harmonischer Erregung
Frequency Response Analysis / Frequenzgang Analyse
AC Power / Leistung in Wechselstromkreisen
Magnetically Coupled Circuits / Magnetisch verkoppelte Kreise
Michael E.Auer
19.03.2010
GET25
Fundamentals of Electrical Engineering
Chapter Content
Mutual Inductance / Gegeninduktion
Transformers / Transformatoren
Applications / Anwendungen
Summary / Zusammenfassung
Michael E.Auer
19.03.2010
GET25
Fundamentals of Electrical Engineering
Chapter Content
Mutual Inductance / Gegeninduktion
Transformers / Transformatoren
Applications / Anwendungen
Summary / Zusammenfassung
Michael E.Auer
19.03.2010
GET25
Fundamentals of Electrical Engineering
Faraday‘s Law
dΦ
v=N
dt
dΦ di dΨ di
v=N
⋅ =
⋅
di dt di dt
di
v = L⋅
L – Self-inductance
dt
Michael E.Auer
19.03.2010
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Fundamentals of Electrical Engineering
Mutual Inductance (1)
•
It is the ability of one inductor to induce a voltage across a
neighboring inductor, measured in henrys (H).
di1
v2 = M 21
dt
di2
v1 = M 12
dt
The open-circuit mutual
voltage across coil 2
Michael E.Auer
In most cases:
M 12 = M 21
The open-circuit mutual
voltage across coil 1
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Fundamentals of Electrical Engineering
Mutual Inductance (2)
•
If a current enters the dotted terminal of one coil, the
reference polarity of the mutual voltage in the second
coil is positive at the dotted terminal of the second
coil.
Right-Hand-Rule
Michael E.Auer
19.03.2010
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Fundamentals of Electrical Engineering
Mutual Inductance (3)
•
The coupling coefficient, k, is a measure of the magnetic coupling
between two coils; 0 ≤ k ≤ 1.
M = k L1 L2
Michael E.Auer
0≤k≤1
19.03.2010
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Fundamentals of Electrical Engineering
Mutual Inductance (4)
Time-domain Analysis
Transformation to the frequency-domain
di1
di2
+M
v1 = i1 ⋅ R1 + L1
dt
dt
V1 = ( R1 + jωL1 ) ⋅ I1 + jωM ⋅ I 2
V2 = jωM ⋅ I1 + ( R2 + jωL2 ) ⋅ I 2
di
di
v2 = i2 ⋅ R2 + L2 2 + M 1
dt
dt
Michael E.Auer
19.03.2010
GET25
Fundamentals of Electrical Engineering
Mutual Inductance (5)
V = (Z1 + jωL1 ) ⋅ I1 − jωM ⋅ I 2
0 = − jωM ⋅ I1 + (Z L + jωL2 ) ⋅ I 2
Michael E.Auer
19.03.2010
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Fundamentals of Electrical Engineering
Mutual Inductance (6)
L = L1 + L2 + 2 M
L = L1 + L2 − 2 M
(series - opposing connection)
(series - aiding connection)
Michael E.Auer
19.03.2010
GET25
Fundamentals of Electrical Engineering
Chapter Content
Mutual Inductance / Gegeninduktion
Transformers / Transformatoren
Applications / Anwendungen
Summary / Zusammenfassung
Michael E.Auer
19.03.2010
GET25
Fundamentals of Electrical Engineering
Ideal Transformer
•
An ideal transformer is a unity-coupled, lossless transformer.
V2 N 2
=
=n
V1 N1
Z in =
Michael E.Auer
19.03.2010
I 2 N1 1
=
=
I1 N 2 n
ZL
n2
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Fundamentals of Electrical Engineering
Linear Real Transformer (1)
•
It is generally a four-terminal
device comprising two (or
more) magnetically coupled
coils
From
V = (Z1 + jωL1 ) ⋅ I1 − jωM ⋅ I 2
0 = − jωM ⋅ I1 + ( R2 + jωL2 + Z L ) ⋅ I 2
we obtain
V
Z in = = R1 + jωL1 + Z R
I1
Michael E.Auer
ω 2M 2
with Z R =
R2 + jωL2 + Z L
19.03.2010
Reflected impedance
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Fundamentals of Electrical Engineering
Linear Real Transformer (2)
Example
In the circuit below, calculate the input impedance and
current I1. Take Z1=60-j100Ω, Z2=30+j40Ω, and
ZL=80+j60Ω.
Ans:
Michael E.Auer
Z in = 100.14∠ − 53.1°Ω; I1 = 0.5∠113.1°A
19.03.2010
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Fundamentals of Electrical Engineering
Chapter Content
Mutual Inductance / Gegeninduktion
Transformers / Transformatoren
Applications / Anwendungen
Summary / Zusammenfassung
Michael E.Auer
19.03.2010
GET25
Fundamentals of Electrical Engineering
Applications (1)
•
Transformer as an Isolation Device to isolate ac supply from a
rectifier
Michael E.Auer
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Fundamentals of Electrical Engineering
Applications (2)
•
Transformer as an Isolation Device to isolate dc between two
amplifier stages.
Michael E.Auer
19.03.2010
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Fundamentals of Electrical Engineering
Applications (3)
•
Transformer as a Matching Device
Using an ideal transformer to match the
speaker to the amplifier
Michael E.Auer
19.03.2010
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Fundamentals of Electrical Engineering
Applications (4)
Example
An ideal transformer is rated at 2400/120V, 9.6 kVA, and has
50 turns on the secondary side.
Calculate:
(a) the turns ratio,
(b) the number of turns on the primary side, and
(c) the current ratings for the primary and secondary windings.
Ans:
(a)
This is a step-down transformer, n = 0.05
(b)
N1 = 1000 turns
(c)
I1 = 4A and I2 = 80A
Michael E.Auer
19.03.2010
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Fundamentals of Electrical Engineering
Applications (5)
Example
Calculate the turns ratio of an ideal transformer required to match a
100Ω load to a source with internal impedance of 2.5kΩ. Find the
load voltage when the source voltage is 30V.
Ans: n = 0.2; VL = 3V
Michael E.Auer
19.03.2010
GET25
Fundamentals of Electrical Engineering
Chapter Content
Mutual Inductance / Gegeninduktion
Transformers / Transformatoren
Applications / Anwendungen
Summary / Zusammenfassung
Michael E.Auer
19.03.2010
GET25
Fundamentals of Electrical Engineering
Summary
•
•
•
•
Michael E.Auer
Two coils are said to be mutually coupled if the magnetic flux Φ emanating from one passes
through the other.
The voltage induced in a coupled coil consists of self-induced voltage and mutual voltage.
A transformer is a four-terminal device containing two or more magnetically coupled coils. It
is used in changing the current, voltage, or impedance level in a circuit.
An ideal transformer is a lossless transformer, where input power and output power are
equal.
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