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Fundamentals of Electrical Engineering
Fundamentals of Electrical Engineering 1
Grundlagen der Elektrotechnik 1
Chapter: Capacitors and Inductors / Kapazitäten und Induktivitäten
Michael E. Auer
Source of figures:
Alexander/Sadiku: Fundamentals of Electric Circuits,
McGraw-Hill
Michael E.Auer
10.12.2009
GET01
Course Content
Basic Concepts / Grundlagen
Basic Laws / Grundgesetze
Circuit Theorems / Zweipoltheorie
Common Methods of Network Analysis / Allgemeine Netzwerkanalyse
Operational Amplifiers / Operationsverstärker
Capacitors and Inductors / Kapazitäten und Induktivitäten
Basic RC and RL Circuits Switching / Einfache Schaltvorgänge
Michael E.Auer
10.12.2009
GET01
Chapter Content
Capacitors / Kapazitäten
Series and Parallel Capacitors / Serien- und Parallelschaltung von
Kapazitäten
Inductors / Induktivitäten
Series and Parallel Inductors / Serien- und Parallelschaltung von
Induktivitäten
Applications / Anwendungen
Summary / Zusammenfassung
Michael E.Auer
10.12.2009
GET01
Chapter Content
Kapazitäten
Induktivitäten
Michael E.Auer
10.12.2009
GET01
Capacitors (1)

A capacitor is a passive element designed to store energy in its
electric field.
 A capacitor consists of two conducting plates separated by an
insulator (or dielectric).
Michael E.Auer
10.12.2009
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Capacitors (2)

Capacitance C is the ratio of the charge q on one plate of a capacitor
to the voltage difference v between the two plates, measured in farads
(F).
q =C⋅v
and
C=
ε⋅A
d
 Where ε is the permittivity of the dielectric material between the
plates, A is the surface area of each plate, d is the distance between
the plates.
 Unit: F, pF (10–12), nF (10–9), and µF (10–6)
Michael E.Auer
10.12.2009
GET01
Capacitors (3)

If i is flowing into the + terminal of C
– Charging
=> + i
– Discharging => - i
 The current-voltage relationship of capacitor according to above
convention is:
dv
i =C⋅
dt
Michael E.Auer
and
10.12.2009
1
v=
C
∫
t
t0
i ⋅ dt + v(t0 )
GET01
Capacitors (4)
dv
i =C⋅
dt
1. v = const. ⇒
2. v = k ⋅ t ⇒
dv
=0 ⇒ i = 0
dt
dv
= k = const. ⇒ i = I
dt
3. v = k ⋅ sin ωt ⇒
dv
= k ⋅ cos ωt ⇒ i = k ⋅ C ⋅ cos ωt
dt
4. Abrupt change of v is not possible.
Michael E.Auer
10.12.2009
GET01
Capacitors (5)
Characteristic of a linear Capacitor
Practical Capacitor
Michael E.Auer
10.12.2009
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Capacitors (6)
dv
p = v ⋅i = v ⋅C ⋅
dt

The energy, w, stored in the capacitor
is:
w = p ⋅ dt = C ⋅ v ⋅ dv
1
2
W = Cv
2
Michael E.Auer
10.12.2009
GET01
Capacitors Example
An initially uncharged 1mF capacitor has the current shown
below across it.
Calculate the voltage across it at t = 2 ms and t = 5 ms.
Michael E.Auer
10.12.2009
GET01
Chapter Content
Kapazitäten
Induktivitäten
Michael E.Auer
10.12.2009
GET01
Parallel Capacitors

The equivalent capacitance of N parallel-connected capacitors is the
sum of the individual capacitances.
C eq = C1 + C 2 + ... + C N
Michael E.Auer
10.12.2009
GET01
Serial Capacitors

The equivalent capacitance of N series-connected capacitors is the
reciprocal of the sum of the reciprocals of the individual capacitances.
1
1
1
1
+
+ ... +
=
CN
C eq C1 C 2
Michael E.Auer
10.12.2009
GET01
Serial and Parallel Capacitors (1)
Find the equivalent capacitance seen at the terminals of the circuit in
the circuit shown below:
Answer:
Ceq = 40µF
Michael E.Auer
10.12.2009
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Serial and Parallel Capacitors (2)
Find the voltage across each of the capacitors in the circuit
shown below:
Answer:
v1 = 30V
v2 = 30V
v3 = 10V
v4 = 20V
Michael E.Auer
10.12.2009
GET01
Chapter Content
Kapazitäten
Induktivitäten
Michael E.Auer
10.12.2009
GET01
Inductors (1)

An inductor is a passive element designed to store energy in
its magnetic field.
 An inductor consists of a coil of conducting wire.
Michael E.Auer
10.12.2009
GET01
Inductors (2)

Inductance is the property whereby an inductor
exhibits opposition to the change of current flowing
through it, measured in henrys (H).
di
v= L⋅
dt
N ⋅µ ⋅ A
L=
l
2
and
 The unit of inductors is Henry (H), mH (10–3) and
µH (10–6).
Michael E.Auer
10.12.2009
GET01
Inductors (3)

The current-voltage relationship of an inductor:
1
i=
L
t
∫ v(t ) ⋅ d t + i(t )
0
t0
 The power stored by an inductor:
1
w = L ⋅ i2
2
 An inductor acts like a short circuit to dc (di/dt = 0) and its current cannot
change abruptly.
Michael E.Auer
10.12.2009
GET01
Inductors (4)
di
v= L⋅
dt
1. i = const. ⇒
2. i = k ⋅ t ⇒
di
=0 ⇒ v = 0
dt
allowed
not allowable
di
= k = const. ⇒ v = V
dt
3. i = k ⋅ sin ωt ⇒
di
= k ⋅ cos ωt ⇒ v = k ⋅ L ⋅ cos ωt
dt
4. Abrupt change of i is not possible.
Michael E.Auer
10.12.2009
GET01
Inductors (5)
Characteristic of a linear Inductor
Practical Inductor
Michael E.Auer
10.12.2009
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Inductors Example 1
The terminal voltage of a 2H inductor is
v = 10(1-t) V
Find the current flowing through it at
energy stored in it within 0 < t < 4 s.
t = 4 s and the
Assume i(0) = 2 A.
Answer:
i(4s) = -18V
w(4s) = 320J
Michael E.Auer
10.12.2009
GET01
Inductors Example 2
Determine vc, iL, and the energy stored in the capacitor and inductor in
the circuit of circuit shown below under dc conditions.
Answer:
iL = 3A
vC = 3V
wL = 1.125J
wC = 9J
Michael E.Auer
10.12.2009
GET01
Chapter Content
Kapazitäten
Induktivitäten
Michael E.Auer
10.12.2009
GET01
Series Inductors

The equivalent inductance of series-connected inductors is the
sum of the individual inductances.
Leq = L1 + L2 + ... + LN
Michael E.Auer
10.12.2009
GET01
Parallel Inductors

The equivalent capacitance of parallel inductors is the reciprocal of
the sum of the reciprocals of the individual inductances.
1
1
1
1
=
+
+ ... +
Leq L1 L2
LN
Michael E.Auer
10.12.2009
GET01
Equivalent Inductance
Calculate the equivalent inductance for the inductive ladder network in
the circuit shown below:
Answer:
Leq = 25mH
Michael E.Auer
10.12.2009
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Chapter Content
Kapazitäten
Induktivitäten
Michael E.Auer
10.12.2009
GET01
Mixed R, L, C Circuits
Under DC conditions, find:
a)
i, vC, vL
b)
the energy stored in C and L
Answer:
i = 2A
wC = 50J
wL = 4J
Michael E.Auer
10.12.2009
GET01
Integrator
Replacing the feedback resistor in
the inverting amplifier by a
capacitor produces an integrator.
t
1
vo = −
vi (t ) ⋅ dt
∫
R ⋅C 0
Michael E.Auer
10.12.2009
GET01
Differentiator
Replacing the input resistor in the
inverting amplifier by a capacitor
produces an differentiator.
vi(V)
vo(V)
dvi
vo = − R ⋅ C
dt
Michael E.Auer
10.12.2009
GET01
Chapter Content
Kapazitäten
Induktivitäten
Michael E.Auer
10.12.2009
GET01
Summary
•
•
•
•
•
•
•
•
Michael E.Auer
The current through a capacitor is directly proportional to the time rate of change of the
voltage across it. The current through a capacitor is zero unless the voltage is changing.
Thus, a capacitor acts like an open circuit for a dc source.
The voltage across a capacitor is directly proportional to the time integral of the current
through it. The voltage across a capacitor cannot change instantly.
Capacitors in series and in parallel are combined in the same way as conductances.
The voltage across an inductor is directly proportional to the time rate of change of the
current through it. The voltage across the inductor is zero unless the current is changing.
Thus, an inductor acts like a short circuit for a dc source.
The current through an inductor is directly proportional to the time integral of the voltage
across it. The current through an inductor cannot change instantly.
Inductors in series and in parallel are combined in the same way as resistors.
The energy stored in a capacitor is directly proportional to the capacitance and grows with
the square of the voltage across it.
The energy stored in an inductor is directly proportional to the inductance and grows with
the square of the current through it.
10.12.2009
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Current and Voltage Relationship for R, L, C
Michael E.Auer
10.12.2009
GET01

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