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Passive components and circuits - CCP
Lecture 3
Introduction
Index
 Theorems for electric circuit analysis




Kirchhoff theorems
Superposition theorem
Thevenin theorem
Norton theorem
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Kirchhoff theorems
http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Kirchhoff.html
 The theorems are applicable in circuit analysis for insulated
circuits (the circuit is not exposed to external factors as
electrical or magnetic fields).
 Kirchhoff’s voltage law :
The algebraic sum of the voltages at any instant around any
loop in a circuit is zero.
 Kirchhoff’s current law
The algebraic sum of the currents at any instant at any node
in a circuit is zero.
TKV :  v  0 TKI :  i  0
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Applying of Kirchhoff’s Theorem
 If a circuit has l branches and n nodes, then the complete
description of its operation is obtained by writing KVL for
l-n+1 loops and KCL for n-1 nodes. The loops must form an
independent system.


Prior to the analysis of an electric circuit, the conventional
directions of the currents in the circuit are not known. So,
before writing the equations (Kirchhoff’s laws) for each loop, a
positive arbitrary direction is selected for each branch of the
circuit.
After performing the analysis of the circuit, if the value of the
current is positive, the arbitrary and conventional directions of
the current flow are identical. If the value of the current is
negative, the conventional direction is opposite to the arbitrary
selected direction.
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Applying of Kirchhoff’s Theorem



Step I – choosing the voltages and currents arbitrary
directions
Step II – choosing the loop’s cover direction
Step III – writing the Kirkhhoff’s theorems
I
I
R1
A
R3
V
R2
V
R1
V1=5 V
R1
330
I
R2
150
V2=9 V
R2
V
R3
R3
1K
 V 1  VR1  VR2  V 2  0

 V 2  VR2  VR3  0
I  I  I  0
 R1 R2 R3
B
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Solving equation systems


In order to solve the equations,
the Ohm’s Law is applied and
the voltage across the resistors
are substituted.
It is obtained a system with
three equations and three
variables, IR1, IR2 and IR3.
VR1  R1  I R1
VR2  R 2  I R2
VR3  R3  I R3
 V 1  R1 I R1  R 2  I R 2  V 2  0

 V 2  R 2  I R 2  R3  I R 3  0
I  I  I  0
 R1 R 2 R 3
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The System Solutions

The solutions are:
 IR1-6 mA


IR2-13 mA
IR37 mA
I
V
R2
V
R1
V1=5 V

The voltages across the
resistances:
 VR1-2 V
 VR2-2 V

I
A R3
R1
R1
330
I
R2
V
R3
R2
150
R3
1K
V2=9 V
B
VR37 V
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Linear and nonlinear circuits
 If the transmittances defined for a circuit are constant (are
represented with linear segments in v-i, v-v or i-i planes), are
called linear transmittances.
 A circuit or a component with only linear transmittances is
called linear circuit or linear component.
 Important: generally, electronics devices and circuits made
with them are nonlinear.
 The method used to approximate a nonlinear circuit
operation with a linear circuit operation is called linearization.
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The Superposition Theorem
The Superposition theorem states that the
response in a linear circuit with multiple
sources can be obtained by adding the
individual responses caused by the separate
independent sources acting alone.
The source passivation  the sources are replaced by
their internal resistance.
By passivation, the ideal voltage source is replaced with
a short-circuit, and the ideal current source is replaced
with an open-circuit.
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The Superposition Theorem

+
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Thevenin’s Theorem
 Any two-terminal, linear network of sources and resistances
can be replaced by a single voltage source in series with a
resistance. The voltage source has a value equal to the opencircuit voltage appearing at the terminals of the network. The
resistance value is the resistance that would be measured at
the network’s terminals for passivated circuit.
 The source passivation= the sources are replaced by their
internal resistance
 By passivation, the ideal voltage source is replaced with
a short-circuit, and the ideal current source is replaced
with an open-circuit.
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Thevenin’s Theorem
Vo and Ro must be determined.
I
A
I
R3
A
V
R3
R1
330
V1=5 V
R2
150
R3
1K
Ro
R3
V
R3
=?
R3
1K
V2=9 V
V =?
O
B
ELECTRONIC CIRCUIT
EQUIVALENT
CIRCUIT
B
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Calculus of open-circuit voltage
 In order to calculate the
open-circuit voltage, the
Kirchhoff’s theorems can
be applied.
 The superposition theorem
will also be applied.
A
R1
330
V1=5 V
R2
150
V
gol
V2=9 V
CIRCUIT ELECTRONIC
B
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The superposition theorem for calculus of
open-circuit voltage
A
R
1
330
R
2
150
V
gol1
V1=5 V
Vgol1 
R2
V 1  1,56 V
R1  R 2
VO  Vgol  Vgol1  Vgol2  7,75 V
B
SUBB-CIRCUIT1
A
R
1
330
Vgol2 
R
2
150
V
gol2
V2=9 V
SUBB-CIRCUIT2
R1
V 2  6,19 V
R1  R 2
B
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The equivalent resistance calculus
 The circuit is passivated.
 A test voltage is applied (VTEST)
 The current through the terminals is determinate (ITEST)
 RO = VTEST / ITEST
ITEST
A
R1
330
I TEST 
R2
150
VTEST
R1 R2
VTEST
RECH
RO  RECH  R1 R 2 
PASSIVATED CIRCUIT
R1 R 2
 103 
R1  R 2
B
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Conclusion
I
A
Ro
CIRCUIT
ECHIVALENT

V
R3
=103
V =7,75
O
R3
R3
1K
V
From the R3 resistance
point of view, the
equivalent circuit will
have the same effect:
I R3
VO
7,75 V


 7 mA
RO  R3 1103 
VR3  R3  I R3  7 mA 1 KΩ  7 V
B
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Norton’s Theorem
Any two-terminal, linear network of sources and
resistances may be replaced by a single current
source in parallel with a resistance.
The value of the current source is the current flowing
between the terminals of the network when they are
short-circuited.
The resistance value is the resistance that would be
measured at the terminals of the network when all
the sources have been replaced by their internal
resistances.
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Norton’s Theorem
Io and Ro must be determined.
I
A
I
R3
R3
A
V
R3
R1
330
R2
150
R3
1K
R3
1K
Ro=?
V
R3
V1=5 V
V2=9 V
I
B
CIRCUIT ELECTRONIC
O
=?
CIRCUIT
ECHIVALENT
B
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Calculus of short-circuit current
 In order to calculate the
short-circuited current, the
Kirchhoff’s theorems can
be applied.
 The Superposition
theorem!
A
R
1
330
R
2
150
I sc
V1=5 V
V2=9 V
ELECTRONIC CIRCUIT
B
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The superposition theorem for calculus of
the short-circuit current
A
R
1
330
R
2
150
Isc1
V1=5 V
I SC1 
V1
 15,15 mA
R1
B
SUBB-CIRCUIT1
I O  I SC  I SC1  I SC2  75,15 mA
A
R
1
330
R
2
150
Isc2
V2=9 V
SUBB-CIRCUIT2
I SC2 
V2
 60 mA
R2
B
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Calculus of equivalent resistance
ITEST
A
R1
330
R2
150
VTEST
RECH
CIRCUIT PASIVIZAT
RO  RECH  R1 R 2 
R1 R 2
 103 
R1  R 2
B
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Conclusion
I
R3
A
R3
1K
Ro =103
V
R3
I
O=75,15
mA
EQUIVALENT
CIRCUIT
 From the R3 resistance
point of view, the
equivalent circuit will
have the same effect:
VR3  R3 RO  I SC
B
I R3 
RO  R3

 I SC  7 V
RO  R3
VR3
 7 mA
R3
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Transfer from Thevenin to Norton
equivalence
 Once having an equivalent circuit (Thevenin or Norton),
the other one is obtained using the relation:
VOThevenin
I ONorton 
RO
 For previous example:
VOThevenin 7,75 V
I ONorton 

 75,15 mA
RO
103 
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Recommendation for individual study
 For the following circuit determine the current through R
resistor and the voltage across it, using:
 Kirchhoff’s theorem
 Thevenin and/or Norton equivalence
(use the superposition theorem)
R3
R4
7K
2K
R
1K
V
9V
I
1 mA
R1
R2
2K
4K
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