Transcript Basic Electronics Ninth Edition Grob Schultz ©2002 The McGraw-Hill Companies Basic Electronics Ninth Edition CHAPTER Network Theorems ©2003 The McGraw-Hill Companies Topics Covered in Chapter 10  Superposition Method  Thevenin’s Theorem  Norton’s.

Basic Electronics
Ninth Edition
Grob
Schultz
©2002 The McGraw-Hill Companies
Basic Electronics
Ninth Edition
CHAPTER
10
Network
Theorems
©2003 The McGraw-Hill Companies
Topics Covered in Chapter 10
 Superposition Method
 Thevenin’s Theorem
 Norton’s Theorem
Topics Covered in Chapter 10
 Conversion of Voltage and
Current Sources
 Millman’s Theorem
 D and Y Conversions
Superposition Theorem
• In a linear, bilateral network that has
more than one source, the current and
voltage in any part of the network can be
found by adding algebraically the effect
of each source separately.
• This analysis is done by:
shorting each voltage source in turn
opening each current source in turn
Superposition Method Applied
15 V
V1
15 V
V1
R1
R2
100 W
20 W
10 W
R3
R1
R2
100 W
20 W
10 W
R3
13 V
V2
V2 shorted
REQ = 106.7 W, IT = 0.141 A and IR3 = 0.094 A
Superposition Method Applied
15 V
V1
V1 shorted
R1
R2
100 W
20 W
10 W
R3
R1
R2
100 W
20 W
10 W
R3
13 V
V2
13 V
V2
REQ = 29.09 W, IT = 0.447 A and IR3 = 0.406 A
Superposition Method Applied
15 V
V1
R1
R2
100 W
20 W
13 V
V2
0.094 A 0.406 A
With V2 shorted
REQ = 106.7 W, IT = 0.141 A and IR3 = 0.094 A
With V1 shorted
REQ = 29.09 W, IT = 0.447 A and IR3 = 0.406 A
Adding the currents gives IR3 = 0.5 A
Checking the Superposition Solution
15 V
V1
R1
R2
100 W
20 W
10 W
R3
0.5 A
13 V
V2
With 0.5 A flowing in R3, the voltage across R3 must
be 5 V (Ohm’s Law). The voltage across R1 must
therefore be 10 volts (KVL) and the voltage across R2
must be 8 volts (KVL). Solving for the currents in R1
and R2 will verify that the solution agrees with KCL.
IR1 = 0.1 A and IR2 = 0.4 A
IR3 = 0.1 A + 0.4 A = 0.5 A
Thevenin’s Theorem
• Any network with two open terminals
can be replaced by a single voltage
source (VTH) and a series resistance (RTH)
connected to the open terminals. A
component can be removed to produce
the open terminals.
A Wheatstone bridge can be
Thevenized.
30 W
45 V
40 W
R3
R1
A R5 B
R 40 W R
60 W
4
2
40 W
Problem: Find the voltage drop across R5.
• The bridge is unbalanced and Thevenin’s theorem
is a good choice.
• R5 will be removed in this procedure making A
and B the Thevenin terminals.
Determining Thevenin
Resistance and Voltage
• RTH is determined by shorting the voltage
source and calculating the circuit’s total
resistance as seen from open terminals A
and B.
• VTH is determined by calculating the
voltage between open terminals A and B.
Applying Thevenin’s Theorem
30 W
40 W
R5
R440 W R2
60 W
40 W
30 W
60 W
A
B
RTH
40 W
45 V
A
R1
B
40 W
R3
R5 is removed. The voltage source is shorted.
RTH is the resistance from A to B.
RTH = 40 W
Applying Thevenin’s Theorem
30 W
40 W
R3
45 V
A
R5
R1
B
R440 W R2
60 W
60
x 45 = 30 V
VA =
90
VB =
40 W
40
80
x 45 = 22.5 V
VAB = 30 – 22.5 = 7.5 V
Choose a reference point.
Remove R5.
VTH is the voltage from A to B. VTH = 7.5 V
Applying Thevenin’s Theorem
RTH
VTH
7.5 V
40 W
A
R5
40 W
B
The Thevenin equivalent circuit
Connect R5 to the equivalent circuit.
40
VR5 =
x 7.5 = 3.75 V
80
Norton’s Theorem
Any network with two terminals can be replaced
by a single current source and parallel resistance
connected across the terminals.
• The two terminals are usually labeled
something such as A and B
• The Norton current is usually labeled IN
• The Norton resistance is usually labeled RN
Determining Norton
Current and Voltage
• IN is determined by calculating the
current through a short placed across
terminals A and B.
• RN is determined by shorting the voltage
source and calculating the circuit’s total
resistance as seen from open terminals A
and B (same procedure as for RTH).
A Wheatstone bridge can be
Nortonized.
IT
IR1
30 W
40 W
R3
A
45 V
R1
R5 I B
N
IT
IT = 1.094 A
IR2 = 0.6563 A
R440 W R2
60 W
REQ = 41.14 W
40 W
IR1 = 0.4689 A
IN = 0.1874 A
IR2
• Replace R5 with a short and determine IN.
• Apply the current divider.
• Apply KCL.
RN = RTH
The Norton Equivalent Circuit
A
IN
0.1874 A
RN
40 W
R5
40 W
B
IR5 = 0.0937 A
VR5 = 3.75 V
The circle is a symbol for a current source. It provides
0.1874 A total flow, regardless of what is connected
across it. With no load, all of the current will flow in
RN. When shorted, all of the current will flow in the
short.
• Connect R5.
• Apply the current divider.
• Use Ohm’s Law.
Conversion of Real Sources
R
40 W
V
7.5 V
A
7.5 V
40 W
A
I
0.1874 A
R
40 W
B
B
0.1874 A x 40 W
• All real voltage sources have a series resistance.
• All real current sources have a parallel resistance.
• Use Ohm’s Law to convert from one to the other.
• The resistance stays the same.
Millman’s Theorem
• The common voltage across parallel
branches with different voltage sources
can be determined by:
VXY
V1 V2 V3


R1 R 2 R 3

 etc
1
1
1


R1 R 2 R 3
Delta-to-Wye Conversion
A delta (D) circuit can be converted to a wye
(Y) equivalent circuit by:
R BR C
R1 
RA  RB  RC
R CR A
R2 
RA  RB  RC
R AR B
R3 
RA  RB  RC
RA
R2
RC
RB
R3
R1
A Wheatstone bridge can be simplified.
40 W
R3
45 V
A
R5
R1
B
R440 W R2
60 W
10.91 W
40 W
14.55 W
REQ = 41.74 W
Apply current divider here.
Apply Ohm’s Law.
0.4687 A
45 V
10.91 W
A
60 W
VA = 28.12 and VB = 24.37
VAB = 3.75 V
IT = 1.078 A
0.6093 A
30 W
It’s possible to
convert this
delta section to
an equivalent wye.
B
40 W
Wye-to-delta Conversion
A wye (Y) circuit can be converted to
a delta (D) equivalent circuit by:
R 1R 2  R 2R 3  R 1R 3
RA 
R1
R 1R 2  R 2R 3  R 1R 3
RB 
R2
RC 
R 1R 2  R 2R 3  R 1R 3
R3
RA
R2
R3
R1
RC
RB