Transcript Basic Concepts - Oakland University

AC Analysis Using
Thevenin's Theorem
and Superposition
Discussion D11.2
Chapter 4
1
AC Thevenin's Theorem
2
AC Thevenin's Theorem
Thevenin’s theorem states that the two circuits given below are
equivalent as seen from the load ZL that is the same in both cases.
I
a
Zth
Linear
Circuit
ZL
b
AC
I
a
ZL
VTh
b
Zin
Zin
VTh = Thevenin’s voltage = Vab with ZL disconnected (= ) = the
open-circuit voltage = VOC
3
Thevenin's Theorem
I
a
Zth
Linear
Circuit
ZL
b
AC
I
a
ZL
VTh
b
Zin
Zin
ZTh = Thevenin’s impedance = the input impedance with all
independent sources turned off (voltage sources replaced by short
circuits and current sources replaced by open circuits). This is the
impedance seen at the terminals ab when all independent sources
are turned off.
4
Problem 4.57 in text: Solve Problem 4.40 using Thevenin's Thm.

-j1
+

j2
+
AC
60 V
-j1
20 A
-
+
VOC
-
AC
60 V
-

20 A
I0

+
8  j2
VOC  6  2(1  j )  8  j 2
-j1
j2

AC
-j1

I0
I0 
ZTH  1  j1
8  j2
8  j 2 8.246  14.04


1  j1  j 2  2 3  j1
3.16218.43
I0  2.608 32.47
5
AC Superposition
6
Superposition Principle
Because the circuit is linear we can find the response of the
circuit to each source acting alone, and then add them up to find
the response of the circuit to all sources acting together. This is
known as the superposition principle.
The superposition principle states that the voltage across (or
the current through) an element in a linear circuit is the
algebraic sum of the voltages across (or currents through) that
element due to each independent source acting alone.
7
Steps in Applying the Superposition Principle
1. Turn off all independent sources except
one. Find the output (voltage or current)
due to the active source.
2. Repeat step 1 for each of the other
independent sources.
3. Find the total output by adding
algebraically all of the results found in
steps 1 & 2 above.
In some cases, but certainly not all, superposition can simplify
the analysis.
8
Example

30sin 5t
+
+
v0(t)
-
-
AC
0.2F
2 cos10t
1H
Note that the voltage source and the current source have two
different frequencies. Thus, if we want to use phasors, the
only way we've solved sinusoidal steady-state problems, we
MUST use superposition to solve this problem. We will
consider each source acting alone, and then find v0(t) by
superposition.
Remember that sin t  cos t  90

9

Example
30sin 5t
+
+
v0(t)
-
-
AC
0.2F
2 cos10t
1H
Consider first the 30sin 5t acting alone.
Since, 30sin 5t  30 cos  5t  90  ,we have  = 5 and
ZC 

30  90 AC
1
1

  j1
jC j 5(0.2)
+
+
V01
-
-
-j1
j5
O.C.
Z L  j L  j 5
10
Example

Use voltage division
30  90 AC
Z1
+
VS AC
ZC 
1
1

  j1
jC j 5(0.2)
+
+
V01
-
-
-j1
j5
Z L  j L  j 5
+
V0
Z2
-
-
O.C.
( j1)( j 5) 5
Z2 

  j1.25
 j1  j5
j4
Z2
V 
VS
Z1  Z 2
1
0
Z1  8
 j1.25
1.25  90
V 
30  90 
30  90
8  j1.25
8.097  8.881
1
0




V01  4.631171.1



v01 (t )  4.631cos 5t  171.12  4.631sin 5t  81.12

11

Example
30sin 5t
+
+
v0(t)
-
-
AC
0.2F
2 cos10t
1H
Now consider first the 2 cos10t acting alone.
We have  = 10 and
S
YC  jC  j10(0.2)  j 2
+
V02
j2
20
-j/10
-
YL 
1
j L

1
  j 10
j10
12
Example
S
+
2
0
V
Yeq
YC  jC  j10(0.2)  j 2
+
I
V02
-
j2
20
-j/10
-
I
V 
YV  I
Y
For a parallel combination of Y's we have
2
0
YL 
1
j L

1
  j 10
j10
Yeq   Yi  1 8  j 2  j0.1  0.125  j1.90
Yeq  1.90486.24
V02 
20
 1.05  86.24
1.90486.24

v02 (t )  1.05cos 10t  86.24

13

Example
30sin 5t
+
+
v0(t)
-
-
AC
0.2F

v01 (t )  4.631sin 5t  81.12

v02 (t )  1.05cos 10t  86.24
By superposition
2 cos10t
1H


v0 (t )  v01 (t )  v02 (t )



v0 (t )  4.631sin 5t  81.12  1.05cos 10t  86.24

14