1 AC Analysis Using Thevenin's Theorem and Superposition Discussion D11.2 Chapter 4.
14
1 AC Analysis Using Thevenin's Theorem and Superposition Discussion D11.2 Chapter 4
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Transcript of 1 AC Analysis Using Thevenin's Theorem and Superposition Discussion D11.2 Chapter 4.
1
AC Analysis Using Thevenin's Theorem
and Superposition
Discussion D11.2
Chapter 4
2
AC Thevenin's Theorem
3
AC Thevenin's TheoremThevenin’s theorem states that the two circuits given below are equivalent as seen from the load ZL that is the same in both cases.
VTh = Thevenin’s voltage = Vab with ZL disconnected (= ) = the open-circuit voltage = VOC
LinearCircuit
b
a
inZ
I
ZL
b
a
ThVACZL
inZ
IZth
4
Thevenin's Theorem
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.
LinearCircuit
b
a
inZ
I
ZL
b
a
ThVACZL
inZ
IZth
5
Problem 4.57 in text: Solve Problem 4.40 using Thevenin's Thm.
0
8 2 8 2 8.246 14.04
1 1 2 2 3 1 3.162 18.43
j j
j j j
I
6 2(1 ) 8 2OC j j V
2 0 A 6 0 V
AC
-j1
+
-
j2
I0
2 0 A 6 0 V
AC
-j1
+
- VOC
+
-
-j1
1 1TH j Z
8 2j AC
-j1
+
-
j2
I0
0 2.608 32.47 I
6
AC Superposition
7
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.
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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.
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Example
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.
sin cos 90t t Remember that
+
-AC
0.2F 1H 2cos10t30sin 5t+
-
v0(t)
10
Consider first the acting alone.Since, ,we have = 5 and
Example
30sin 5 30cos 5 90t t 30sin 5t
+
-AC
0.2F 1H 2cos10t30sin 5t+
-
v0(t)
+
-AC
-j1 j530 90
+
-
O.C.
1 11
5(0.2)CZ jj C j
5LZ j L j
10V
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Example
+
-AC
-j1 j530 90
+
-
O.C.
1 11
5(0.2)CZ jj C j
5LZ j L j
10V
AC
Z1
Z2VS V0
+
-
+
-
Use voltage division
1 20
1 2S
Z
Z Z
V V
2
( 1)( 5) 51.25
1 5 4
j jZ j
j j j
1 8Z
10
1.25 1.25 9030 90 30 90
8 1.25 8.097 8.881
j
j
V
10 4.631 171.1 V
10 ( ) 4.631cos 5 171.12 4.631sin 5 81.12v t t t
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Now consider first the acting alone.We have = 10 and
Example
2cos10t
+
-AC
0.2F 1H 2cos10t30sin 5t+
-
v0(t)
S
j2 -j/10 2 0 +
-
10(0.2) 2CY j C j j
1 110
10LY jj L j
20V
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Example
For a parallel combination of Y's we have
YV I
1 8 2 0.1 0.125 1.90eq i j j j Y Y
20
2 01.05 86.24
1.904 86.24
V
20 ( ) 1.05cos 10 86.24v t t
Yeq
+
-
20V I
20
IV
Y
S
j2 -j/10 2 0 +
-
10(0.2) 2CY j C j j
1 110
10LY jj L j
20V
1.904 86.24eq Y
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By superposition
Example
+
-AC
0.2F 1H 2cos10t30sin 5t+
-
v0(t)
10 ( ) 4.631sin 5 81.12v t t
20 ( ) 1.05cos 10 86.24v t t
0 ( ) 4.631sin 5 81.12 1.05cos 10 86.24v t t t
1 20 0 0( ) ( ) ( )v t v t v t
