inc 111 basic circuit analysis week 5 thevenin’s theorem
TRANSCRIPT
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INC 111 Basic Circuit Analysis
Week 5
Thevenin’s Theorem
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Special Techniques
• Superposition Theorem
• Thevenin’s Theorem
• Norton’s Theorem
• Source Transformation
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Linearity Characteristic
RL
6Ω
42V
4Ω
10V
IL+
VRL
-
If RL change its value , how will it effect the current and voltage across it?
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I
VVOC
ISC
For any circuit constructed from only linear components
Not just RL, all resistors have this property.
Voc = Voltage open-circuit Isc = Current short-circuit
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Thevenin’s TheoremWhen we are interested in current and voltage across RL, we can simplify other parts in the circuit.
RL
6Ω
42V
4Ω
10V
RL
RTH
VTH
Equivalent circuit
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RL
6Ω
42V
4Ω
10V
RL
RTH
VTH
I
V
VOC
ISC
Voc = Voltage open-circuitIsc = Current short-circuitRth = Rth equivalent
Slope =thR
1
RLchangevalue
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Thevenin’s Equivalent Circuit
RL
RTH
VTH
Thevenin’sequivalentcircuit
VTH = Voc (by removing RL and find the voltage difference between 2 pins)
RTH (by looking into the opened connections that we remove RL, see how much resistance from the connections. If we see a voltage source, we short circuit. If we see a current source, we open circuit.)
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Why do we needequivalent circuit?
• To analyze a circuit with several values of RL
• For circuit simplification (source transformation)
• To find RL that gives maximum power (maximum power transfer theorem)
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Procedure
1. Remove RL from the circuit
2. Find voltage difference of the 2 opened connections. Let it equal VTH.
3. From step 2 find RTH by3.1 short-circuit voltage sources3.2 open-circuit current sources3.3 Look into the 2 opened connections. Find equivalent resistance.
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Example
3Ω
2Ω
10V
10Ω
RL
2Ω
Find Thevenin’s equivalent circuitand find the current that passes through RL when RL = 1Ω
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3Ω
2Ω
10V
10Ω
2Ω
0V
10V
0V 0V
6V 6V
VVTH 61032
3
Find VTH
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3Ω
2Ω
10V
10Ω
2Ω
3Ω
2Ω 10Ω
2Ω
Short voltage source
RTH
2.13
232
3210
23||210THR
Find RTH
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13.2Ω
6V RL
Thevenin’s equivalent circuit
If RL = 1Ω, the current is A423.012.13
6
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Example
3Ω
2Ω 10Ω
RL
2Ω
1A
Find Thevenin’s equivalent circuit
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Find VTH
3Ω
2Ω 10Ω
2Ω
1A
0V
5V
0V 0V
3V 3V
VVTH 331
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3Ω
2Ω 10Ω
2Ω
Open circuitcurrent source
RTH
15
2310THR
Find RTH
3Ω
2Ω 10Ω
2Ω
1A
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15Ω
3V RL
Thevenin’s equivalent circuit
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R3=4K
R2=8K
R1=2K
R4=1K
RL=1K10V
+ -
Example: Bridge circuit
Find Thevenin’s equivalent circuit
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Find VTH
R3=4K
R2=8K
R1=2K
R4=1K
10V
0V
10V
8V 2V
VTH = 8-2 = 6V
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Find RTH
R3=4K
R2=8K
R1=2K
R4=1K
RTH
R3=4K
R2=8K
R1=2K
R4=1K
R3=4K
R2=8K
R1=2K
R4=1K
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R3=4K
R2=8K
R1=2K
R4=1K
KKK
KKKKRTH4.28.06.1
1||48||2
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2.4K
6V RL
Thevenin’s equivalent circuit
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Special Techniques
• Superposition Theorem
• Thevenin’s Theorem
• Norton’s Theorem
• Source Transformation
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I
VVOC
ISC
For any point in linear circuit
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Thevenin’s Equivalent Circuit
RL
RTH
VTH
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Norton’s Equivalent Circuit
RLRNIN
In= Isc from replacing RL with an electric wire (resistance = 0) and find the currentRn = RTH (by looking into the opened connections that we remove RL, see how much resistance from the connections. If we see a voltage source, we short circuit. If we see a current source, we open circuit.)
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Example
3Ω
2Ω
10V
10Ω
RL
2Ω
Find Norton’s equivalent circuitand find the current that passes through RL when RL = 1Ω
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3Ω
2Ω
10V
10Ω
2Ω
Isc
Find In
Find R total
Find I total
Current divider
4.4123
1232)210(||32
AR
VI 27.2
4.4
10
AISC 45.027.2123
3
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3Ω
2Ω
10V
10Ω
2Ω
3Ω
2Ω 10Ω
2Ω
Short voltage source
RTH
2.13
232
3210
23||210THR
Find Rn
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Norton’s equivalent circuit
If RL = 1Ω, the current is A418.045.012.13
2.13
RL13.20.45
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Relationship BetweenThevenin’s and Norton’s Circuit
THNTH
NTH
RIV
RR
I
VVOC
ISC
Slope = - 1/Rth
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RL13.20.45
Norton’s equivalent circuit
13.2
6V RL
Thevenin’s equivalent circuit
Same R value
2.1345.06 THNTH
NTH
RIV
RR
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Example
3Ω
2Ω 10Ω
RL
2Ω
1A
Find Norton’s equivalent circuit
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Find In
3Ω
2Ω 10Ω
2Ω
1A Isc
Current divider AISC 2.01123
3
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3Ω
2Ω 10Ω
2Ω
Open circuitcurrent source
RTH
15
2310THR
Find RTH
3Ω
2Ω 10Ω
2Ω
1A
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Norton’s equivalent circuit
RL150.2
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15
3V RL
Thevenin’s equivalent circuitNorton’s equivalent circuit
RL150.2
0.2 x 15 = 3
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Equivalent Circuits withDependent Sources
We cannot find Rth in circuits with dependent sources usingthe total resistance method.
But we can use
SC
OCTH I
VR
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Example
1V 4K
2K
80
250
RL+
Vx- -
+ 100Vx
+
-
Find Thevenin and Norton’s equivalent circuit
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1V 4K
2K
80
250
+Vx- -
+ 100Vx
+
-
Find Voc
I1 I2
12400014250
0)21(400012501
II
IIIKVLloop1
024060801404000
)21(4000
010028022000)12(4000
II
IIVx
VxIIIIKVLloop2
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1V 4K
2K
80
250
+Vx- -
+ 100Vx
+
-
I1 I2
Solve equations
I1 = 3.697mA I2 = 3.678mA
V
mA
IIIVxIVOC
3.7
)678.3697.3(400000)678.3(80
)21(400000280100280
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1V 4K
2K
80
250
+Vx- -
+ 100Vx
Isc
Find Isc
I1 I2 I3
12400014250
0)21(400012501
II
IIIKVLloop1
038024060801404000
)21(4000
0100)32(8022000)12(4000
III
IIVx
VxIIIIIKVLloop2
KVLloop3
038024000801400000
0100)23(80
III
VxII
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1V 4K
2K
80
250
+Vx- -
+ 100Vx
Isc
Find Isc
I1 I2 I3
I1 = 0.632mAI2 = 0.421mAI3 = -1.052 A
Isc = I3 = -1.052 A
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94.6052.1
28.7
SC
OCTH I
VR
6.94
-7.28V RL
Thevenin’s equivalent circuit Norton’s equivalent circuit
RL6.94-1.052