circuit analysis methods chapter 3 mdm shahadah ahmad
TRANSCRIPT
![Page 1: CIRCUIT ANALYSIS METHODS Chapter 3 Mdm shahadah ahmad](https://reader036.vdocuments.us/reader036/viewer/2022062517/56649f2a5503460f94c44c45/html5/thumbnails/1.jpg)
CIRCUIT ANALYSIS METHODS
Chapter 3
Mdm shahadah ahmad
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CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS
• Node-Voltage method• Mesh-current method• Source transformation• Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle
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INTRODUCTION OF NODE-INTRODUCTION OF NODE-VOLTAGE METHODVOLTAGE METHODINTRODUCTION OF NODE-INTRODUCTION OF NODE-VOLTAGE METHODVOLTAGE METHOD
• Use KCL.
• Important step: select one of the node as reference node
• Then define the node voltage in the circuit diagram.
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Node-voltage example
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• In the diagram, node 3 is define as reference node and node 1 and 2 as node voltage V1 and V2.
• The node-voltage equation for node 1 is,
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• In the diagram, node 3 is define as reference node and node 1 and 2 as node voltage V1 and V2.
• The node-voltage equation for node 1 is,
251
100 2111 VVVV
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• Node-voltage equation of node 2,
2102
0 212
VVV
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• Solving for V1 and V2 yeilds
VV
VV
91.1011
120
09.911
100
2
1
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THE NODE-VOLTAGE METHOD AND DEPENDENT SOURCES
• If the circuit contains dependent sources, the node-voltage equations must be supplemented with the constraint equation imposed by the presence of the dependent sources.
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example…
Use the node-voltage method to find the power dissipated in the 5Ω resistor.
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• The circuit has 3 node. • Thus there must be 2 node-voltage
equation.• Summing the currents away from node 1
generates the equation,
05202
20 2111
VVVV
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• Summing the current away from node 2 yields,
02
8
1052212
iVVVV
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• As written, these two equations contain three unknowns namely V1, V2 and iØ.
• To eliminate iØ, express the current in terms of node-voltage,
521 VV
i
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• Substituting this relationship into the node 2 equation,
06.1
102.075.0
21
21
VV
VV
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• Solving for V1 and V2 gives,
VV 161 VV 102
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• Then,
Ai 2.15
1016
W
Rip
2.7
544.12
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SPECIAL CASE
• When a voltage source is the only element between two essential nodes, the node-voltage method is simplified.
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Example…
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• There is three essential nodes, so two simultaneous equation are needed.
• Only one unknown node voltage, V2 where as V1=100V.
• Therefore, only a single node-voltage equation is needed which is at node 2.
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055010
212 VVV
Using V1 =100V, thus V2=125V.
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SUPERNODESUPERNODE
• When a voltage source is between two essential nodes, those nodes can be combine to form a supernode (voltage sourse is assume as open circuit).
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Supernode example…
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• Nodes chosen,
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• Node-voltage equation for node 2 and 3,
0505
212
iVVV
04100
3 iV
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• Summing both equation,
04100505
3212 VVVV
Above equation can be generates directly using supernode approach
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Supernod
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• Starting with resistor 5Ω branch and moving counterclockwise around the supernode,
04100505
3212 VVVV
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• Using V1 =50V and V3 as a function of V2,
iVV 1023
5
502 V
i
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• Substituded into the node-voltage equation,
1410500
10
100
1
5
1
50
12
V
VV
V
60
15)25.0(
2
2
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• Using V2 value, gives
Ai 25
5060
VV 8020603
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CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS
• Node-Voltage method• Mesh-current method• Source transformation • Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle
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INTRODUCTION OF MESH-INTRODUCTION OF MESH-CURRENT METHODCURRENT METHOD
• A mesh is a loop with no loop inside it.
• A mesh current is the current that exist only in the perimeter of a mesh.
• Mesh-current method use KVL to generates equation for each mesh.
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Mesh-current example…
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• Mesh-current circuit with mesh current ia and ib.
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• Use KVL on both mesh,
311 RiiRiV baa
232 RiRiiV bab
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• Solving for ia and ib, and you can compute any voltages or powers of interest.
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THE MESH-CURRENT METHOD AND DEPENDENT SOURCES
• If the circuit contains dependent sources, the mesh-current equations must be supplemented by the appropriate constraint equations.
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Example…
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• Use the mesh-current method to determine the power dissipated in the 4Ω resistor.
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• Using KVL,
iiiii
iiiii
iiii
154200
4150
20550
2313
32212
3121
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• But
• Substituting into the mesh-current equation,
31 iii
321
321
321
9450
41050
2052550
iii
iii
iii
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• Using Cramer rule, the values of i2 and i3 can be determine,
945
4105
20525
905
405
205025
2i
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45
5254
95
202510
94
2055
95
4550
2i
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Ai
i
26125
32505001250625
3250
)125(4)125(10)125(5
)65(50
2
2
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A
i
28125
3500125
45
10550
125
045
0105
50525
3
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• Power dissipated by 4Ω resistor is
W
Rip
16
4)2628( 2
2
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SPECIAL CASE (SUPERMESH)
• When a branch includes a current source, the mesh-current method can be simplified.
• To create a supermesh, remove the current source from the circuit by simply avoiding the branch when writing the mesh-current equations.
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• Supermesh equation,
06450
23100
ac
bcba
ii
iiii
cba iii 65950
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• Mesh 2 equation,
cb
bab
ii
iii
2
1030
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• From the circuit,
ic –ia= 5A• Using Cramer rule, the three
mesh current can be obtain.
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CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS
• Node-Voltage method• Mesh-current method• Source transformation • Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle
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SOURCE SOURCE TRANSFORMATION TRANSFORMATION
• Source transformation allows a voltage source in series with a resistor to be replaced by a current source in parallel with the same resistor or vice versa.
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Sorce transformation
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Example…
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• Source transformation procedure
s
ss R
VI
sp RR
From To methodUse,
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pss RIV
ps RR
From To method
Use,
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CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS
• Node-Voltage method• Mesh-current method• Source transformation • Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle
![Page 60: CIRCUIT ANALYSIS METHODS Chapter 3 Mdm shahadah ahmad](https://reader036.vdocuments.us/reader036/viewer/2022062517/56649f2a5503460f94c44c45/html5/thumbnails/60.jpg)
THEVENIN EQUIVALENT THEVENIN EQUIVALENT CIRCUITCIRCUIT
• Thevenin equivalent circuit consist of an independent voltage source, VTh in series with a resistor RTh.
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Thevenin equivalent circuit
ThV
ThRa
b
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• Thevenin voltage, VTh = open circuit voltage in the original circuit.
• Thevenin resistance, RTh is the ratio of open-circuit voltage to the short-circuit current.
sc
ThTh
Th
Thsc i
VR
R
Vi
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Example…
V25
5a
b
20
4
A3
1V
abV
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• Step 1: node-voltage equation for open-circuit:
ThVVV
VV
32
03205
25
1
11
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• Step 2: short-circuit condition at terminal a-b
V25
5 a
b
20
4
A3
2V scI
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• Node-voltage equation for short-circuit:
VV
VVV
16
04
3205
25
2
222
![Page 67: CIRCUIT ANALYSIS METHODS Chapter 3 Mdm shahadah ahmad](https://reader036.vdocuments.us/reader036/viewer/2022062517/56649f2a5503460f94c44c45/html5/thumbnails/67.jpg)
84
32
sc
ThTh I
VR
AI sc 44
16
Short-circuit current:
Thevenin resistance:
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Thevenin equivalent circuit
V32
8a
b
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CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS
• Node-Voltage method• Mesh-current method• Source transformation• Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle
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NORTON EQUIVALENT NORTON EQUIVALENT CIRCUITCIRCUIT
• A Norton equivalent circuit consists of an independent current source in parallel with the Norton equivalent resistance.
• Can be derive from a Thevenin equivalent circuit simply by making a source transformation.
• Norton current, IN = the short-circuit current at the terminal of interest.
• Norton resistance, RN = Thevenin resistance, RTh
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Example
V25
5a
b
20
4
A3
![Page 72: CIRCUIT ANALYSIS METHODS Chapter 3 Mdm shahadah ahmad](https://reader036.vdocuments.us/reader036/viewer/2022062517/56649f2a5503460f94c44c45/html5/thumbnails/72.jpg)
Step 1: Source transformation
A5 5
a
b
20
4
A3
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Step 2: Parallel sources and parallel resistors combined
a
b
4
A8 4
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Step 3: Source transformation, series resistors combined, producing the Thevenin equivalent circuit
V32
8a
b
THEVENINEQUIVALENT
CIRCUIT
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Step 4: Source transformation, producing the Norton equivalent circuit
a
b
A4 8
NORTONEQUIVALENT
CIRCUIT
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CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS
• Node-Voltage method• Mesh-current method• Source transformation• Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle
![Page 77: CIRCUIT ANALYSIS METHODS Chapter 3 Mdm shahadah ahmad](https://reader036.vdocuments.us/reader036/viewer/2022062517/56649f2a5503460f94c44c45/html5/thumbnails/77.jpg)
MAXIMUM POWER MAXIMUM POWER TRANSFERTRANSFER
• Two basic types of system:– Emphasizes the efficiency of the power
transfer– Emphasizes the amount of power
transferred.
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• Maximum power transfer is a technique for calculating the maximum value of p that can be delivered to a load, RL.
• Maximum power transfer occurs when RL=RTh.
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Example…Example…
ThV
ThR a
b
LRi
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• Power dissipated by resistor RL
LLTh
Th
L
RRR
V
Rip2
2
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• Derivative of p with repect to RL
4
22 2
LTh
LThLLThTh
L RR
RRRRRV
dR
dp
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• Derivative is zero and p is maximum when
)(22LThLLTh RRRRR
LTh RR
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• The maximum power transfer occurs when the load resistance, RL = RTh
• Maximum pwer transfer delivered to RL:
L
Th
L
LTh
R
V
R
RVp
42
2
2
2
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CIRCUIT ANALYSIS METHODSCIRCUIT ANALYSIS METHODS
• Node-Voltage method• Mesh-current method• Source transformation• Thevenin equivalent circuit• Norton equivalent circuit• Maximum power transfer• Superposition principle
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PRINSIP PRINSIP SUPERPOSISISUPERPOSISIPRINSIP PRINSIP SUPERPOSISISUPERPOSISI
• In a circuit with multiple independent sources, superposition allows us to activate one source at a time and sum the resulting voltages and currents to determine the voltages and currents that exist when all independent sources are activate.
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Step of Superposition principle
1. Deactivated all the sources and only remain one source at one time. Do circuit analysis to find voltages or currents.
2. Repeat step 1 for each independent sources.
3. Sum the resulting voltages or currents.
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1. Independent voltage source will become short-circuit with 0Ω resistance.
2. Independent current source will become open-circuit.
3. Dependent sources are never deactivated when applying superposition.
REMEMBER!!!
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Example…Example…
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• Step 1: deactivated all sources except
voltage source
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• V0 is calculated using voltage divider:
Vk
kV 54
1020
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• Step 2: Deactivated all sources except
current source
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• V0 is calculated by using current divider:
VkmV
mAmk
ki
2)2)(1(
1)2(4
2
0
0
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V0 =2+5=7V.
• Step 3: Sum all the resulting voltages:
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Question 1 (node-voltage)
• Calculate the value of Io
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Solution
• Node 1:
622
3
4221
21
211
VV
VVV
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• Node 2:
44
5
2
44
1
2
1
2
1
2
4422
21
21
2212
VV
VV
VVVV
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846.1625.1
36
4
6
45
21
21
23
21
23
2
V
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AI 923.02
846.10
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Question 2 (mesh-current)
• Determine the value of currents, I1, I2 and I3.
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• Supermesh:
• Mesh 3:
0)(510 321 III
125510
012555
23
233
II
III
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• Dependent current source
• Vo
021 2VII
)(5 320 IIV
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• Substitute V0
01011
)(10
321
3221
III
IIII
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• Use Cramer rule
10111
1050
5510
10110
105125
550
1I
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A1625
625
111
505
101
1005
1011
10510
1011
55125
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• Current I2:
A
I
21625
13125625
101
510125
625
1001
101250
5010
2
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• Current I3:
A
I
23625
14375625
101
510125
625
0111
12550
0510
3
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Question 3 (thevenin)
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• Open-circuit voltage, Voc:
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• Node-voltage equation for Voc
VV
V
VV
VV
oc
oc
ococ
ococ
10
202
0424
0222
24
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• Thevenin resistance, RTh:
5422THR
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• Thevenin equivalent circuit:
VV 88.6)10(16
110
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Question 4 (norton)
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• Open-circuit current, Isc:
AI sc 64
123
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• Norton resistance, RN:
RN = 4Ω
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• Norton equivalent circuit:
VV 18)3(612460
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Question 5 (superposition)
• Use superposition principle to determine the voltage Vo.
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• Deactivated current source
V412
224V0
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• Deactivated voltage source
V4212
46)2(iV oo
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• Summing the voltage V0
V8VV 00
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Question 6 (node-voltage)
• Determine the value of Vo.
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node-voltage equation:
020
80
10
5
2003 000
ViVV
20
800 V
iCurrent iΔ:
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• Thus:
V0 =50V