Chapter 1tl
Transcript of Chapter 1tl
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Circuit Theorems
Chapter 1 :
Resistive circuit analysis techniques
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Nodal analysis (revision)
Based on Kirchhoff Current Law (KCL).
Every point at the junction can be used as a
node. One of the nodes will be acting as a
reference node where this reference node
voltage will always equal to zero, V = 0.
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Nodal analysis for
circuit with currentsource
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Example
R1 R3
R2
IB2IB1
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a) Label all the nodes including reference node
and all the currents flow in the circuit.
R2
R3R1
I1 I2 I5
I3 I4
IB2IB1
a b
c
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b) Write down KCL equation at node a and b.
KCL at node a: I1= I2 +I3 (1)
12
1
R
V
R
VVIaba
B
!
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KCL at node b: I2 + I5 = I4 (2)
Equations (1) and (2) can be solved by usingsimultaneous equations or Cramer rule to
obtain the value of Va and Vb
32
2
R
VI
R
VV bba!
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Nodal analysis for
circuit with current andvoltage source
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Example
I1 I2
I3
I5
I4
R1 R3
R2
IBVB
a b
c
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a) Nodal analysis equation for the above circuit.
Node a: Va = VB (1)
Node b: (2)32 R
VIR
VV bbba
!
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a) Substitute equation (1) into equation (2).
IB = -VB(1/R2) + VB((1/R2) + (1/R3)) (3)
b) From equation (3), the value of Vb can be
obtained.
))/1()/1((
))/1((
22
2
RR
RVIV
BB
b
!
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Supernode
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Occur when the existence of voltage
source between two nonreference nodesand any elements connected parallel with
it.
The first step is to write a KCL equation
for the nodes that being unite and thisequation is known as supernode equation.
Write down the equation that linked the
voltage for the unite node with the voltagesource. This equation is known as
supporting equation.
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supernode
I1
I2 I3
I4
IB2IB1
a
c
bVB
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Node a and b.
a) Supporting equation,
Va Vb = VB
b) KCL equation at supernode,
Ientering supernode = Iout of supernode
I1 +I4 = I2 +I3
IB1 +IB2 = Va /R1 + Vb /R2
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Nodal analysis for the
circuit with thedependent source
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For the circuit that exist dependent
sources, the nodal analysis can be doneby the same method. The solution for the
circuit that deal with this dependent
sources are shown as in example.
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Example
2 6
3
9V16V
Ia
6Ia
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a) Label all the nodes (including reference node)
and the currents of each branch.
2 6
3
9V16V
Ia
6Ia
I1
I2
I3
b
a
Vb = 0
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b) Write down KCL equation at node a.
((16 Va) /2) + ((6Ia - Va) /6) = ((Va +9) /3)
6Va 6Ia = 30 (1)
Where Ia = I1 = ((16 - Va) /2) (2)
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c) Solve the equations (1) and (2).
Va = 8.67 V
Ia = ((16 - Va) /2) = ((16 8.67) /2) = 3.67 A
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MESH
ANALYSIS(revision)
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Step to Determine Mesh Currents :
1. Assign mesh currents i1, i2, i3.in to the
n meshes
2. Apply KVL to each of the n meshes. Use
Ohms law to express the voltages in
terms of the mesh currents.
3. Solve the resulting simultaneous
equations to get the mesh currents.
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Mesh Analysis for circuit with voltage
source
Figure 1 : Circuit with 2 mesh and 1 voltage source
1
1
2
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Solutions :-
a) Assign mesh currents i1 and i2 in to the 2 meshes
1
1
2
I1
I1 I2
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22211
22111
21
)(
)(
RIRRI
RIIRI
VVVB
!
!
!
b) Apply KVL to each of the n meshes. Use Ohms law
to express the voltages in terms of the mesh
currents.
mesh 1 :
mesh 2 :
0)(
0)(
0
32221
32221
34
!
!
!
RRIRI
RIRII
VV
c) Solve the resulting simultaneous equations to get the mesh
currents.
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Figure 2 : Circuit with 3 mesh and 2 voltage source
R5
R4 2V
R1 R3
R2
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R5
R4 V2V1
R1 R3
R2
mesh 1 :
mesh 2 :
mesh 3 :
122211
122111
)(
)(
VRIRRI
VRIIRI
!
!
0)(
0)()(
43432221
43232212
!
!
RIRRRIRI
RIIRIRII
254342
253423
)(
)(
VRRIRI
VRIRII
!
!
c) Solve the resulting simultaneous equations to get the mesh
currents.
I1 I2 I3
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Mesh Analysis for circuit with
Voltage and Current Source
Applying mesh analysis to circuits
containing current sources ( dependent or
independent ) , the presence of the
current sources reduces the numberof
equations.
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Example :
Find current that flow in 3 resistor for the circuit below :
1 V 1
5
3
2
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Solutions :
a) Assign mesh currents i1 and i2 in to the 2 meshes
18V 1A
5
3
2
I1 I2
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b) Apply KVL to each of the n meshes. Use Ohms law to express the
voltages in terms of the mesh currents.
Mesh 1 :
Mesh 2 : amp
c) Solve the resulting simultaneous equations to get the mesh currents I1
I1 = 1.875 amp ; I3 = (I1-I2)= 2.875 amp
1
1838
2
21
!
!
I
II
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Supermesh
A supermesh results when two meshes have a dependent or
independent current source between them.
We have to treat it as supermesh because mesh analysis
applies KVL, which wedo
no
t kno
w the vo
ltage acro
ss acurrent source in advance.
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Figure below shows a current source ( IB ) exists between mesh 1
and mesh 2, therefore supermesh analysis is needed to solve the
circuit..
Mesh 1 & 2 :
Supermesh equation : R1I1 +R3I2 = VB1 + VB2
Supporting equation : I1 I2 = IB
R2
R1 R3
VB1 VB2
IB
I1 I2
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b) Write KVL equations for each mesh.
Mesh 1 : I1 = 2A (1)
Mesh 2 and 3 :
Supporting equation : ( I3 I2) = 5A (2)
Supermesh equation : -5I1 +4I2 +4I3 = 38 (3)
c) Solve equation (1), (2) dan (3) to get the value ofI2 and I3
I2 = 3.5 A ; I3 = 8.5 A
d) From circuit above :
V3 = 4 ( I2 I1 ) = 6V
Power absorb by the 4 resistor, P4 = = 9 W
@
@
4
2
4;V@
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Mesh Analysis for circuit with
dependent source
To analyze circuit that have dependent source, mesh analysis
can be done as same as before.
Below is an example for circuit with dependent source
Example :
For circuit in figure below, get the value ofIa
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a) Assign mesh currents I1 and I2 in to the 2 meshes
b) Write KVL equations for each mesh
Mesh 1 : 5I1 3I2 = 25 (1)
Mesh 2 : -3I1 +9I2 = -9 6Ia (2)
where Ia = I1 (3)
Substitude equation (3) into equation (2)
3I1 +9I2 = -9 (4)@
I1 I2
16V 6Ia
3
2 6
9V
Ia
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c) From equation (1) and (4), using Cramers Rule to get the value ofI1.
I1 = 3.67 A Ia = I1 = 3.67 A
@ @
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SUPERPOSITION
THEOREM
The superposition principle statesthat the voltage across (or
current through) an element in a
linear circuit is the algebraic sumof the voltage across (or currentsthrough) that element due to eachindependent source acting alone.
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Steps to Apply SuperpositionPrinciple :
Turn off all independent source exceptone source. Find the output (voltage orcurrent) due to that active source.
Repeat step 1 for each of the otherindependent source.
Find the total contribution by adding
algebraically all the contributions due tothe independent source.
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Example 1:
Find the value of V.
Figure 1.1
Let, V = V1 + V2
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Current Division : i3 = 8 (3) = 2 A4 + 8
V2 = 4i3 = 8 V
The Value of V :
V = V1 + V2 = 2 + 8 = 10 V
Figure 1.1 (b)
Find V2 :
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Example 2:
Find the value of io.
Figure 1.2
Let, i 0 = i0 + i0
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Solution:
Figure 1.2 (a)
i1 = 4 A ....(1)
-3i1 +6 i2 - 1i3 - 5i0 = 0 ....(2)
-5 i1 - 1i2 + 10 i3 + 5i0 = 0 ......................................( 3)
At node 0 : i3= i1 - i0 = 4 - i0 . .(4)
Substitute (1) & (4) into (2) & (3) :3 i2 - 2i0 = 8 .(5)
i2 +5i0 = 20 ....(6)
i0 = 52 A ...(7)17
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Figure 1.2 (b)
6i4 - i5 - 5i0 = 0..(8)
Loop 5 : - i4 + 10 i5 - 20 + 5i0 = 0 ... (9)
But , i5= - i0 substitute into (8) & (9) :
6i4 - i0 = 0 .(10)
i4 +5i0 = 0 ...(11)
i0 = - 60 A .....(12)
17The value ofi0 : i0= - 8 = - 0.4706 A
17
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Example 3:
Figure 1.3
Find the value of i.
Let, i = i1 + i2 + i3
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Solution:
i1 = 12 = 2 A6
Figure 1.3 (a)
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Figure 1.3 (b)
RT = 8+ 4 + (12/7)= 13.714
I= 24/13.714 = 1.75A
i2 = -(4/7)(1.75) = -1 A
DC
I
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Figure 1.3
(c)
I= (8/(5.714+8))x3 = 1.75 A
i3 = (4/7)x1.75 = 1 A
The value ofi: i = i1 + i2 + i3 = 2 - 1 + 1 = 2 A
I
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THEVENINS
THEOREM
Thevenins theorem states that alinear two terminal circuit can bereplaced by an equivalent circuit
consisting of a voltage source VTh inseries with a resistor RTh ,where
VTh is the open circuit voltage atterminals a & b and RTh is the inputor equivalent resistance at theterminals when the independent
sources are turned off.
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This complex circuit can be replacedby the simpler circuit known as
Thevenin equivalent circuit.
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Thevenin Equivalent circuit
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Assignment 1
Question
Prove that any complex linear electriccircuit can be replaced by itsThevenin
equivalent circuit consisting of a voltage
source in series with a resistor.
The deadline: 20th February 2008
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Finding VTh and RTh
15 Volt
3 Ohm a
b
RL= 6 Ohm
I
Vab
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Finding VTh and RTh
15 Volt
3 Ohm a
b
RL= 6 Ohm
I
Vab
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Two cases when finding theThevenin resistance
Case 1 : Circuit with independent sources only.
RTh is the equivalent resistance of the
network looking between terminals a and bby turned off all the independent sources.
Voltage source Short circuit
Current source Open circuit
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Case 2 :Circuit with independent and dependentsources.
Turn off all the independent sources asmentioned above but the dependent
sources are cannot be turned off. Apply the voltage source, V0 at terminals
a & b and determine the resultingcurrent i0. Then RTh = V0
i0
Fi di R h i i h
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Finding RTh when circuit hasdependent source
Figure 1.6 (a) Figure 1.6 (b)
RTh = Voio
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A circuit with a load
IL = VTh(RTh +RL)
VL = RL IL
= RL x VTh
(R
Th+
R
L)
Figure 1.7 (a) : Original circuit
Figure 1.7 (b) : Thevenin equivalent
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Example 1:
Find the Thevenin equivalent circuit of the circuit shown in Fig.1.30
to the left of the terminals a-b. Then find the current through
RL = 6, 16 and 36.
Figure 1.8
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Finding VTh :
Figure1.8 (b)
Loop i1 : - 32 + 4 i1 + 12 ( i1 - i2 ) = 0 .(1)
Loop i2 : i2 = - 2 A ...(2)
Substitute (2) into (1) :
i1 = 0.5 A
VTh = 12 ( i1 - i2) = 12 ( 0.5 + 2.0 ) = 30 V
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IL = VTh = 30
RTh +RL 4+RL
When RL = 6, IL = 30 = 3A
10
When RL = 16, IL = 30 = 1.5A
20
When RL = 36, IL = 30 = 0.75A
40
Figure 1.8 (c)
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Example 2 :
Find the Thevenin equivalent of the circuit.
Figure 1.9
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Figure 1.9 (a)
Solution:
To find RTh, we set the independent sources equal to zero but the
dependent sources remains in the circuit. We may set Vo = 1V to ease
calculation.
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Loop i1 : - 2 Vx + 2 ( i1 i2 ) = 0 or Vx = i1 i2
But, - 4i2 = Vx = i1 - i2
Hence, i1 = - 3i2
Loop i2& i3 : KVL : 4 i2 + 2 ( i2 - i1 ) + 6 ( i2 i3 ) = 0
6 ( i3 i2) + 2 i3 + 1 = 0
i3 = -1 A6
But, io = - i3 = 1 A6
Hence, RTh = 1V = 6io
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Finding VTh :
Figure 1.9 (b)
Loop i1 : i1 = 5A
Loop i3 : - 2 Vx + 2 ( i3 i2 ) = 0
Vx = i3 i2
Loop i2 : 4 ( i2 i1) +2 ( i2 i3 ) +6 i2 = 0
Or 12i2 4i1 2i3 = 0
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But, 4 ( i1 i2 ) = Vx
i2 = 10 A3
Hence, VTh = Voc = 6 i2 = 20V
Figure 1.9 (c) : Thevenin equivalent circuit
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Example 3 :
Determine the Thevenin equivalent of the circuit.
Figure 1.10
Solution:
Since there is no dependent source in the circuit,
Vab = VTh = 0 V
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Substitute (2) into (1) :
i0 = ix + V0 = - V0 + V0 = - V0
4 2 4 4
Or , V0 = - 4i0
Thus , RTh = V0 = - 4
i0
Nodal Analysis : i0 + ix = 2ix + V0 (1)4
But , ix = 0 - V0 = - V0 ........(2)2 2
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NORTONSTHEOREM
Nortons theorem states that alinear two terminal circuit can bereplaced by an equivalent circuit
consisting of a current source IN inparallel with a resistor RN where IN
is the short circuit current throughthe terminals and RN is the input orequivalent resistance at the
terminals when the independentsources are turned off.
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Norton equivalent circuit
Figure 1.11 (a) : Original circuit Figure 1.11 (b) : The Norton equivalent circuit
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To find the Norton current, IN, We determine the short-circuit-
current flowing from terminal a to b in both circuit.
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Example 1:
Find the Norton equivalent circuit of the circuit
Figure 1.12
Solution:
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RTh = RN
Solution:
RN = 5 // ( 8 +4+ 8 )= 5//20
= 20 5 = 4
25
Figure 1.12 (a)
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Short circuit terminal a & b
Ignore 5 resistor
i1 = 2A
20 i2 4i1 12 = 0
i2 = 1A = isc = IN
Figure 1.12 (b)
Finding IN :
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Example 2 :
Using the Nortons theorem,find RN and IN at terminals a-b
Figure 1.13
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Fi di I
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Short circuit terminals a & b and find the current isc
Note the 4 resistor, the 10V voltage source, the 5
resistor, and the dependent current source are all in parallel
Hence,
ix = 10 = 2.5 A
4 At node a, KCL : isc = 10 + 2ix
5
= 2 + 2 (2.5) = 7A
Thus, isc = IN = 7A
Figure 1.13 (b)
Finding IN :
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MAXIMUM POWERTRANSFER
To determine the value of loadresistance which the maximum
power is transferred to the
load.
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Maximum power transfer
RTh
VTh
P = I RL = ( VTh ) RL( RTh +RL )
Figure 1.14
I = ( VTh )
( RTh +RL )
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Maximum power transfer
RL
P
P = I RL = ( VTh ) RL( RTh +RL )
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Maximum power transfer
RL
P
P = I RL = ( VTh ) RL( RTh +RL )
The slope = dP
dRL
dRLdP
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Maximum power transfer
RL
P
P = I RL = ( VTh ) RL( RTh +RL )
The slope = dP
dRL
dRLdP = 0
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Maximum power transfer
dP = (RTH +RL)2.VTH 2 2VTH
2RL(RTH +RL)
P = I RL = ( VTh ) RL( RTh +RL )
The slope = dP
dRL= 0
dRL (RTH +RL)4 = 0
or
(RTH +RL)2.VTH 2 2VTH
2RL(RTH +RL) = 0
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Maximum power transfer
(RTH +RL)2.VTH 2 2VTH
2RL(RTH +RL) = 0
(RTH +RL)2.VTH 2 = 2VTH
2RL(RTH +RL)
(RTH +RL) = 2RL
RL-2RL = -2RTH
RL = RTH
When RL = RTH , RL absorbing the maximum power
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Maximum power transfer
RTh
VTh
P = I RL = ( VTh ) RL( RTh +RL )
Figure 1.14
For RL = RTh
P max = VTh
4RTh
I = ( VTh )
( RTh +RL )
E l 1
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Example 1:
Find the value ofRL for maximum power transfer in the circuit. Find the
maximum power.
6 3 2
12 RL12V 2 A
Figure 1.15
S l ti
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Solution:
Finding RTh :6 3 2
12
Figure 1.15 (a)
RTh = 2 + 3 + 6 //12
= 5 + 6 12
18= 9
RTh
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Finding VTh :
Figure 1.15 (b)
Mesh analysis : - 12 +18 i1 12 i2 = 0,
i2= -2A
Get, i1 = - 2 A3
-12 + 6 i1 +3 i2 + 2 (0) + VTh = 0VTh = 22 V
RL = RTh = 9
Pmax = VTh = 22 = 13.44 watt
4RL (49)
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