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Method of (Circuit)
Analysis
Chapter 8
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If two or more sources are present
in the circuit, what will you do?
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Methods for two or more souces
Branch-current Analysis
Mesh analysis and
Nodal analysis
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All the methods will be described
for linear bilateral networks only Liner indicates that the characteristics of
the of the network elements are
independent of voltage across or currentthrough them.
e. g. Resistor
A 10k resistor always shows 10kwhether the voltage across it is 10V or100V
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So resistor is a linear element
but diode is not
Bilateral: it refers to the fact that
there is no change in the behaviour
or characteristics of an element if
the current through or voltageacross the element is reversed
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II
I I
Bilateral ????
Bilateral ????
R R
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Current Source
It is the dual of the voltage source.
For a fixed dc voltage source, voltageacross its
terminal is constant but current may vary
For a fixed dc current source, current through it, is
constant but voltage may vary
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Current Source
IL
VL
+
+
E I
vL
IL
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Example 8.1. Find the Vs and I1 for the circuit of Fig 8.3
Soln: I1 = I = 10mAVs = V1 = I1R1 = 1010-320103 = 200 V
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Example: 8.2. Calculate V1, V2 and Vs for the circuit of Fig.
8.4.
Soln: V1 = I1 R1 = IR1 = 52 = 10 V (Ans)
V2 = I2 R2 = IR2 = 53 = 15 V (Ans)
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Example: 8.2. Continued
(Ans.)V251510
0
21
21
VVV
VVV
S
S
Applying KVL, we obtain
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Example: 8.3. Consider the series- parallel circuit of Fig. 8.5
Soln: Using current divider rule, we get
(Ans.)V422Law,sOhm'By
(Ans.)A421
26
(Ans.)A221
16
22333
23
23
23
32
RIRIV
RR
RII
RR
RII
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Example: 8.3. Continued
(Ans.)V1843640
0
directionCWin theKVLApplying
31
31
VIREV
VVEV
S
S
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Source Conversion
LS
SL
LS
S
LS
S
S
LS
S
S
L
LS
SS
LSLS
L
RR
RII
RRRI
RR
R
R
E
RR
E
R
R
I
RRERR
RRE
RREI
/)1(
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Source Conversion
LS
SL
LS
S
S
LS
SS
LSLS
L
RR
RII
RR
ERR
RR
ERR
RR
E
RR
EI
/)1(
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Example 8.4. Convert the voltage source of Fig. 8.9 to a
current source and calculate the current through the load for
each source.
(Ans.)A142
6
LS
LRR
EI
Soln:
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Example 8.4. Continued..
(Ans.)A142
23
LS
SL
RRRII
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Example 8.5. Convert the current source of Fig. 8.10 to a
voltage source and find the current through the load for each
source.
Soln:
(Ans.)mA3k6k3
k3109 3
LS
SL
RR
RII
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Example 8.5. Continued.
(Ans.)mA3k)63(
27
LS
LRR
EI
V27k3m9 SIRE
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Current Sources in Parallel
Two or more current sources can be connected in parallel
and they may be replaced (represented) by one current
source having the magnitude and direction of the resultant.
Sum if the currents enters a node and subtract if they haveopposite directions.
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Parallel Current Source
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Example 8.6. Reduce the left side Fig. 8.11 to a minimum
number of elements.
Soln.
(Ans.)A461012
III
Ohm263
63||
21
2121
RR
RRRRR
R
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Example 8.7. Reduce the network of Fig. 8.13 to a single
current source and calculate the current through RL.
Soln: Converting voltage source into a current source
A48
32
1
1
1
R
EI
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Here thevoltage
source is
converted
into a current
source
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A
A106421 IIIS
Ohm6248
248||
21
2121
RR
RRRRRS
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The reduced circuit can be represented as
(Ans.)A3146
6
10 LS
S
SL RR
R
II
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8.5 Current Sources in Series
Applying KCL at the node a, we get
I1= I2
So I1 and I2 must be same.
Current Sources of different ratings are not connected in series.
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8.6 Determinants
333
222
111
333
222
111
333
222
111
333
222
111
333
222
111
333
222
111
3333
2222
1111
cba
cba
cba
dba
dba
dba
z
cba
cba
cba
cda
cda
cda
y
cba
cba
cba
cbd
cbd
cbd
x
dzcybxa
dzcybxadzcybxa
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Determinant continued
1221
2112
22
11
22
11
22
11
22
11
222
111
baba
cbcbx
ba
ba
ca
ca
y
ba
ba
bc
bc
x
cybxa
cybxa
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8.7 Branch Current Method
Steps
Assign a distinct current of
arbitrary direction to each
branch of network
I1
I2
I3
Indicate the polarities for
each resistor as
determined by the
assumed current
direction.
Apply KVL around each
closed loop and KCL at
the node which includes
all the branch currents. Solve the linear
simultaneous equations.
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Branch Current Method Continueda
loop1
loop2
Applying KVL in loop 1, we get
1...........0
0
133211
33111
EIRIIR
IRIRE
3...........0321
321
III
III
Applying KVL in loop 2, we get
Applying KCL at the node a, we get
2...........0
0
233221
22332
EIRIRI
IRIRE
Solve equation 1, 2 and 3 to get I1,
I2 and I3.
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Example 8.15. Apply the branch-current method to the
network of Fig. 8.19
Soln.
Assign a distinct current of
arbitrary direction to each
branch of network
Indicate the polarities for
each resistor as
determined by the
assumed current
direction.
Apply KVL around each
closed loop and KCL at
the node which includes
all the branch currents. Solve the linear
simultaneous equations.
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Example 8.15 continued ..
)1.......(2402,
0,
0
321
133211
33111
IIIor
EIRIIRor
IRIRE
)2.......(6410,
0,
0
321
233221
22332
IIIor
EIRIRIor
IRIRELoop 1 Loop 2
Applying KVL around loop 1, we get
Applying KVL around loop 1, we get
Applying KCL at node a, we get
)3..(..........0-, 321
213
IIIor
III
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Example 8.15 continued ..
)3.(....................0
)2..(..........6410
)1..(..........2402
321
321
321
III
III
III
(Ans.)A114
14
14
2410
)10(40)41(2
)06(40)41(2
111
410
402
110
416
402
1
I
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A