Module 1 Basic Concepts

Chapter 1
Module 1 Basic Concepts
1.1 Introduction
Resistors in series:
Consider two resistors R1 and R2 in series.
R1 R2
V1 V2
I
Figure 1.1
V = V1 + V2 = I(R1 + R2 )
V
= Req = R1 + R2
I
If n number of resistors R1 , R2 ....., Rn are connected in series then the equivalent resistance Req is
Req = R1 + R2 ....., Rn
Resistors in parallel:
Consider two resistors are connected in parallel.
I1 R1
I I
I2 R2
Figure 1.2
Current in each branch is
V
I1 =
R1
V
I2 =
R2
1
1.1. Introduction Chapter 1. Module 1 Basic Concepts
The current I is
I = I1 + I2
V V
= +
R1 R2

1 1
= V +
R1 R2

I 1 1 1
= + =
V R1 R2 Req
If n number of resistors are connected in parallel then

1 1 1 1
= + ...
Req R1 R2 Rn
If only two resistors are connected in parallel then Equivalent resistance Req is
1 1 1 R1 + R2
= + =
Req R1 R2 R1 R2
R1 R2
Req =
R1 + R2
KIRCHHOFF’S RULES
1. Current Law or Junction Rule or Kirchhoff ’S Current Law (KCL): The algebraic sum
of electric currents at any junction in electrical network is always zero.
Xn
In = 0
i=1
or The sum of incoming currents towards the junction are equal to sum of outgoing
currents at a junction.
This law is a statement of conservation of charge. If current reaching a junction is not equal to
the current leaving the junction, charge will not be conserved.
2. II Law or Loop Law or Junction Rule: Kirchhoff ’S Voltage Law (KVL): The algebraic
sum of changes in potential around any closed loop involving resistors and cells in the
loop is zero.
This law represents conservation of energy. If the sum of potential changes around a closed loop
is not zero, unlimited energy could be gained by repeatedly carrying a charge around a loop.
Sign convention for the application of Kirchoff ’s law
1. While traversing in a loop the direction of current is in the same path then the potential
drop at a resistance is -IR while in the in opposite direction it is +IR.
2. The emf is taken negative when we traverse from positive to negative terminal of the
cell. The emf is taken positive when we traverse from negative to positive terminal of
the cell.
Dr. Manjunatha P Prof., Dept of ECE, JNN College of Engg Shimoga manjup.jnnce@gmail.com 2
Example
b 4Ω c
I1 1
9V
I3 3Ω
a d
I2 2
5Ω
f e
8V
Figure 1.3
KCL for the junction at node ’a’ is
Incoming current at node ’a’ is I3 and outgoing currents are I1 and I2 .
I3 = I1 + I2
sum of the currents meeting at node ’a’ is zero OR
I3 − I1 − I2 = 0
For the node ’d’
I1 + I2 = I3
I1 + I2 − I3 = 0
For the loop 1 abcda
−4I1 + 9 − 3I3 = 0
−4I1 + 9 − 3(I1 + I2 ) = 0
7I1 + 3I2 = 9 (1.1)
For the loop 2 afeda
8 − 5I2 + 9 − 3I3 = 0
17 − 5I2 − 3(I1 + I2 ) = 0
3I1 + 8I2 = 17 (1.2)
From Equation 1.1 and 1.2
7I1 + 3I2 = 9
3I1 + 8I2 = 17
Solving the above equations
I1 = 0.446A
I2 = 1.95A
Applying Node voltage method
Va Va − 9 Va + 8
+ + = 0
4 3 5
1 1 1 8
Va + + −3+ = 0
4 3 5 5
Va = 1.787
Va 1.787
I1 = = = 0.4464
4 4
Va + 8 1.787 + 8
I2 = = = 1.954
5 5
Branch Current Rule Using Method 1

I1 R1 4
R3 i1 = 2A
4 + 12
= 0.5A
12
i2 = 2A = 1.5A
I I2 4 + 12
R2
Using Method 2
V 3
i1 = 2A = 0.5A
Figure 1.4 12
3
When two resistors are connected in parallel: i2 = 2A = 1.5A
4
Branch Current is
Example 2
Resistance of other branch
= M ain Current Find the magnitude of I in ampere
Sum of resistances
I 60
R2 1A 5
I1 = I 15 1A
R1 + R2
R1 10
I2 = I
R1 + R2
Also it is given by Figure 1.6
RP Solution:
I1 = I Using Method 1
R1
RP I 60 I 60
I2 = I
R2
1A 20 1A 1A 1A
where I is the main current and RP is the parallel
branch effective resistance. 10 6.666
R1 R2
RP =
R1 + R2 Figure 1.7
V
I =
RP + R3
6.6666
I = 1A ' 0.1A
60 + 6.6666
Example 1
Find the current i1 and i2 for the circuit shown in Using Method 2
Figure When the Resistors 10, 15 and 60 Ω, are
i1 12 connected in parallel hence
2
1 1 1 1
= + +
i 2 4 RT 10 20 60
6+3+1 1
= =
60 6
12 V 1 RT = 6
Figure 1.5 Current I2 is

Solution: 6
12Ω and 4Ω are in parallel I = 1A
60
12 × 4 = 0.1
RT = +2=3+2
12 + 4
= 5Ω
Total Current I is
E 12
I = =
RT + r 5+1
= 2A
Find the power dissipated in the 3 Ω resistor

4.5 V
1 For the circuit shown in Figure 1.11 find the value
3
of current I2
I1 10
2 6
1.2 A I2 15
4 I3 30
Figure 1.8
Figure 1.11
Solution: Ans (b): The given circuit is redrawn.
Solution: The total Resistance of the network is
4.5 V
1 1 1 1 1
= + +
RT 10 15 30
I 3 6 1
= =
I1 2 30 5
6 RT = 5
Current I2 is
4
5
I2 = 1.2A
15
Figure 1.9 = 0.4
3 and 6 Ω are in parallel which is in series with 2 Ω
6×3
2 + (3||6) = 2 + = 2 + 2 = 4Ω
6×3 Find the current I flowing in the circuit as shown in
4Ω and 4Ω are in parallel which is in series with 1 Ω Figure 1.12
4
4×4
1 + (4||4) = 1 + = 1 + 2 = 3Ω 4
4×4 4
The current I is I
4
4.5
= 1.5A
3 1.6
4V
1 4.5 V
Figure 1.12
I Solution:
I1 The 4 Ω and 4 Ω are in parallel which combination
is in series with 4 Ω
4
4×4
= 2Ω
4+4
4
4
Figure 1.10
The current I1 is 2 4
4 I
I1 = 1.5A = 0.75A
4+4
The current through 3 Ω is 4V 1.6
6
I3 = 0.75A = 0.5A Figure 1.13
3+6
Again 4 Ω and 6 Ω are in parallel
The power dissipated in the 3 Ω is
4×6
(I3 )2 × 3 = (0.5)2 × 3 = 0.75W = 2.4Ω
4+6
2.4 I 60
1A 15 5 1A
10
4V 1.6
Figure 1.15
Figure 1.14
Solution: When the Resistors 10, 15 and 30 Ω, are
Current from battery is
connected in parallel hence
4
I= = 1A 1 1 1 1
2.4 + 1.6 = + +
RT 10 20 60
The current I is 6+3+1 1
= =
60 6
4 RT = 6
= 1A = 0.4A
4+6
Current I2 is
6
Find the magnitude of the current I for the circuit I = 1A
60
shown in Figure 1.15 is = 0.1
1.2. Source Transformation Technique Chapter 1. Module 1 Basic Concepts
1.2 Source Transformation Technique
R
a a
Vs +- Is R
b b
Voltage Current
Source Source
Figure 1.16
VS = IS R
VS
IS =
R
4 and 8 Ω are in parallel

Q 1) In the circuit below, use a source 4×8
R= = 2.66 Ω
transformation to determine vO . 4+8
Solution:
2.66Ω
2
3A 2.66Ω 8Ω vO 8V + 8Ω vO
vO
-
4 3A 8
Figure 1.21
Figure 1.17
The current in the circuit
The transformed circuit is as shown in Figure 1.258.
8
The current in the circuit is 0.857 A. The voltage I= = 0.75
across the resistor 8 Ω is 10.667
The voltage across the resistor 8 Ω
0.857 × 8 = 6.857V 0.75 × 8 = 6V
4Ω 2Ω
Q 3) For the circuit shown in Figure determine the

12 V + 8Ω vO
- V0 using source transformation.
2Ω 3Ω
Figure 1.18
3A +- 12V
Q 2) For the circuit shown in Figure. 1.256, use a 4Ω 8Ω V0
source transformation to determine vO .
Solution:
Figure 1.22
4Ω
Solution:
12 V +
- 8Ω 8Ω vO Replace the current source and parallel resistance by
voltage source in series with resistor
Figure 1.19 V = I × R = 3 × 4 = 12V

4Ω 2Ω 3Ω
3A 8Ω vO
4Ω 8Ω
12V +- 8Ω V0
+- 12V
Figure 1.20
Figure 1.23 5Ω 10V 10Ω

b c e
Now 4 Ω and 2 Ω are in series which are replaced by
+-
single resistance 6 Ω i1
+- 5Ω
5Ω
6Ω 3Ω 10V
a d f
12V +- 8Ω V0
+- 12V
Figure 1.28
Figure 1.24 Now the 10 Ω and 5 Ω are in series
Replace the voltage sources by current source in b 5Ω 10V
c e
+-
parallel with resistance 6 Ω and 3 Ω i1
+ 15Ω
- 5Ω
2A 10V
6Ω 8Ω V0 3Ω 4A
a d f
Figure 1.25 Figure 1.29

Replace two current sources by single current source Replace the voltage source by current source. First
and two parallel resistors 6 Ω and 3 Ω with current source is I = VR = 10 5 = 2A and the
single resistance. Current sources are in opposite second source is and each resistors are in parallel
directions. Again replace current source by voltage with respective current sources I = VR = 10 2
15 = 3 A
source in series with resistor 2 Ω c
6×3 18 2A i1 2
R= = = 2Ω 5Ω 15Ω A
6×3 9 5Ω 3
The current in the circuit is
d
4 4
I= = = 0.4A
8+2 10 Figure 1.30
The voltage drop across 8 Ω is Now two current sources are in parallel they can be
added and total current is
V = 0.4A × 8 = 3.2V
2 8
I =2+ = A
2Ω 3 3
The parallel resistances are added
8Ω V0 2Ω +- 4V V0 8Ω 5 × 15 15
2A R= = Ω
5 + 15 4
The equivalent circuit is as shown in Figure. The
Figure 1.26 current source is replaced by voltage source which is
as shown in Figure. The new voltage source is
Q 4) In the circuit shown in Figure 1.257 determine 8 15
the current i1 through 5 Ω resistor by source V =I ×R= × = 10V
3 4
transformation.
1A
The total resistance in the circuit is
15 35
b 5Ω c e R= +5= = 3.75 + 5 = 8.75Ω
4 4
i1 10Ω
+- 5Ω 5Ω Current i1 through 5 Ω resistor is
10V
a
V 10
d f I= = = 1.142A
R 35/4
Figure 1.27
Solution: KVL cannot be applied due to the c c
presence of current source. Transform the current 8 15
A 15 i1 10V Ω i1
source to voltage source. 3 Ω +- 4
4 5Ω 5Ω
d d
V = I × R = 1 × 10 = 10V
Figure 1.31 3Ω 4Ω
The other method to find current in the above circuit I

is +- +- 16/3 V
10V
Resistance in other baranch Figure 1.35

i1 = Currrent ×
T otal Resistance Current I is
8 3.75 16
i1 = × = 1.142A + 10 − 7 × I = 0
3 3.75 + 5 3
16 + 30
− 7I = 0
3
46
7I =
Q 5) In the circuit shown in Figure 1.32 determine 3
46
the current I I = = 2.19A
3×7
2A
3Ω Q 6) In the circuit shown in Figure 1.257 determine

8Ω current I by source transformation.
I
+- 6Ω 4Ω 5Ω a
10V
I
+- 0.1A
Figure 1.32 5V 20Ω 30Ω
Solution: b
KVL cannot be applied directly due to the presence
of current source. Replace the current source into Figure 1.36
voltage source in series with 8 Ω and current source Solution:
as First the parallel resistance is replaced by single
resistor which is as shown in Figure
V = 2 × 8 = 16V olts
20 × 30 600
Now the 8 Ω and 4 Ω are in series which are placed R= = = 12Ω
20 + 30 50
in series with voltage source.
3Ω 16 V 5Ω a
-
+
8Ω I
I +- 0.1A
+- 6Ω 4Ω
5V 12Ω
10V
b
Figure 1.33
Replace the voltage source into current source in Figure 1.37
parallel with 12 Ω Now replace the current source in parallel with
3Ω
resistor by voltage source in series with resistor,
which is as shown in Figure
I
+- 6Ω 12Ω 4/3 A 5Ω 12Ω
10V
I
5V +- + 1.2V
-
Figure 1.34
12 Ω and 6 Ω are in parallel. Current source
can be replaced by voltage source the details are as Figure 1.38
shown in Figure The current I in the circuit is
6 × 12 5 − 1.2 3.8
R= = 4Ω I= = = 0.224A
6 + 12 5 + 12 17
4 16 Q 7) In the circuit shown in Figure 1.39 determine
V =I ×R= × 4 = V olts the voltage v0 across 100 Ω resistor
3 3
100Ω 8V
- Q 8) In the circuit shown in Figure 1.44 determine
+
30 mA the current in the 12 Ω resistor using source
+
-
v0 100Ω 100Ω transformation method
10 V 60 V 2Ω
+-
Figure 1.39 24Ω
12Ω 48 A 4Ω 30 A
Solution:
240 V +-
Replace the current source by voltage source in series
with 100 Ω resistor which is as shown in Figure. Figure 1.44
100Ω 8V- 100Ω Solution:
+
+ v0 100Ω + 3V Replace the current source and parallel resistor 4 Ω

- -
10 V by voltage source in series with resistor 4 Ω
Figure 1.40 V = I × R = 30 × 4 = 120V

Voltage sources of 8 and 3 are in series which are
replaced by single voltage source. 60 V 2Ω
+-
100Ω 100Ω
24Ω +- 120 V
12Ω 48 A
+- v0 100Ω +- 11 V 240 V +- 4Ω
10 V

Replace the voltage source of 11 volts with current Replace the voltage sources and source resistors by
source in parallel with 100 Ω resistor. single voltage source in series with single resistor 4
Ω
11
I= = 110mA
100 V = 60 + 120 = 180V
100 mA v0 100Ω 100Ω 110 mA R = 2 + 4 = 6Ω

100Ω
24Ω +- 180 V
Figure 1.42 12Ω 48 A
100 mA current source with 100 Ω resistor and 240 V +- 6Ω

110 mA current source with 100 Ω resistor are in
parallel which are replaced by single current source Figure 1.46
and single resistor as
Replace 240 V voltage source in series resistor 24
Ω by a current source and 180 V voltage source in
I = 100 + 110 = 210mA
series resistor 6 Ω by a current source
100 × 100
R= = 50Ω
100 + 100 V 240
I= = = 10A
R 24
210 mA v0 100 V 180

50 I=
R
=
6
= 30A
Figure 1.43 12Ω 24Ω

6Ω
10 A 48 A 30 A
Current through 100 Ω resistor is
50
I = 210 = 70mA
50 + 100 Figure 1.47
Voltage across the 100 Ω resistor is Current sources 10 A, 48A, and 30 A are in parallel.
Replace these by single current source. Also replace
V = I × R = 70 × 10−3 × 100 = 7V parallel resistor by a single resistor.
5Ω
I = 10 + 48 − 30A = 28A
5A
24 × 6 144 4Ω
R= = = 4.8Ω 2Ω 3Ω
24 + 6 30 6 V +-
4.8Ω Figure 1.51

12Ω 28 A 4.8Ω 12Ω
+- 134.4 V 5 Ω and 2 Ω resistors are in parallel and are replaced
by single resistor.
2×5
Figure 1.48 R= = 1.428Ω
2+5
The current in 12 Ω resistor is sources 10 A, 48A,
and 30 A are in parallel. Replace by single current
5A
source. Replace parallel resistors by a single resistor. 4Ω
1.428 Ω 3Ω
6 V +-
134.4
I= = 8A
12 + 4.8
Figure 1.52
Q 9) In the circuit shown in Figure 1.52 determine
the current in the 3 Ω resistor using source 7.14 V
-
+
transformation method
1.428Ω
1A 5Ω 4Ω I
3Ω
2Ω - 2Ω 6 V +-
+
10 V I
2A 2Ω 3Ω Figure 1.53
The current through 3 Ω is determined by applying
KVL in the loop as:
Figure 1.49
Solution: 7.14 − 6
I=
= 0.135 A
8.428
Replace the 2 A current source and parallel resistor
of 2 Ω by voltage source in series with resistor Q 10) In the circuit shown in Figure 1.54 determine
2 Ω similarly replace the 1 A current source and the currentIx using source transformation method
parallel resistor of 2 Ω by voltage source in series 200Ω 30Ω 50Ω
with resistor 2 Ω 10 mA I x
- 60Ω 10Ω
+ 80Ω
V = 2 × 2 = 4V 10 V
10Ω
V = 1 × 2 = 2V
Figure 1.54
2 V 2Ω 5Ω
- Solution:
+
- 2Ω Replace the voltage source of 10 V by current source

+
2Ω 10 V I in parallel with resistor of 200 Ω

3Ω
4 V +- 10
I= = 50 mA
200
4 volts and 2 volts are in series and are added 10 mA Ix
50
similarly 2 Ω 2 Ω resistors are in series and are 200Ω 60Ω 10Ω
mA 80Ω
added. Replace the 10 V voltage source by current
source in parallel resistor of 2 Ω.
10Ω
10
I= = 5A Figure 1.55
2
Replace the parallel resistors of 200 and 80 Ω by a 50Ω

single resistor
19.41 Ix
200 × 80 mA 37 Ω 10Ω
I= = 57.1 Ω
200 + 80
30Ω 50Ω Figure 1.60

50 10 mA Ix By Branch Rule
mA 57.14Ω 60Ω 10Ω
37
Ix = 19.41 = 7.42mA
10Ω 37 + 60
Figure 1.56
the currentIx using source transformation method
Replace the current source of 50 mA and parallel
resistor of 57.14 Ω by voltage source 10Ω
Ix 8Ω
−3
V = 50 × 10 × 57.14 = 2.857V 3I x
60 V +- 50Ω 40Ω
Figure 1.61
30Ω 50Ω
Solution:
- 10 mA Ix
2.857 V +
60Ω 10Ω Replace the current source and parallel resistor of
57.14Ω
10Ω 10 Ω by voltage source in series with resistor of 10
Ω.
Figure 1.57 Ix 8Ω 10Ω
-+
30 Ω, 57.14 Ω and 10 Ω are in series replace by single
30I x
resistor 60 V +- 50Ω 40Ω
97.14Ω 50Ω
10 mA Ix
- 60Ω 10Ω Figure 1.62
2.857 V +
10 Ω and 40 are in series which are replaced by single
resistor. Replace voltage source by current source.
Figure 1.58
Ix 8Ω
Replace the voltage source of 2.857 V by current
source in parallel with resistor of 97.15 Ω
60 V +- 50Ω 50Ω
0.6I x
2.857
I= = 29.41 mA
97.15
Figure 1.63
50Ω
Ix 8Ω
10 mA Ix
29.41
mA 97.15Ω 60Ω 10Ω
60 V +- 0.6I x 25Ω
Figure 1.59
Figure 1.64
29.41 mA and 10 mA are in opposite directions
Replace the current source and parallel resistor of
and are in parallel, replace by single current source.
25 Ω by voltage source in series with resistor of 25
Resistors 97.15 Ω 60 Ω are in parallel with single
Ω.
resistor of 37 Ω
Ix 8Ω 15I x
-+
I = 29.41 − 10 = 19.41 mA
60 V +- 25Ω
97.15 × 60
R= = 37 Ω
97.15 + 60
Figure 1.65 3Ω
The current through Ix is determined by applying
1A
KVL in the loop as:
60 − 8Ix + 15Ix − 25Ix = 0
5i1
18Ix = 60 5Ω
- +
60 i1
Ix = = 3.333A 5Ω
18
Q 12) In the circuit shown in Figure ?? determine
the current I1 6/5A
3A
1Ω 2Ω Figure 1.70
5Ω
- +
i1
Current source 1 A and 65 A are in opposite
2Ω 5i1 3Ω directions, replace by single current source and also
replace parallel resistors 3 Ω and 5 Ω by a single
Figure 1.66 resistor
Solution: 6 1
Ieq = − 1 = = 0.2A
For the given circuit there is a current source of 3A. 5 5
Shift the current source between resistors 1 Ω 2 Ω. 3×5 15
The modified circuit is as shown in Figure 1.67 Req = = = 1.875Ω
3+5 8
1Ω 2Ω
3A 5Ω 5i1
- + 5Ω
i1 - +
2Ω 5i1 3Ω
3A 1.875Ω
Figure 1.67
Convert current sources into voltage sources in series 0.2A
with resistor 1 Ω and 2 Ω.
Figure 1.71
1Ω
2Ω
Replace the 0.2A current source and parallel resistor
3V +-
- +
5Ω 1.875 Ω by voltage source in series with 1.875 Ω
i1
6V +- 5i1 3Ω resistor
5i1
2Ω
5Ω i1
- +
Figure 1.68
1.875Ω
3V 3Ω
-
+
+-
0.375 V
5Ω
- + Figure 1.72
i1
5i1
Apply KVL for the loop
-
+
6V 5Ω
0.375 − 6.875i1 − 5i1 = 0
Figure 1.69
0.375 − 11.875i1 = 0
Now convert voltage sources into current sources in 11.875
parallel with resistors as shown in Figure i1 = = 31.67A
0.375
1.3. Question Papers Chapter 1. Module 1 Basic Concepts
1.3 Question Papers

2019 Dec (2018 Scheme ) 1 a). Using source 8Ω 4Ω
transformation technique find the current through
5 Ω resistor for the circuit shown in Figure 1.73.
9 V +- 9 V +-
4Ω
5 30
+-
20 4Ω
5 V +- + 3V
-
8Ω 8Ω
Figure 1.73: 2019-Dec-Question Paper
Solution:
5 9 V +- 9 V +-
20 V +-
20 30 0.1 A
5 V +-
8 8 2.5 A
1.125 A -
+
Figure 1.74 9V
5
4Ω 1.375 A
20 30 0.1 A -
5 V +- +
9V
Figure 1.75 4Ω 4Ω
5 12
+- +- 5.5 V +-
9V 3.5 V
5 V +- + 1.2 V
-
The equivalent current source is

Figure 1.76
The current through 5 Ω resistor is 4
0.875 A
12 − 5
I= = 0.4117 A
17 JULY-2017 Calculate the current through 2 Ω
resistor for the circuit as shown in Figure 1.78 using
source transformation
JAN-2018 Use source transformation to convert as 17Ω
shown in Figure 1.77 to a single current source in 4Ω +Vx −
parallel with single resistor
3Vx I 2Ω
5A 3Ω 7Ω 9Ω 1A
4Ω 4Ω
a
5A 4Ω
9 V +- 9 V +- Figure 1.78: JULY-2017-Question Paper
b Solution: Replace the each current source 5 A in
4Ω parallel with 3 Ω resistor by voltage source, 1 A in
parallel with 9 Ω resistor by voltage source and 3 Vx
Figure 1.77: JAN-2018-Question Paper
current sources in parallel with resistor of 17 Ω by
Solution: voltage source which is as shown in Figure 1.79
JULY-2016 Using source transformation find the

current through RL in the circuit as shown in Figure
V1 = I × R = 5 × 3 = 15V
1.83
V2 = I × R = 9 × 1 = 9V 22 V 5Ω
V1 = I × R = 3Vx × 17 = 51V
48 V 12 Ω
4Ω
51Vx
17Ω +Vx −
- + 12 V 4Ω
3Ω I 2Ω 9Ω
7Ω
15 V +- + 9V
-
R L = 10Ω
Figure 1.79: JULY-2017-Question Paper

From the figure it is observed that Vx = 2 × I
Solution:
Replace the voltage source of 15 V in series (3+4) Ω
resistor by current source Solution:
15 Replace the each voltage sources into current sources
I= = 2.14 A
7 in parallel with resistor which is as shown in Figure
1.84
51Vx
17Ω +Vx −
- + 22 / 5 A
I 2Ω 9Ω
2.14 A 7Ω 7Ω
+ 9V 5Ω
-
4A
7 Ω and 7 Ω are in parallel replace by single resistor 12 Ω
7×7
R= = 3.5 Ω 3A
7+7
4Ω
51Vx
17Ω +Vx −
- +
I 2Ω 9Ω R L = 10Ω
2.14 A 3.5 Ω
+- 9 V
Figure 1.81: JULY-2017-Question Paper Each current sources are in parallel, add these
current source and replace the parallel resistor by
Replace the current by source voltage source in series
single resistor, the modified circuit is as shown in
3.5 Ω resistor
Figure 1.85
V = 2.14 × 3.5 = 7.49V
22
I = I1 + I2 + I3 = + 4 + 3 = 11.4A
17Ω +Vx −
51Vx 5
- +
3.5Ω I 2Ω 9Ω 1 1
= +
1 1
+ = 0.533
R 5 12 4
7.49 V +- +- 9 V
1
R= = 1.875
0.533
From the figure it is observed that Vx = 2 × I 11.4A
Apply the KVL for the circuit by using Vx = 2 × I
1.875 Ω
7.49 − 51Vx − 9 − 29.5I = 0
−1.51 − 51 × 2I − 29.5I = 0 R L = 10Ω
−1.51 − 102I − 29.5I = 0

−1.51 − 131I = 0
1.51 Current source in parallel with resistor is replaced
I= = 11.5 mA voltage source which is as shown in Figure 1.86
131
21.375 V Figure 1.89

1.875 Ω
Replace current sources 10 A and in parallel with 10
3
A by single current source and parallel resistors by
R L = 10Ω single resistor
Figure 1.86: JULY-2016-Question Paper 10 40

I = 10 + =
Current through RL is, by KVL 3 3
21.375 − I × 11.875 = 0
2×3 6
21.375 R= = Ω
I= = 1.8A 2+3 3
11.875
JULY-2014 Using source transformation find the
power delivered by 50 V source i given network of 6
as shown in Figure 1.87 5Ω Ω
40 5
A
50 V +- 3
5Ω 3Ω
10 A
50 V +- +- 10 V
2Ω Figure 1.90
Replace the current source 403 A in parallel with 3
10
Figure 1.87: JULY-2014-Question Paper resistor by a voltage source in series with resistor
Solution: 6
R= = 1.2Ω
The redrawn circuit is as shown in Figure 3
5Ω 2Ω 3Ω
10 A 5Ω 1.2 Ω
50 V +- +- 10 V
50 V +- +- 16 V
Figure 1.88
Replace the 10 V voltage source in series with 3 Ω Figure 1.91
by current source in parallel with 3 Ω resistor. By applying KVL in the circuit is
V 10 50 − 16 − I × 6.2 = 0
I= = A
I 3
34
I= = 5.48A
5Ω 2Ω 3Ω 6.2
10
A
50 V +-
10 A 3 The power delivered by 50 V source is
P = 50 × I = 50 × 5.48 = 274W
1.4. Source Shifting & Transformation Techniques Chapter 1. Module 1 Basic Concepts
1.4 Source Shifting & Transformation Techniques

1 Use source shifting and transformation techniques
to find voltage across 2 Ω resistor as shown in Figure 2A
1 3
2 A 4 A 4
1.92 4 4
3
Figure 1.96
1 4
3A 2 2
2A +
2 - 3V
Figure 1.97
The voltage across 2 Ω resistor is
Figure 1.92: 2018-DEC
2
I = 3 = 1.5 A
2+2
Solution: V = 1.5 × 2 = 3V
3 3V
+-
to find voltage across a, b resistor as shown in Figure
1 4 3V 1.98
+- 3
2A 2
2A
Figure 1.93 6 9
3 3V 5A
+- b a
I1
2V 1 3V 2 8
4
+- +-
2A 2
Solution:
3
Figure 1.94
4 1V
+-
6 9
4 3V
5A
+-
b 5A a
2A 2
2 8
3 60 V
c a
-+
6 3 6
30 V 9
+
12  15 
-
b + a 8
10 V
- 8
2
b
Figure 1.100
3
Solution:
2A 60 V
c a
-+
- + 60 V
15 
b a 3
12  6
15 
1A
8
10  b
3 a
2  10 A
15 
6 12 
b a 4A
1A 8
b
Figure 1.102
Figure 1.105
The voltage across a, b resistor is
a
I = 1
3
= 0.3333 A 2
6+3 -
V = 0.3333 × 6 = 2V 20 V +
15 
12 
4A
8
b
1.103
Figure 1.106 18 
Vx
a 15 V 15 V
3 7
-+ +- -+
3
6 15 V
+ 30 V
10  15  - 6
12  4
2A 4A
18 
b
Vx
Figure 1.107
10 
a 3A 5A
4
6A
10  2
b
The voltage across a, b 18 
resistor is
Vx
V = 6 × 4 = 24V
5 2
+- -+
15 V 10 V
4 Use source shifting and transformation techniques The voltage across 18Ω resistor is
1.103 5
i = = 0.2A
18  25
V = 0.2 × 15 = 3.6V
Vx
3 7 3
DEC 2018 (2017 scheme) DEC 2018 (2015 scheme)
15 V 1 a) Reduce the network shown in Figure 1.108 to a
5A +
- 6 single voltage source in series with a resistance using
6 source shift and source transformations..
4 Solution:
8
2 3
Solution: A B
45 A 30 V +-
18 
Vx 5 6
15 V 15 V 10 
3 7
+- -+
3
6
5A 5A
6 The modified circuit diagram is as shown in Figure
4
1.109 by shifting the voltage and current sources.
45 A 8  30 V 30 V 90 A 16  60 V 60 V
+-
+-
3
-+
-+
6
A 2 B A 4 B
5 10 
6 12
45 A 10  90 A 20 
16  60 V 60 V
8  30 V 30 V
360 V -
+
90 V
-+
6
-
+-
+
-+ 3
+-
A 4 B
A 2 B
-+
10 
900 V
-+
5 12
225 V 6 20 
10 
Figure 1.115
Figure 1.110
20  6
10  15 A
3 A 10 A B
6A 30 
A 10 A B
15  12
30 A
6
15 A
Figure 1.116
Figure 1.111
A 12  4
-+ B
+-
A 6 2 180 V 40 V
-+ B
+-
54 V 20 V
A 140 V 16  B
-+
A 34 V
-+ 8 B
Figure 1.117
Figure 1.112
Aug 2020 (2018 scheme EE) 2 c) Apply source

transformation and shifting method to reduce the
Aug 2020 (2018 scheme EE) 2 c) Apply source network shown in Figure 1.118 to a single voltage
transformation and shifting method to reduce the source in series with a resistance.
network shown in Figure 1.113 to a single voltage
source in series with a resistance. Solution:
3
Solution:
16 
6 4 1
4
A B
90 A 60 V +-
18 V +- 2 6 Vx
10  12 
20 
Figure 1.113: 2018-DEC Figure 1.118: 2018-DEC
18 V 3 6V
-+
-+ 1.333 1
6A 3
6 Vx
18 V 4 1
-+
Figure 1.121
2 6 Vx
6A 3 2.57 A 2.333 Vx
Figure 1.119 6
Figure 1.122
6
8.57 A 1.312 Vx
1
6A 3 4.5 A 4 2
6 Figure 1.123
Vx
1.312
Figure 1.120 I = 8.57 = 1.352 A
1.312 + 7
1.5. Star Delta Transformation Chapter 1. Module 1 Basic Concepts
1.5 Star Delta Transformation
Rxy Ryz Rzx (Rxy + Ryz + Rzx )

x Rx Ry +Ry Rz +Rz Rx =
( Rxy )2
P
x
Rxy Ryz
Rx Rx Ry + Ry Rz + Rz Rx = P Rzx = Rx Ryz
Rxy Rzx Ryz
Rx Ry + Ry Rz + Rz Rx
Ry y Ryz =
Rz z Rx
y Ryz Ry Rz
Ryz = Ry + Rz + (1.8)
z Rx
Rx Ry
Rxy = Rx + Ry + (1.9)
Figure 1.124 Rz
Rz Rx
Rxy (Ryz + Rzx ) Rzx = Rz + Rx + (1.10)
Rx + Ry = (1.1) Ry
Rxy + Ryz + Rzx Q 1) 2020-Aug (2018 scheme ECE) 2020-Aug
Ryz (Rzx + Rxy ) (2018 scheme EE) 2017-June Determine the
Ry + Rz = (1.2) Input resistance between PQ using star-delta
Rxy + Ryz + Rzx
transformation
Rzx (Rxy + Ryz )
Rz + Rx = (1.3) 5Ω A
Rxy + Ryz + Rzx P
Equation (1.1)-(1.2)
6Ω 18Ω
6Ω
Rxy Rzx − Ryz Rzx B C
Rx − Rz = (1.4)
Rxy + Ryz + Rzx 6Ω 18Ω
Q
Equation (1.3)+(1.4) D
6Ω E
2Rxy Rzx Figure 1.125

2Rx =
Rxy + Ryz + Rzx
Solution:
Rxy Rxz In the given circuit 6 Ω, 6 Ω and 18 connected
Rx = (1.5) between A,B and C are in delta connection, convert
Rxy + Ryz + Rzx
this into star network. The details are as shown in
Similarly Figure 1.126
Ryz Ryx
Ry = (1.6) RAB × RAC
Rxy + Ryz + Rzx RA = P
RAB
Rzy Rzx 6 × 18
Rz = (1.7) RA = = 3.6Ω
Rxy + Ryz + Rzx 30
6×6
RB = = 1.2Ω
30
Star to Delta Transformation 6 × 18
RC = = 3.6Ω
R2 xy Rxz Ryz 30
Rx Ry = P
( Rxy )2
5Ω A
2 R R
Ryz P
yx zx
Ry Rz = P
( Rxy )2 3.6Ω
2 R R
Rzx 3.6Ω
Rz Rx = P yx yz B C
( Rxy )2 1.2Ω
6Ω 18Ω
Q
Rxy Ryz Rzx (Rxy + Ryz + Rzx ) D
Rx Ry +Ry Rz +Rz Rx = 6Ω E
( Rxy )2
P
Figure 1.126 y
In the circuit 5 Ω and 3.6 Ω are in series, 6 Ω

and 1.2 Ω are in series and 3.6 Ω 18 Ω and 6 Ω are
in series. 4Ω
4Ω
8.6Ω 12 Ω
12 Ω
P
12 Ω
7.2Ω 27.6Ω z
x
4Ω
Q y
7.2 Ω 27.6 Ω are in parallel, which is in series

with 8.6 Ω. The net resistance is.
x z
3Ω
7.2 × 27.6 Figure 1.129

RP Q = 8.6 +
7.2 + 27.6
= 8.6 + 5.71 By looking from XZ terminal the 3 Ω resistance
= 14.31Ω is in parallel with series connection of 3 Ω and 3 Ω
resistance.
Q 2) For the circuit shown in Figure shown in Figure
1.128 determine the equivalent resistance between
3×6
any two terminals. RXZ = = 2Ω
3+6
Solution:
y
4Ω 4Ω 4Ω
4Ω
4Ω
x z
4Ω
Figure 1.128 Q 3) For the circuit shown in Figure shown in Figure

Solution: 1.130 determine the equivalent resistance between
In the given 4, 4, 4, and Ω are in star network, (1)A and B and (2) A and N.
convert this star network to delta network
A
Rx × Ry
Rxy = Rx + Ry + = 8 + 4 = 12Ω
Rz
4×4 6Ω 2Ω 10 Ω
Rxy = 4+4+ = 8 + 4 = 12Ω
4 4Ω 30 Ω
4×4 N
Ryz = 4+4+ = 8 + 4 = 12Ω
4
4×4 C
Rzx = 4+4+ = 8 + 4 = 12Ω B
4 15 Ω
As shown in the Figure 1.129, by looking by any Figure 1.130

two terminals 12 Ω is parallel with 4 Ω
Solution:
12 × 4
= = 3Ω Convert the internal star network to delta
12 + 4
network A
RA RB
RAB = RA + RB +
RC
RAB = 2 + 30 +
2 × 30
= 32 + 15 = 47Ω 6Ω 2Ω 10 Ω
4
4 × 30 4Ω 30 Ω
RBC = 4 + 30 + = 34 + 60 = 94Ω N
2
4×2
RCA = 4+2+ = 6 + 0.266 = 6.266Ω C B
30 15 Ω
A Figure 1.133
From the Figure 1.133 by looking from the
terminal resistance between A and N, there is a
N delta network between NCB which is replaced by
6Ω
10 Ω
6.266 Ω
star network.
47 Ω
94 Ω Rxy Rxz
Rx =
Rxy + Ryz + Rzx
B
C 30 × 15
15 Ω RB = = 9.18
30 + 15 + 4
Figure 1.131 4 × 15
RC = = 1.225
In the circuit 10 Ω is parallel with 47 Ω 30 + 15 + 4
30 × 4
RN = = 2.5
10 × 47 30 + 15 + 4
= = 8.245Ω
10 + 47
Similarly 94 Ω is parallel with 15 Ω A
94 × 15
= = 12.93Ω
94 + 15 2Ω
6Ω 10 Ω
Similarly 6 Ω is parallel with 6.266 Ω N
2.5 Ω
6 × 6.266
= = 3.06Ω
6 + 6.266
C 1.225 Ω 9.18 Ω B
The details are as shown in Figure ??
A Figure 1.134
In the network of 6 Ω and 1.225 Ω are in series,
10 Ωand 9.18 Ω are in series. The details are as
3.06 Ω shown in Figure 1.135
8.245 Ω
A
6Ω 2Ω 10 Ω
C 12.93 Ω B N
1.225 Ω 2.5 Ω 9.18 Ω
Figure 1.132
While looking from terminal AB the resistances Figure 1.135
12.93 and Ω and 3.06 Ω are in series and this
A
resistance is parallel with 8.245 Ω
7.225 Ω 2Ω 19.18 Ω
= 12.93 + 3.06 = 16Ω
N
2.5 Ω
8.245 × 16
RAB = = 5.44Ω
8.245 + 16 Figure 1.136
(ii)The resistance between A and N is The resistances 7.225 Ω and 19.18 Ω are in
parallel Figure 1.140

7.225 × 19.18
= = 5.25Ω
7.225 × 19.18
The details are as shown in Figure 1.137 12.44 × 18.667
RAB = 1.2 + + 3.3Ω
A 12.44 + 18.667
232.217
5.25 Ω
A = 1.2 + + 3.3Ω
2Ω 12.44 + 31.187
N 2Ω 7.75 Ω = 1.2 + 7.446 + 3.3 = 11.94Ω
2.5 Ω
N
Figure 1.137 Q 5) 2018-Dec (2010 scheme) Find the resistance of

The resistances 2.5 Ω and 5.25 Ω are in series. the circuit shown in Figure shown in Figure 1.128
The resistances 2 Ω 7.75 Ω are in in parallel and the between A and B
net resistance between A and N is A 10 Ω X 20 Ω Y 5 Ω
2 × 7.75
= = 1.59Ω
2 + 7.75 10 Ω 5Ω 15 Ω 25Ω
Q 4) Determine the equivalent resistance between

Z
terminals A and B
2Ω
6Ω
5Ω 30Ω
B
4Ω 10 Ω
20 Ω B
A
10 Ω Figure 1.141
Solution:
6Ω 9Ω 15 Ω
In the given network 1.141 there is delta network
between xyz and it is replaced by star network.
Figure 1.138
Solution: 20 × 5
RX = = 2.5Ω
40
20 × 15
4×6 RY = = 7.5Ω
RA = = 1.2Ω 40
20 15 × 5
4 × 10 RZ = = 1.875Ω
RB = = 2Ω 40
20
6 × 10
RC = = 3Ω A 10  5
20 X Y
2.5 
10  7.5  25
10 × 15
RA = = 3.33Ω N
45
1.875
10 × 20
RB = = 4.44Ω Z
45 2
20 × 15 5 30
B
RC = = 6.667Ω
45
Figure 1.142
6Ω
2Ω 4.44 Ω
In the circuit 1.142, 10 Ω 2.5 Ω are in series,
1.2 Ω 3.33 Ω similarly 5 Ω 7.5 Ω are in series. The details are as
A B
shown in Figure 1.143
3Ω 9Ω 12.5Ω N 37.5Ω
6.667 Ω A
Figure 1.139
15 Ω 3.875Ω
12.44 Ω
B
A 1.2 Ω 3.33Ω B
P 30Ω
18.667 Ω Figure 1.143
In the circuit 1.143 at NPB 37.5 Ω 3.875 Ω and

30 are in delta connection, convert into star network. 1×1
The details are as shown in Figure 1.144 RA = = 0.333Ω
1+1+1
A 12.5Ω N 1×1
RB = = 0.333Ω
2.04Ω 1+1+1
15.76Ω 1×1
15 Ω RC = = 0.333Ω
B 1+1+1
1.63Ω
P B
Figure 1.144 2Ω
0.333 Ω
In the circuit 1.144, 12.5 Ω, 2.045 Ω are in series, A 0.333 Ω C
similarly 1.635 Ω 15 Ω are in series. The details are
as shown in Figure 1.145
0.333 Ω 1Ω
14.54Ω
D
15.76Ω i 1Ω
1V
16.63Ω
Figure 1.147
Figure 1.145 The resistances 0.333 Ω and 2 Ω are in series and
0.333 Ω and 1 Ω are in series.
In the circuit 1.145 14.54 Ω, 16.63 Ω are in The resistances 2.333 Ω and 1.333 Ω are in
parallel, which is in series with similarly 15.76 Ω. parallel its equivalent resistance is
The net resistance between AB is.
2.333 Ω
14.54 × 16.63 0.333 Ω
RAB = 15.76 +
14.54 + 16.63
= 15.76 + 7.76
= 23.52Ω 1.333 Ω
1Ω
Q 6) For the circuit shown in Figure shown in
Figure 1.128 determine the current i using star delta 1V
transformation.
Figure 1.148
Solution:
B 2.333 × 1.333 3.109
1Ω = = = 0.848Ω
2.333 + 1.333 3.666
1Ω
2Ω Total network resistance is 1+0.333+0.848=2.181Ω
A C 1
1Ω I= = 0.458A
2.181
Q 7) Calculate the current in the 40 Ω resistance of
1Ω 1Ω the circuit shown in Figure shown in Figure 1.128
D using star delta transformation.
i 1Ω 20 Ω
A C
1V
5Ω B 5Ω
Figure 1.146
4V
As shown in Figure 1.146 there is delta network 40 Ω
between ABC, convert this to star network

Solution: Solution:
As shown in Figure 1.146 there is delta network As shown in Figure 1.152 there is delta network
between ABC, convert this to star network between ABC, convert this to star network
20 × 5 4×6
RA = = 3.333Ω RA = = 1.6Ω
30 15
5×5 4×5
RB = = 0.833Ω RB = = 1.3Ω
30 15
20 × 5 6×5
RC = = 3.333Ω RC = = 2Ω
30 15
A 3Ω 2Ω
C
3.33Ω
1.6 Ω
3.33Ω 15 V 1.3 Ω 20 V
4V N
0.833 Ω
2Ω
B
40 Ω
Figure 1.153
In the circuit 1.153, 3 Ω and 1.6 Ω are in series
Figure 1.150 and also 2 Ω and 1.3 Ω are in series. The details are
3.33Ω as shown in Figure 1.154
3.33Ω
3 Ω 1.6 Ω 1.3 Ω 2Ω
I1 I2
40.833Ω
4V 2Ω
15 V 20 V
I3

The total resistance of the network is
4.6 Ω I2 3.3 Ω
40.833 × 3.333 I1
= 3.333 + = 3.333Ω
40.833 + 3.333 I3
136
= 3.333 + = 3.333 + 3.08 + Ω 15 V 2Ω 20 V
44.163
= 6.41Ω
The total current flowing in the network is Figure 1.155

4
I1 = = 0.624A I3 = I1 + I2
6.41
Using branch current division method the current in KVL for loop 1
40 Ω is
15 − 4.6I1 − 2I3 = 0
3.333
I3 = 0.624A = 0.047A 15 − 4.6I1 − 2I1 − 2I2 = 0
44.163
15 − 6.6I1 − 2I2 = 0
Q 8) Determine the currents supplied by the each
6.6I1 + 2I2 = 15
battery using star-delta transformation
3Ω 4Ω KVL for loop 2
A B 2Ω
20 − 3.3I2 − 2I3 = 0
6Ω 5Ω 20 − 3.3I2 − 2I1 − 2I2 = 0
15 V 20 V
20 − 5.3I2 − 2I1 = 0
C
2I1 + 5.3I2 = 20
Simultaneous equations are Figure 1.156

6.6I1 + 2I2 = 15 (1.11) Solution:
40
2I1 + 5.3I2 = 20 (1.12) 3
20 18
A
50 80
6.6 2 6
∆ = = 6.6 × 5.3 − (2 × 2) = 34.98 − 4
2 5.3 B
=30.98 Figure 1.157

15 2
43

20 5.3 79.5 − 40 39.5
i1 = = = = 1.275A 38 80
∆ 32.96 30.98 A

6.6 15
56

2 20 132 − 30 102
i2 = = = = 3.292A B
∆ 30.98 30.98
The current supplied by each battery are 1.275A and
3.292A respectively 38 80
A B
9). Find the equivalent resistance between A and B 24.3

as shown in Figure 1.156 Figure 1.158
30 40
A
10
20 60 50 80 6.94 × 3.55
RAB = 1.87 +
6.94 × 3.55
B
= 38 + 24.3 + 80 = 142.3Ω
1.6 Question Papers network
Question Papers
6 × 5.17
2019-Jan (2017 scheme ECE) Find the equivalent RA = = 1.87Ω
6 + 5.17 + 5.35
resistance between a and b as shown in Figure 1.159
5.17 × 5.35
using star delta transformation RB = = 1.67Ω
16.52
A 6 × 5.35
RC = = 1.94Ω
6 4 16.52
3 5
A
5 8 4 1.87
1.94 1.674
B
5 1.88
Figure 1.159: JAN-2019-Question Paper
Solution:
As shown in Figure 1.159 there is delta network B
between 8Ω 5Ω and 4 Ω, convert this to star network
8×5 Figure 1.162
RA = = 2.35Ω
8+4+5
5×4 A
RB = = 1.17Ω 1.87
17
8×4
RC = = 1.88Ω
17
A 6.94 3.55
6 4
B
3
2.35 1.17 Figure 1.163
5
1.88
6.94 × 3.55
B RAB = 1.87 +
6.94 × 3.55
= 1.87 + 2.34 = 4.21Ω
Figure 1.160
A
6 5.178 2019-DEC Determine the resistance between A and

B of the network shown in Figure 1.164.
10  12.5 
5.35
5 1.88 10 
A 12.5 
30
10  12.5 
B
B
Figure 1.161
As shown in Figure 1.161 there is delta network
between 6Ω 5.17Ω and 5.35 Ω, convert this to star Solution:
E A
10  12.5 
1.74 
10 
A 12.5 
C D 3.41  7.12
30
10  12.5  E D
B 6 12.5 
Figure 1.165
B
Figure 1.168
10 × 30 A
RA = = 6Ω
10 + 10 + 30
RB =
10 × 30
= 6Ω 1.74 
50
10 × 10
RC =
50
=2 9.41  19.62 
22.5 
B
A 12.5  Figure 1.169

D
6
12.5 
C 9.41 × 19.62
2 RAB = 1.74 + = 1.74 + 6.36
6
9.41 × 19.62
= 8.1Ω
B
Figure 1.166 Also by observation its a wheatstone bridge because

the ratio is
A 12.5 6
= =1
12.5 6
12.5 
6 Hence no current flows between E D
24.5 
E D RAB =
25 × 12
= 8.1Ω
25 × 12
6 12.5 
B
2019-DEC Determine the resistance between A and
Figure 1.167 B of the network shown in Figure 1.164.
6 8
between 6Ω 12.5Ω and 24.5 Ω, convert this to star 5
A
network 2
6 × 12.5 3 10
RA = = 1.74Ω 4
6 + 12.5 + 24.5
B
6 × 24.5
RB = = 3.41Ω
43 Figure 1.170: 2019-June-(2015-scheme-ECE)1
12.5 × 24.5
RC = = 7.12Ω Solution:
43
6 D 8 6 8
C 5 E
A A 2.77
2
2.83
4
3 10
1.67 4
B B

As shown in Figure 1.171 there is star network
6
D
2×3 10.77
RAB = 2 + 3 + = 5 + 1.2 = 6.2Ω A
5 2.83
2×5 E
RAE = 2+5+ = 7 + 3.3 = 10.3Ω 1.67 4
3
5×3 B
RBE = 5+3+ = 8 + 7.5 = 15.5Ω
2
Figure 1.175
As shown in Figure 1.175 there is a delta network
6 8 between 2.83Ω 6Ω and 10.77Ω, convert this to star
network
10.33
A
2.83 × 6
RA = = 0.87Ω
19.6
6.2 15.15 10 4 2.83 × 10.77
RE = = 1.55
B 19.6
10.77 × 6
RD = = 3.3
Figure 1.172 19.6
3.3
10 × 15.5
=
10 × 15.5
= 6.07Ω
A
0.87
1.55
6 8 1.67 4
D B
10.33 E
A
Figure 1.176
6.2 6.07 4 A
B
0.87
Figure 1.173
As shown in Figure 1.173 there is a delta network 3.22 7.3
between 6.2Ω 10.33Ω and 6.07 Ω, convert this to star B
network
Figure 1.177
6.2 × 10.33
RA = = 2.83Ω
6.2 + 10.33 + 6.07
6.2 × 6.07
RB = = 1.67Ω 3.22 × 7.3
22.6 = 0.87 + = 0.87 + 2.23
6.07 × 10.33 3.22 + 7.3
RE = = 2.77Ω = 3.1Ω
22.6
2018-Dec,2014-Dec Using star/delta transformation network

determine the resistance between M and N of the
1.1 × 5.636
network shown in Figure 1.178. RB = = 0.528Ω
11.736
1 3 1.1 × 5
M RN = = 0.468Ω
11.736
5.636 × 5
2 RC = = 2.4Ω
4 11.736
6 7
1 A
N M
5 0.667 0.528
7
Figure 1.178: 2018-DEC-(2015-scheme-ECE)1
0.468 2.433
Solution:
1 3 B N
A
M C
2 4 Figure 1.182
7
6
1 A 1 A
M M
N
5 C
1.06
1.195 9.43
Figure 1.179
As shown in Figure 1.179 there is delta network 0.468 0.468
between 2Ω 3Ω and 6 Ω, convert this to star network N N C
2×3
RA = = 0.545Ω Figure 1.183
2+3+6
3×6
RB = = 1.636Ω
11 1.195 × 9.43
6×2 RB = = 1.06Ω
RC = = 1.1Ω 1.195 + 9.43
11
B RM N = 1 + 1.06 + 0.468 = 2.528Ω

1 A
M
1.636
2015-Dec Find the equivalent resistance between A
0.545
4 7 and B of the network shown in Figure 1.184 using
Star-Delta transformation.
1.1 8 10 9
B
N
7 5 9 6
5 C
A
Figure 1.180
8 10 3
1 A
M
Solution:
0.545 C
10 E 9
B 7 B
1.1 5.636 15 5 9
A
9
N D F
5 C 8 10

As shown in Figure 1.182 there is delta network As shown in Figure 1.185 there is delta network
between 1.1Ω 5.636Ω and 5 Ω, convert this to star between 15Ω 8Ω and 5 Ω, and there is delta network
between 9Ω 9Ω and 9 Ω, convert this to star network A
30 30
15 × 8
RA = = 4.28Ω 30
28 C D
5×8
RD = = 1.42Ω 30 30 30
28 30
15 × 5 30 30
RC = = 2.67Ω B
E F
28
9×9
RB = = 3Ω Figure 1.189: 2013-DEC1
27
9×9 A
RE = = 3Ω
27 10
9×9
RF = = 3Ω 10 10
27
10
10 60
30
2.67 3 10 10
4.28 3
B
E F
B
A 3
1.42
10 Figure 1.190: 2013-DEC1
Figure 1.186 A
10
15.67
4.28 3
A B 20 10
14.42
10 20
10
B
4.28 7.51 3
A B Figure 1.191: 2013-DEC1
Figure 1.187 A A
10 10
15.67 and 14.42 Ω are in parallel
20 40 13.33
15.67 × 14.42
= = 7.51Ω
15.67 × 14.42
10
10
B B
RAB = 4.28 + 7.51 + 3 = 14.79Ω

Figure 1.192: 2013-DEC1
2013-DEC1 Using star/delta transformation deter- 2008-June Using star/delta transformation deter-
mine the resistance between M and N of the network mine the resistance between M and N of the network
shown in Figure 1.188. shown in Figure 1.193.
A A
30 30 5 5
30 10
30 30 30 5 10 5

30 10
30 30 5 5
B B
Figure 1.188: 2013-DEC1 Figure 1.193: 2008-June

Solution: Solution:
A Figure 1.197: 2008-June24

5 5
10
C D RAB = 1.25 + 3.33 + 1.25
5 10 5 = 5.83Ω
10
5 5
B
E F
2007-July Using star/delta transformation deter-
Figure 1.194: 2008-June mine the resistance between A and B of the network
As shown in Figure 1.194 there is delta network shown in Figure 1.198.
between A C D 5Ω 5Ω and 10 Ω, and there is delta
18
network between B C E 5Ω 5Ω and 10 Ω, convert C
A
this to star network 6 6
5×5
RA = = 1.25Ω
20 6 54 54 54
5 × 10
RC = = 2.5Ω B
20
5 × 10 Figure 1.198: 2007-July
RD = = 2.5Ω
20
5×5 Solution:
RB = = 1.25Ω As shown in Figure 1.198 there is star network
20
5 × 10 between 6Ω 6Ω and 6 Ω, convert this to star to delta
RE = = 2.5Ω network
20
5 × 10 6×6
RF = = 2.5Ω RAB = 6 + 6 + = 12 + 6 = 18Ω
20 6
6×6
A
RAC = 6 + 6 + = 12 + 6 = 18Ω
6
1.25 6×6
RBC = 6 + 6 + = 12 + 6 = 18Ω
2.5 2.5 6
2.5
10
10 18
1.25
A C
B 2.5
E F 18
18 18 18

Figure 1.195: 2008-June22
B
A
1.25 Figure 1.199: 2007-July
18 Ω between AC and AB are in parallel
5 2.5
18 × 18
= = 9Ω
36
1.25 5
2.5 A
B
9
Figure 1.196: 2008-June23 18 9
A A
B
1.25 1.25
Figure 1.200: 2007-July
5 10 3.33
9Ω and 9Ω are in series which is in parallel with
1.25
18 Ω
1.25
B B 9×9
AB = 18 + = 18 + 4.5 = 22.5Ω
18
Using star/delta transformation determine the Figure 1.202

resistance between A and B of the network shown in
Figure 1.201. 7kΩ
P
4kΩ A
P
14kΩ 18kΩ
6kΩ 18kΩ
12kΩ 2kΩ
B C Q
12kΩ 12kΩ
Q Figure 1.203
2kΩ D E
Figure 1.201 7kΩ

P
Solution:
4kΩ A
7.875kΩ
P
3kΩ Q
2kΩ
2kΩ 6kΩ
B C Figure 1.204
12kΩ 12kΩ
Q
2kΩ D E
RP Q = 7 + 7.785 + 2 = 16.785kΩ
1.7. Mesh Analysis Chapter 1. Module 1 Basic Concepts
1.7 Mesh Analysis The two simultaneous equations are:
Steps to find a current flowing in a circuit using 14i1 − 5i2 = 50 (1.1)

Mesh Analysis −5i1 + 15i2 = −25 (1.2)
1. Identify loops or meshes in a circuit and Label Multiply eqn 1.1 by 3 and adding with equation 1.2
a mesh current to N meshes 42i1 − 15i2 = 150
2. Apply KVL to each mesh with the correspond- −5i1 + 15i2 = −25
ing mesh current to generate N equations. −−−−− = −−
37i1 = 125
3. Solve the resulting simultaneous linear equa-
tions for the unknown mesh currents using i1 = 3.3784A i2 = −0.541A
Cramer’s Rule. iab = 3.3784A ieb = −i2 = 0.541A
ibc = i1 − i2 = 3.3784 − (−0.541) = 3.9194A
Q 1) In the circuit shown in Figure 1.257 determine voltage across the 5 Ω resistor is 5ibc = 19.597V
all branch currents and the voltage across the 5 Ω
resistor by loop current analysis.
3Ω 2Ω Q 2) In the circuit shown in Figure 1.207 determine
the mesh currents i1 , i2 , i3
+ b c f
+ -
- 5Ω 25V
50V
1Ω i2 2Ω
6Ω 8Ω 7V +- i1 d g
3Ω
Figure 1.205 6V +-
2Ω i3 1Ω
Solution: a h
e
In the given circuit there are two meshes and named
as i1 and i2 as shown in Figure 1.206.
Figure 1.207
a 3Ω b 2Ω e Solution:
+ Applying the KVL for the loop abcdea
+ i1 i2 -
- 5Ω 25V
50V 1(i1 − i2 ) + 2(i1 − i3 ) + 6 − 7 = 0
d 6Ω c 8Ω f 3i1 − i2 − 2i3 = 1
Figure 1.206 For the loop cfgdc

Applying the KVL for the loop abcda. The loop is 2i2 + 3(i2 − i3 ) + 1(i2 − i1 ) = 0
passing through 3, 5 and 6 Ω resistors and it touches −i1 + 6i2 − 3i3 = 0
negative terminal of the battery of 50 volts. The 5
Ω resistor is common to loop currents i1 and i2 . For For the loop dghed
loop1 the current i2 in 5 Ω resistor is opposite to the
3(i3 − i2 ) + 2(i3 − i1 ) + i3 − 6 = 0
i1 current. Hence loop1 equation is.
−2i1 − 3i2 + 6i3 = 6
3i1 + 5(i1 − i2 ) + 6i1 − 50 = 0 The three mesh equations are,
14i1 − 5i2 = 50
3i1 − i2 − 2i3 = 1
Similarly for the loop befcb , The loop2 is passing −i1 + 6i2 − 3i3 = 0
through 2, 8 and 5 Ω resistors and it touches positive −2i1 − 3i2 + 6i3 = 6
terminal of the battery of 25 volts. The 5 Ω resistor
is common to loop currents i1 and i2 . For loop2 Solving these equations Using Cramer’s rule
the current i1 in 5 Ω resistor is opposite to the i2     
3 −1 −2 i1 1
current. Hence loop2 equation is.  −1 6 −3   i2  =  0 
−2 −3 6 i3 6
2i2 + 8i2 + 5(i2 − i1 ) + 25 = 0
−5i1 + 15i2 = −25 ZI = V
 
i1 Figure 1.208
I =  i2  Solution:
i3
  Applying the KVL for the loop abcda
v1
V =  v2  8i1 + 4(i1 − i2 ) − 42 = 0
v3 12i1 − 4i2 = 42
 
Z11 Z12 Z13
Similarly for the loop befcb
Z =  Z21 Z22 Z23 
Z31 Z32 Z33 6i2 + 4(i2 − i1 ) − 10 = 0
 
v1 −1 −2 −4i1 + 10i2 = 10
 v2 6 −3 
v3 −3 6 Simultaneous equations are
i1 =
∆
12i1 − 4i2 = 42
where ∆ is
−4i1 + 10i2 = 10
3 −1 −2

∆ = −1 6 −3

12 −4
−2 −3 6 ∆ = = 12×10−(−4×−4) = 120−16 = 104
−4 10
3[6 × 6 − (−3 × −3)] + 1[−1 × 6 − (−2 × −3)]
−2[−1 × −3 − (−2 × 6)]

42 −4

=3(36-9)+1(-6-6)-2(3+12)=81-12-30=39 10 10 420 + 40 460
i1 = = = = 4.42A
∆ 104 104

1 −1 −2
12 42
0 6 −3
−4 10 120 + 168 288
6 −3 6 1(36 − 9) + 1(18) − 2(−36) i2 = = = = 2.769A
i1 = = ∆ 104 104
∆ 39
27 + 18 + 72 the current Ix
= 3A
39 8A
8Ω c I x 2Ω e

3 1 −2 b 10Ω
g
−1 0 −3

−2 6 6 3(18) − 1(−6 − 6) − 2(−6) +
- 4Ω 3Ω 5Ω
i2 = = 100V
∆ 39
a d f h
54 + 12 + 12
= 2A
39 Figure 1.209
Solution: For the given circuit there is current

3 −1 1
source, to apply KVL current source has to be
−1 6 0
converted into voltage source, the modified circuit
−2 −3 6 3(36) + 1(−6) + 1(3 + 12)
i3 = = is as shown in Figure 1.262
∆ 39
b 8Ω c I x 2Ω e 10Ω 80V
g
+-
108 − 6 + 15
= 3A
39 +- i1 4Ω i2 3Ω i3 5Ω
i1 = 3A i2 = 2A i3 = 3A 100V
Q 3) In the circuit shown in Figure 1.255 determine a d f h
all branch currents by mesh current analysis.
8Ω b 6Ω Figure 1.210
a e
Applying the KVL for the loop abcda
+-
+- i1 4Ω i2 10V
42V 8i1 + 4(i1 − i2 ) − 100 = 0
d c f
12i1 − 4i2 = 100
For the loop cefdc 5Ω 5Ω 5Ω
2i2 + 3(i2 − i3 ) + 4(i2 − i1 ) = 0 10 A 10Ω 20Ω 30Ω +

- 100 V
−4i1 + 9i2 − 3i3 = 0
For the loop eghfe Figure 1.211

Solution:
10i3 + 5i3 + 3(i3 − i2 ) + 80 = 0
Converting current source to voltage source and
0i1 − 3i2 + 18i3 = −80
then 10 Ω is in series with 5 Ω. The redrawn circuit
The three mesh equations are, is as shown in Figure 1.212
15 Ω 5Ω 5Ω
12i1 − 4i2 + 0i3 = 100
−4i1 + 9i2 − 3i3 = 0 100 V +- I1 20Ω I 2 30Ω
I3 +
- 100 V
0i1 − 3i2 + 18i3 = −80

12 −4 0 Figure 1.212

∆ = −4 9 −3 Applying the KVL for the loop
0 −3 18
35i1 − 20i2 + 0i3 = 100
12[162 − 9] + 4(−72 − 0)] + 0[12 − 0]
= 1836 − 288 = 1548 −20i1 + 55i2 − 30i3 = 0
0i1 − 30i2 + 35i3 = −100
100 −4 0

0
9 −3
−80 −3 18
35 −20

i1 = 0

∆ ∆ = −20 55 −30 = 35(1925−900)+20(−700)

0 −30 35
100(162 − 9) + 4(0 − 240)
=
1548 ∆ = 25875 − 14000 = 21875
15300 − 960 + 0 14340
= = 9.26A

100 −20 0
1548 1548
0
55 −30

12 100 0
−100 −30 35

−4
i1 =
0 −3 ∆
0 −80 18 100(1925 − 900) + 20(−3000)
i2 = =
∆ ∆
42500
= = 1.9428A
12(0 − 240) − 100(−72 − 0) + 0(0 + 720) 21875
=
1548
35 100 0

−2880 + 7200 + 0 4320 −20 0 −30
= = 2.79A
1548 1548 0 −100 35
i2 =
∆
12 −4 100

−4 9
−105000 + 2070000)
0 =

0 −3 −80
∆
i3 = −35000
∆ = = −1.6A
21875

35 −20 100
12(−720 − 0) + 4(320 − 0) + 100(12 + 0)
=

−20 55 −30
1548
0 −30 −100
−8640 + 1280 + 1200 −6160 i3 =
= = −3.97A ∆
1548 1548 −192500 + 40000 + 60000
i1 = 9.26A i2 = 2.79A i3 = −3.97A =
∆
Q 1) In the circuit shown in Figure 1.211 determine −92500
all branch currents by mesh current analysis. = = −4.2285A
21875
Q 2) In the circuit shown in Figure 1.213 determine Figure 1.216

all branch currents by mesh current analysis. Applying the KVL for the loop
10 Ω 10 Ω
30i1 − 10i2 − 10i3 = 200
−10i1 + 30i2 − 10i3 = 0
10 Ω
10 A −10i1 − 10i2 + 44i3 = 0
10 Ω
+
100 V -
10 Ω 24Ω i1 = 8.997A i2 = 3.97A i3 = 2.941A
Figure 1.213 Q 3) In the circuit shown in Figure 1.217 determine

the current I by mesh current analysis.
Solution:
Converting 100V voltage source to current 1 1
source and then 10 Ω is in parallel with current
source. The redrawn circuit is as shown in Figure
1A
1
1.214 2
10 Ω +
-+
2V - 2V I
1
10 Ω 1
10 A
10 Ω
10 A
10 Ω Figure 1.217
10 Ω 24Ω Solution:
Converting 2 V voltage source to current source
Figure 1.214 and then 1 Ω is in parallel with current source. The
10 A current source and 10 A current source are redrawn circuit is as shown in Figure 1.218
in parallel which are added and become 20 A 1
10 Ω
1A
1
2
10 Ω 2A -+
1
10 Ω 2V
20 A 1
10 Ω 1
10 Ω 24Ω
Figure 1.218
Figure 1.215 2 A current source and 1 A current source are in

parallel which are added and become 3 A
Converting 20 A current source to voltage source
and then 10 Ω is in series with voltage source. The 1
redrawn circuit is as shown in Figure 1.216
10 Ω 1
10 Ω 2
3A -+
1
2V
I2 1
10 Ω 1
10 Ω
I1
Figure 1.219
10 Ω
+
200 V -
Converting 3 A current source to voltage source
I3 24Ω
and then 1 Ω is in series with voltage source. The
redrawn circuit is as shown in Figure 1.216
1 1
Vx = 3(i3 − i2 )
1 I2 i1 = 10A
2 Vx
-+ i3 − i1 = = 0.125Vx
+ 8
3V - I1 2V I i3 − 10 = 0.125 × 3(i3 − i2 )
I3 1
1 i3 − 10 = 0.375i3 − 0.375i2
0.375i2 + 0.625i3 = 10
Figure 1.220
Applying KVL to mesh i2
Applying the KVL for the loop
4i2 + 3(i2 − i3 ) + 5(i2 − i1 ) = 0
3i1 − i2 − i3 = 3 −5i1 + 12i2 − 3i3 = 0
−i1 + 4i2 − 2i3 = −2 −5 × 10 + 9i2 − 4i3 = 0
−i1 − 2i2 + 4i3 = 2 12i2 − 3i3 = 50
The simultaneous equations are

fx-100MS/fx-115MS/fx-570MS/fx-991MS/
0.375i2 + 0.625i3 = 10
Press MODE MODE MODE 1 EQN 3 12i2 − 3i3 = 50

0.375 0.625
Then a1 3= b1 -1= c1 -1= d1 3= ∆= = −1.125 − 7.5 = −8.6255
12 −3

Then a2 -1= b2 4= c2 -2= d2 -2= 10 0.625

50 −3 −30 − 31.25 −61.25
i2 = = = = 7.1A
Then a3 -1= b3 -2= c3 4= d1 2= ∆ −8.6255 −8.6255
x=1.5

0.375 10
y=0.4166

12 50 18.75 − 120 −101.25
z=1.0833 i3 = = = = 11.74A
∆ −8.6255 −8.6255
The current through 2 Ω resistor I is The three loop currents i1 = 10A, i2 = 7.1A, i3 =
11.74A
I = I3 − I2 = 1.0833 − 0.4166 = 0.6667A
Q 5) Find the loop currents i1 , i2 , i3 in the circuit

i2 4Ω
5Ω
i1 3Ω
10 A + Vx -
Vx
i3 2Ω
8
6Ω
Figure 1.221
Solution:
1.8. Supermesh Analysis Chapter 1. Module 1 Basic Concepts
1.8 Supermesh Analysis 10 V

-
+
• If there is a current source between meshes 2 2
which is common for meshes, then we have to
combine meshes which is called a supermesh. 2A
4 8
Ix
• In supermesh write a KVL equation combining
meshes.
Figure 1.225
Q 1) In the circuit shown in Figure 1.222 determine Solution:
the current through 2 Ω resistor. 10 V
-
4Ω 6Ω
+
I1
+
10 V - 2A + 2 2
- 6V
2 A I3
2Ω 4 I2 8
Figure 1.222
Solution: Figure 1.226
4Ω 6Ω KVL for I1 is
+ 2A I2
+
10 V - I1 - 6V
2(I1 − I2 ) + 2(I1 + I3 ) − 10 = 0
2Ω
4I1 − 2I2 + 2I3 = 10
Figure 1.223
KVL for supermesh is
I1 and I2 are the two meshes in which 2 A current
source is common to I1 and I2 meshes which is as 4I2 + 2(I2 − I1 ) + 2(−I3 − I1 ) − 8I3 = 0
shown in Figure 1.223. −4I1 + 6I2 − 10I3 = 0
Draw a supermesh by combining two meshes I1 and
I2 as shown in Figure 1.224. For current source
4Ω 6Ω
I2 + I3 = 2
+ 2A I2
+
10 V - I1 - 6V
2Ω 4I1 − 2I2 + 2I3 = 10

−4I1 + 6I2 − 10I3 = 0
Figure 1.224
0I1 + I2 + I3 = 2
Apply KVL to the supermesh as
Solving the above equations
4I1 + 6I2 + 6 + 2I2 − 10 = 0
4I1 + 8I2 = 4 I1 = 3.667A I2 = 2.167A, I3 = 0.167A
I2 − I1 = 2 Current through 8 Ω resistor is I3 = 0.167A

4I1 + 8I2 = 4 the current Ix
I1 − I2 = −2 1A
Solving the above equations 2 2
I1 = −1A I2 = 1A 2A
6
+
10 V -
Current through 2 Ω resistor is I2 = 1A
Q 2) In the circuit shown in Figure 1.225 determine Figure 1.227

the current Ix Solution:
1A
i2 − i1 = 0.5i2
2 I1 2
i1 − 0.5i2 = 0
I2 2 A I3 6
+
10 V -
4i1 + 8i2 + 2i2 = 20

Figure 1.228 i1 − 0.5i2 = 0
I1 = −1A

KVL for Supermesh is 4 8
∆= = −2 − 8 = −10
1 −0.5
2(I2 − I1 ) + 2(I3 − I1 ) + 6I3 − 10 = 0
20 8
−4I1 + 2I2 + 8I3 = 10
0 −0.5 −10
−4(−1) + 2I2 + 8I3 = 10 i1 = = = 1A
∆ −10
2I2 + 8I3 = 10 − 4 = 6
i1 − 0.5i2 = 0
I3 − I2 = 2
1 − 0.5i2 = 0
I2 − I3 = −2
i2 = 2A

2I2 + 8I3 = 6 all the loop currents
I2 − I3 = −2A 3
Solving the above equations I3

2 4
I1 = −1A I2 = −1A, I3 = 1A
5A I2
+
1 I1 - 10 V
the current Ix
4 6 Figure 1.231
vx Solution:
+
I2
+
20 V - I1 4 vx 2
3
-
Figure 1.229 I3
2 4
Solution:
4 6 5A I2
+
1 I1 - 10 V
vx
+
I2
+
20 V - I1 4 vx 2
-
Figure 1.232
Figure 1.230: Example KVL for Supermesh
1i1 + 2(i1 − i3 ) + 4(i2 − i3 ) + 10 = 0
4i1 + 6i2 + 2i2 − 20 = 0 3i1 + 4i2 − 6i3 = −10
4i1 + 8i2 + 2i2 = 20 KVL for i3
3i3 + 4(i3 − i2 ) + 2(i3 − i1 ) = 0
vx = 2i2 −2i1 − 4i2 + 9i3 = 0
i2 − i1 = 5
vx 2i2 i2
i2 − i1 = = = i1 − i2 = −5
4 4 2
3i1 + 4i2 − 6i3 = −10 Q 7) In the circuit shown in Figure 1.235 determine
−2i1 − 4i2 + 9i3 = 0 the current Ix
i1 − i2 + 0i3 = −5 4Ω
I3 Ix
-j2 Ω
Solving the above equations 5∠0o A j10 Ω
+
I2 20∠90o V
-
I1 = 5.556A I2 = −0.556A, I3 = −1.4814A -j2 Ω
8Ω I1

the all the mesh currents Figure 1.235
Solution: Applying KVL to mesh I1
I2
2 2
4 8I1 + j10(I1 − I3 ) − j2(I1 − I2 ) = 0
I1 d
+ (8 + j8)I1 + j2I2 − j10I3 = 0
6V -
3A
8
1  I3 Applying KVL to mesh I2
Figure 1.233 −j2(I2 − I1 ) + 4I2 − j2(I2 − I3 ) + 20∠90 = 0

Solution: j2I1 + (4 − j4)I2 + j2I3 = −20∠90
For the given circuit there is current source, to apply For mesh I3 , I3 = 5A
KVL current source has to be converted into voltage
source, the modified circuit is as shown in Figure (8 + j8)I1 + j2I2 − j10I3 = 0
1.234 (8 + j8)I1 + j2I2 − j50 = 0
b c
f (8 + j8)I1 + j2I2 = j50
I1 I2
2 2
4
d g j2I1 + (4 − j4)I2 + j2I3 = −20∠90
+
6V - 3A j2I1 + (4 − j4)I2 + j10 = −j20
I3 8
1 j2I1 + (4 − j4)I2 = −j30
a h
e
Figure 1.234 (8 + j8)I1 + j2I2 = j50

KVL for I2 is j2I1 + (4 − j4)I2 = −j30
2I2 + 4(I2 − I3 ) + 2(I2 − I1 ) = 0

8 + j8 j2
∆ = = 32−j32+j32+32+4 = 68
−2I1 + 8I2 − 4I3 = 0 j2 4 − j4

KVL for supermesh I2 is j50 j2

−j30 4 − j4 j200 + 200 − 60
2(I1 − I2 ) + 4(I3 − I2 ) + 8I3 − 6 = 0 i1 = =
∆ 68
2I1 − 6I2 + 12I3 = 6
140 + j200
=
68
I1 − I3 = 3
214.13∠55
I3 = I1 − 3 = = 3.15∠55
68

8 + j8 j50

j2 −j30 −j240 + 240 + 100 340 − j240
−2I1 + 8I2 − 4I3 = 0 I2 = = =
∆ 68 68
2I1 − 6I2 + 12I3 = 6
I1 + 0I2 − I3 = 3
416.17∠ − 35.2
= = 6.12∠ − 35.21 = 6.12∠144.78
Solving the above equations 68
I1 = 3.4736A I2 = 1.105A, I3 = 0.4736A Ix = −I2 = 6.12∠144.78
Q 8) Use loop analysis to find Vx in the circuit 10 Ω 8+j8 Ω 15 Ω

shown in Figure 1.236 such that the current through
2 + J3 Ω is zero.
50∠0o V I1 j10 Ω I 2 10 Ω I 3
5Ω 2+j3 Ω 4Ω Vx
30∠0o V I1 j5 Ω I 2 6Ω I3
Vx Figure 1.237
Solution: Applying KVL to mesh I1
Figure 1.236 105I1 + j10(I1 − I2 ) − 50∠0 = 0
Solution: Applying KVL to mesh I1 (10 + j10)I1 − j10I2 = 50∠0
5I1 + j5(I1 − I2 ) − 30∠0 = 0
Applying KVL to mesh I2
(5 + j5)I1 − j5I2 = 30∠0
(8 + j8)I2 + 10(I2 − I3 ) + j10(I2 − I1 ) = 0
Applying KVL to mesh I2
−j10I1 + (18 + j18)I2 −10I3 = 0
(2 + j2)I2 + 6(I2 − I3 ) + j5(I2 − I1 ) = 0
−j5I1 + (8 + j8)I2 −6I3 = 0 Applying KVL to mesh I3
Applying KVL to mesh I3 15I3 + 10(I3 − I2 ) + Vx = 0

0I1 − 10I2 + 25I3 = −Vx
4I3 + 6(I3 − I2 ) + Vx = 0
0I1 − 6I2 + 10I3 = −Vx

10 + j10 −j10 0
5 + j5 −j5 0

∆ = −j10 18 + j18 −10 =
∆ = −j5 8 + j8 −6 0 −10 25
0 −6 10
= (5 + j5)(80 + j80 − 36) + j5(−j50) (10 + j10)(450 + j450 − 100) + j10(−j250)
(5+j5)(44+j80)+250 = 220+j400+j220+400+250 (10 + j10)(350 + j450) + 2500

870 + j620 = 1068∠35.47 3500 + j4500 + j3500 − 4500 + 2500 = 1500 + j8000

5 + j5 30
0 1500 + j8000 = 8139.4∠79.38
−j5 0 −6

10 + j10 50 0
0 −Vx 10
I2 = = (5+j5)(−6Vx )−30(−j50) −j10 0 −10
∆
0 −Vx 25
Since I2 I2 =
∆
0 = (5 + j5)(−6Vx ) + j1500 Since I2
j1500 1500∠90
6Vx = = 0 = (10 + j10)(−10Vx ) − 50(−j250)
(5 + j5) 7.07∠45
Vx = 35.36∠45 (10 + j10)(−10Vx ) + j12500 =
j12500 12500∠90
Q 9) Use loop analysis to find Vx in the circuit 10Vx = =
shown in Figure 1.264 such that the current through (10 + j10) 14.14∠45
2 + J3 Ω is zero. Vx = 88.4∠45
1.9 Question Papers

Q 2019-DEC 1 b) Use mesh analysis determine the KVL for supermesh is
current I1 , I2 , I3 as shown in Figure 1.238
1(I1 − I2 ) + 3(I3 − I2 ) + 2I3 − 10 = 0
1
1I1 − 4I2 + 5I3 = 10
I3 I1 − I3 = 5
2 1
I1 + 6I2 − 3I3 = 0
1A
+
2V - I1
I2 2 I1 − 4I2 + 5I3 = 10
I1 + 0I2 − I3 = 5
Figure 1.238: 2019-DEC-1b(2018-scheme) Solving the above equations
Solution: I1 = 5.357A I2 = −0.714A, I3 = 0.357A
1
JAN-2018-CBCS 2-a Determine the loop currents
I3 I1 , I2 , I3 and I4 for the network shown in Figure
2 1 1.242.
+
1A 4V 4V
2V - I1
I2 2 - -
+
+
Figure 1.239 1Ω I1 3 A I2 1Ω
Q 2019-JAN) Using mesh analysis determine the 1Ω 1A

current I1 , I2 , I3 shown in Figure 1.240
I3 1 Ω I4 1Ω
I -
+
1 2 2
3 2A 5V
I1
+
10 V -
5A
2 Figure 1.242: 2018-CBCS-Question Paper
1  I3
Solution:
Figure 1.240 4V 4V
- -
+
+
Solution:
For the given circuit there is current source between 1Ω I1 3 A I2 1Ω

mesh1 and mesh2. Convert the modified circuit is
as shown in Figure 1.241
1Ω 1A
b c
I3 1 Ω 1Ω
f
I4
I2
I1 1 2
3 -
+
d g
10 V
+
- 5A 2A 5V
I3 2
1
h Figure 1.243
a e
Figure 1.241 I3 = 2
KVL for I2 is I2 − I1 = 3
−I1 + I2 = 3
2I2 + 3(I2 − I3 ) + 1(I2 − I1 ) = 0
−1I1 + 6I2 − 3I3 = 0 I2 − I4 = 1
Applying the KVL for the Supermesh The three mesh equations are,
−4 − 4 + 1I2 + 1I4 − 5 + 1(I4 − I3 ) 12i1 − 2i2 + 0i3 = 5

+1(I1 − I3 ) + I1 = 0 −2i1 + 34i2 − 2i3 = 0
2I1 + I2 − 2I3 + 2I4 = 13 0i1 − 2i2 + 12i3 = 10
2I1 + I2 − 2 × 2 + 2I4 = 13 Solving the above equations
2I1 + I2 + 2I4 = 17
I1 = 0.429A I2 = 0.075A, I3 = 0.8453A
Simultaneous equations are
Voltage across 20 Ω is
−I1 + I2 + 0I4 = 3
i2 × 20 = 0.075× = 1.56V
0I1 + I2 − I4 = 1
2I1 + I2 + 2I4 = 17
JULY 2017 CBCS) Use mesh analysis to determine
Solving the above equations the three mesh currents i1 , i2 , i3 as shown in Figure
1.246
I1 = 2A I2 = 5A, I3 = 2A, I4 = 4A b c f
JAN-2017-CBCS Find the voltage across 20 Ω 1Ω i2 2Ω

resistor in the network shown in Figure ?? by mesh 7V +- i1 d g
analysis. 3Ω
6V +-
b 10  c 10  e 10 g 2Ω i3 1Ω
a h
e
-
5 V +- 2 2 + 10 V
Figure 1.246
a d f h Solution:
20
Applying the KVL for the loop abcdea
Figure 1.244
1(i1 − i2 ) + 2(i1 − i3 ) + 6 − 7 = 0
Solution:
3i1 − i2 − 2i3 = 1
b 10  c 10  e 10 g
For the loop cfgdc
5 V +- i1 2 i2 2 i3 +- 10 V 2i2 + 3(i2 − i3 ) + 1(i2 − i1 ) = 0
−i1 + 6i2 − 3i3 = 0
a d f h
20
For the loop dghed
Figure 1.245
3(i3 − i2 ) + 2(i3 − i1 ) + i3 − 6 = 0
Applying the KVL for the loop abcda
−2i1 − 3i2 + 6i3 = 6
10i1 + 2(i1 − i2 ) − 5 = 0
The three mesh equations are,
12i1 − 2i2 = 5
3i1 − i2 − 2i3 = 1
For the loop cefdc −i1 + 6i2 − 3i3 = 0
−2i1 − 3i2 + 6i3 = 6
10i2 + 2(i2 − i3 ) + 20i2 + 2(i2 − i1 ) = 0
−2i1 + 34i2 − 2i3 = 0 Solving these equations
i1 = 3A i2 = 2A i3 = 3A
For the loop eghfe
July-2016 1-b Using mesh current method find
10i3 + 2(i3 − i2 ) − 10 = 0 current through 10 Ω resistor in the network shown
0i1 − 2i2 + 12i3 = 10 in Figure 1.247.
5Ω 1Ω 3Ω
I3
5Ω 5A 10 Ω
2Ω
4Ω
50 V +- I1 I2 1A
4Ω + Vx -
10 A Vx
A 5Ω
7
-
Figure 1.247: July-2016-Question Paper
+
Solution: Applying KVL to mesh I1 10 V
5Ω Figure 1.249
Solution:
I3 1Ω 3Ω
5Ω 5A 10 Ω
I2 i2
50 V +- I1 2Ω
4Ω 10 A 4Ω
i1
1A + Vx -
Figure 1.248: July-2016-Question Paper Vx
A i3 5Ω
7
-
+
10 V
I1 − I3 = 5
I2 = 10 Figure 1.250
Applying the KVL for the Supermesh Vx = 4(i3 − i2 )

i1 = 1A
5I3 + 10(I3 − I2 ) + 4(I1 − I2 ) − 50 + 5I1 = 0 Vx
i3 − i1 = = 0.142Vx
7
9I1 − 14I2 + 15I3 = 50
i3 − 1 = 0.142 × 4(i3 − i2 )
9I1 − 14 × 10 + 15I3 = 50
i3 − 1 = 0.568i3 − 0.568i2
9I1 + 15I3 = 190
0.568i2 + 0.432i3 = 1
Applying KVL to mesh i2

3i2 + 4(i2 − i3 ) + 2(i2 − i1 ) = 0
−2i1 + 9i2 − 4i3 = 0
I1 − I3 = 5
−2 × 1 + 9i2 − 4i3 = 0
9I1 + 15I3 = 190
9i2 − 4i3 = 2
Solving above equations The simultaneous equations are
0.568i2 + 0.432i3 = 1
9i2 − 4i3 = 2

I1 = 11.04A, I2 = 10A, I3 = 6.04A 0.568 0.432
∆= = 2.272 − 3.888 = −1.616
9 −4

0.568 1

JAN-2015) For the circuit shown in Figure 1.264 find 9 2 1.136 − 9 −7.864
the power supplied 10 V source using mesh analysis. i3 = = = = 4.866A
∆ −1.616 −1.616
The power dissipated by 10V source is Simultaneous equations are

10 × i3 = 10 × 4.866A = 48.66W 30i1 − 10i2 − 15i3 = 100
−10i1 − 440i2 + 450i3 ) = 0
−15i1 − 50i2 + 90i3 = 0

30
−10 −15

∆ = −10 −440 450

−15 −50 90
30[−39600+22500]+10[−900+6750]−15[500−6600]
= 30[−17100] + 10[5850] − 15[−6100]

= −513000 + 58500 + 91500 = −363000
DEC-2015) find the power delivered by the
dependent voltage source in the circuit shown in Bridge Network) Use mesh analysis to determine
Figure 1.251 by mesh analysis. the three mesh currents i1 , i2 , i3 also determine the
voltage drop across 2Ω for the circuit shown in
5Ω Figure 1.253
c
+
100 V - 15Ω
10Ω 3 i2 6
30 V +- i1 a b
10I x I 2
+
- x 50Ω 25Ω
6 i3 4
d
Figure 1.251
5Ω Solution:
Applying the KVL for the loop cadc
+ i1
100 V - 15Ω
3(i1 − i2 ) + 6(i1 − i3 ) − 30 = 0
10Ω i3
9i1 − 3i2 − 6i3 = 30
10ix ix
+
- i2 50Ω 25Ω For the loop cabc
6i2 + 2(i2 − i3 ) + 3(i2 − i1 ) = 0

Figure 1.252 −3i1 + 11i2 − 2i3 = 0
KVL for loop i1
For the loop abda
5i1 + 15(i1 − i3 ) + 10(i1 − i2 ) − 100 = 0
2(i3 − i2 ) + 6(i3 − i1 ) + 4i3 = 0
30i1 − 10i2 − 15i3 = 100
−6i1 − 2i2 + 12i3 = 0
ix = 50(i2 − i3 ) = 50i2 − 50i3
KVL for loop i2 The three mesh equations are,
10(i2 − i1 ) + 50(i2 − i3 ) − 10ix = 0 9i1 − 3i2 − 6i3 = 30

−10i1 + 60i2 − 50i3 − 10ix = 0 −3i1 + 11i2 − 2i3 = 0
−10i1 + 60i2 − 50i3 − 10(50i2 − 50i3 ) = 0 −6i1 − 2i2 + 12i3 = 0
−10i1 − 440i2 + 450i3 = 0 i1 = 6.6667A i2 = 2.5A i3 = 3.75A The voltage
drop across 2Ω resistor is
KVL for loop i3 V3 = (i3 − i2 )2
25i3 + 50(i3 − i2 ) + 15(i3 − i1 ) = 0 = (3.75 − 2.5)2
−15i1 − 50i2 + 90i3 = 0 = 2.5V
Bridge Network) Use mesh analysis to determine the Applying the KVL for the Mesh2
mesh currents i1 , i2 , i3 , i4 for the circuit shown in
Figure 1.254 10i2 + 10(i2 − i1 ) + 80(i2 − i3 ) − 30 = 0
−10i1 + 100i2 − 80i3 = 0
100i2 − 80i3 = −20
10 
2A
I1 10 
i4 − i3 = 10A
I2
20 V
-
Applying supermesh
+
80 
I3 80(i3 − i2 ) + 20i4 + 20 − 20 = 0
I4
10 A −80i2 + 80i3 + 20i4 = 0
20 
-
+
20 V
Solving simultaneous equations
100i2 − 80i3 + 0i4 = −20

Figure 1.254
0i2 − i3 + i4 = 10A
−80i2 + 80i3 + 20i4 = 0
i1 = −2A i2 = −5A i3 = −6A i4 = 4A
Important: All the diagrams are redrawn and solutions are prepared. While
preparing this study material most of the concepts are taken from some text books
or it may be Internet. This material is just for class room teaching to make
better understanding of the concepts on Network analysis: Not for any commercial
purpose
1.10. Node Analysis Chapter 1. Module 1 Basic Concepts
1.10 Node Analysis Q 2) In the circuit shown in Figure 1.257 determine

all branch currents and the voltage across the 5 Ω
Steps to find a current flowing in a circuit using Node resistor by node analysis.
Analysis
3Ω 2Ω
1. Identify nodes in a circuit and Label all the
nodes.
+-
+-
2. Select one of node as reference node. 5Ω 25V
50V
3. Apply KCL to each node.
6Ω 8Ω
4. Solve the resulting simultaneous linear equa-
tions for the unknown node voltages using Figure 1.257
Cramer’s Rule. Solution:
The circuit is labeled by nodes which is as shown in
5. Branch currents can be calculated using node Figure 1.258
voltages
3Ω V1 2Ω
+
all branch currents by node analysis. +- -
5Ω 25V
50V 6Ω 8Ω
8Ω 6Ω
2 (Ref)
+-
+- 4Ω 10V Figure 1.258
42V
Apply KCL to node V1
Figure 1.255
V1 − 50 V1 − 25 V1

+ + = 0
Solution: The circuit is labeled by nodes which 3+6 2+8 5
is as shown in Figure 1.256 [0.111 + 0.1 + 0.2] V1 − 5.556 − 2.5 = 0
8Ω V1 6Ω [0.4111] V1 = 8.056
V1 = 19.59
+-
+- 4Ω 10V Current through 3 Ω and 6 Ω resistor is
42V
V1 − 50 19.59 − 50
2 (Ref) I3 = = = −3.37A
3+6 3+6
Figure 1.256 Current through 2 Ω and 8Ω resistor is
Apply KCL to node V1 V1 − 25 19.59 − 25
I5 = = = −0.541A

V1 − 42 V1 V1 + 10
2+8 2+8
+ + = 0
8 4 6 Current through 5 Ω resistor is
[0.125 + 0.25 + 0.167] V1 − 5.25 + 1.67 = 0 8.056
[0.5416] V1 = 3.58 I5 = = 3.918A
5
V1 = 6.61 Voltage across 5 Ω resistor is
Current through 8 Ω resistor is V = I5 × 5 = 3.918 × 5 = 19.59V
42 − 6.61
I8 = = 4.42A Q 3) In the circuit shown in Figure 1.260 determine
8 all branch currents and the voltage across the 5 Ω
Current through 6 Ω resistor is resistor by node analysis.
10 + 6.61 1Ω 2Ω
I6 = = 2.77A
6
Current through 4 Ω resistor is 2A
10 V +- 5Ω 10Ω
6.61
I4 = = 1.65A
4

Solution: Solution:
1Ω V1 2Ω V2 For the given circuit there is current source, to apply
KCL current source has to be converted into voltage
source, the modified circuit is as shown in Figure
10 V +- 5Ω 10Ω 2A 1.262
8Ω V1 I x 2Ω V2 10Ω 80V
3 (Ref)
+-
Figure 1.260 +- 4Ω 3Ω 5Ω
Apply KCL at node V1 100V

V1 − 10 V1 V1 − V2
+ + = 0 Figure 1.262
1 5 2
V1 [1 + 0.2 + 0.5] − 10 − 0.5V2 = 0

V1 [1 + 0.2 + 0.5] − 0.5V2 = 10 V1 − 100 V1 V1 − V2
+ + = 0
1.7V1 − 0.5V2 = 10 8 4 2
[0.125 + 0.25 + 0.5] V1 − 12.5 − 0.5V2 = 0
Apply KCL at node V2 0.875V1 − 0.5V2 = 12.5

V2 V2 − V1
−2 = 0

+ V2 V2 − 80 V2 − V1
10 2 + + = 0
3 15 2
−0.5V1 + V2 [0.1 + 0.5] = 2
−0.5V1 + [0.33 + 0.067 + 0.5] V2 − 5.33 = 0
−0.5V1 + 0.6V2 = 2
−0.5V1 + 0.897V2 = 5.33
The simultaneous equations are The simultaneous equations are
1.7V1 − 0.5V2 = 10 0.875V1 − 0.5V2 = 12.5
−0.5V1 + 0.6V2 = 2 −0.5V1 + 0.897V2 = 5.33

1.7 −0.5
0.875 −0.5
∆ = = 1.02 − .25 = 0.25 = 0.77 ∆= = 0.7848 − .25 = 0.5348
−0.5 0.6 −0.5 0.897

10 −0.5
12.5 −0.5

2
5.33 0.897 11.212 + 2.665
0.6 6+1 V1 = =
V1 = = = 9.09 ∆ 0.5348
∆ 0.77
13.877
= = 25.94V
0.5348
1.7 10
0.875 12.5
−0.5 2 3.4 + 5
−0.5 5.33 4.66 + 6.25 10.91
V2 = = = 10.91
∆ 0.77 V2 = = = = 20.4V
∆ 0.5348 0.5348
V1 is the voltage across 5 Ω resistor which is 9.09 V
Q 5) In the circuit shown in Figure 1.261 determine V1 − V2 25.94 − 20.4
Ix = = = 2.77A
the current Ix 2 2
8A Q 6) Find the loop currents i1 , i2 , i3 in the circuit
b 8Ω c I x 2Ω e 10Ω
g 1Ω i1 2Ω
3Ω
+- 4Ω 3Ω 5Ω 15 A
+ Vx -
100V 3i1
a d f h
Figure 1.263 4Ω
V1
2Ω V2 8Ω
V1 V3
1Ω i1 2Ω ix
3Ω
V2 4Ω 2ix
15 A 3A
+ Vx -
3i1
3 (Ref) Figure 1.265

Solution:
Solution: Applying KCL to node V1
Figure 1.264
Applying KCL to mesh V1 V1 − V2 V1 − V3
+ −3 = 0
V1 − V2 V1 2 4
+ − 15 = 0 [0.5 + 0.25] V1 − 0.5V2 − 0.25V3 = 3
1 2
(1 + 0.5)V1 − V2 = 15 0.75V1 − 0.5V2 − 0.25V3 = 3
1.5V1 − V2 = 15 3V1 − 2V2 − V3 = 12
V1 Applying KCL to node V2

i1 =
2
Applying KCL to mesh V2 V2 − V1 V2 − V3 V2
+ + = 0
2 8 4
V2 − V1 V2 −0.5V1 + 0.875V2 − 0.125V3 = 0
+ − 3i1 = 0
1 3 −4V1 + 7V2 − V3 = 0
−V1 + [1 + 0.333]V2 − 3i1 = 0
−V1 + 1.333V2 − 3i1 = 0 V1 − V2
ix =
−V1 + 1.333V2 − 3(0.5V1 ) = 0 2
−2.5V1 + 1.333V2 = 0 Applying KCL to node V3
V3 − V1 V3 − V2
+ + 2ix = 0
1.5V1 − V2 = 15 4 8
−2.5V1 + 1.333V2 = 0 V3 − V1 V3 − V2 V1 − V2
+ +2 = 0
4 8 2
1.5 −1
= 2 − 2.5 = −0.5 0.75V1 − 1.125V2 + 0.375V3 = 0
∆ =
−2.5 1.333 2V1 − 3V2 + V3 = 0

15 1

0 1.333 Linear equations are
20
V1 = = = −40V
∆ −0.5 3V1 − 2V2 − V3 = 12
−4V1 + 7V2 − V3 = 0
1.5 15 2V1 − 3V2 + V3 = 0

−2.5 0 37.5
V2 = = = −75V
3 −2 −1

∆ −0.5
∆ = −4 7 −1 = 3(7−3)+2(−4+2)−1(12−14)
Power delivered by dependent current source is 2 −3 1
V1 −40 = 12 − 4 + 2 = 10
i1 = = = −20
2 2
V2 × 3i1 = −75 × 3 × −20 = 4.5kW

12 −2 −1

0 7 −1
Q 7) In the network shown in Figure 1.265 find the

0 −3 1 12(7 − 3)
node voltages using nodal analysis V1 = = = 4.8V
∆ 10

3 12 −1 = 133 − 60 − 7 = 66

−4 0 −1

2 0 1 −12(−4 + 2)
V2 = = = 2.4V
∆ 10
0 −4 −1
3 −2 12
100 3 −1

−4 7 0
0 −2 7

2 −3 0 12(12 − 14) V1 =
V3 = = = −2.4V ∆
∆ 10
node voltages using nodal analysis 4(700) − 1(−200) 2800 + 200)
= = = 45.45V
66 66
20 Ω

7 0 −1
5Ω 10 Ω

V2 −2 100 −1
V1 V3
−1 0 7

ix V2 =
∆
10 Ω 5Ω
10 A 7(700) − 1(100) 4900 − 100)
= = = 72.73V
66 66

7 −4 0
Figure 1.266
−2 3 100

−1 −2 0
V3 =
V1 − V2 V1 − V3 V1 ∆
+ + = 0
5 20 10
[0.1 + 0.2 + 0.05] V1 − 0.2V2 − 0.05V3 = 0 7(200) + 4(100) 1400 + 400)
= = = 27.27V
0.35V1 − 0.2V2 − 0.05V3 = 0 66 66
7V1 − 4V2 − V3 = 0
Q 9) In the network shown in Figure ?? find the
Applying KCL to node V2 current I by node voltage method.
V2 − V1 V2 − V3 5Ω A 2+j3 Ω B 4Ω
+ − 10 = 0
5 10
−0.2V1 + [0.2 + 0.1] V2 − 0.1V3 = 10
30∠0o V j5 Ω 6Ω
−0.2V1 + 0.3V2 − 0.1V3 = 10 20∠0o V
I
−2V1 + 3V2 − V3 = 100
Applying KCL to node V2

Figure 1.267
V3 − V1 V3 − V2 V3 Solution: Applying KCL to node VA
+ + = 0
20 10 5
−0.05V1 − 0.1V2 + [0.2 + 0.1 + 0.05] V3 = 0
VA − 30∠0 VA − VB VA
−0.05V1 − 0.1V2 + 0.35V3 = 0 + + = 0
5 2 + j3 j5
−V1 − 2V2 + 7V3 = 0
1 1 1 1 30
VA + + − VB − = 0
5 j5 2 + j3 2 + j3 5
7V1 − 4V2 − V3 = 0

1 VB 30∠0
.2 − j0.2 + VA − − = 0
−2V1 + 3V2 − V3 = 100 3.6∠56.3 3.6∠56.3 5
−V1 − 2V2 + 7V3 = 0 [.2 − j.2 + .277∠ − 56.3] VA − .277∠ − 56.3VB = 6

7 −4 −1
[.2 − j.2 + .153 − j.23] VA − .277∠ − 56.3VB = 6

∆ = −2 3 −1 = 7(21−2)+4(−14−1)−1(4+3) [.353 − j.43] VA − .277∠ − 56.3VB = 6
−1 −2 7 [.556∠ − 50.6] VA − .277∠ − 56.3VB = 6
Applying KCL to node VB Applying KCL to node VA

VA − 50 VA
VB − 20∠0 VB − VA VB + = 0
+ + = 0 10 j10
4 2 + j3 6
1 1 50
1 1 1 1 20∠ + VA − = 0
VB + + − VA − = 0 10 j10 10
6 4 2 + j3 2 + j3 4
[0.1 − j0.1] VA = 5
1 VA 20∠0
.16 + .25 + VA − − = 0 0.141∠ − 45VA = 5
3.6∠56.3 3.6∠56.3 4
[.16 + .25 + .277∠ − 56.3] VB − .277∠ − 56.3VA = 5 VA = 35.46∠45
−.277∠ − 56.3VA + [.16 + .25 + .153 − j.23] VB = 6
−0.277∠ − 56.3VA + [0.563 − j0.23] VB = 5Applying KCL to node VB
−0.277∠ − 56.3VA + 0.6∠ − 22.3VB = 5 VB − E2 VB
+ = 0
15 10
1 1 E2
0.556∠ − 50.6VA − 0.277∠ − 56.3V2 = 6 + VB − = 0
10 15 15
−0.277∠ − 56.3VA + 0.6∠ − 22.3VB = 5 E2
[0.1 + 0.066] VB − = 0
15
E2
= 0.166VB
15

0.556∠ − 50.6 −0.277∠ − 56.3
∆= E2 = 2.5VB
−0.277∠ − 56.3 0.6∠ − 22.3
= 0.336∠ − 72.9 − 0.076∠ − 112.6

VA = VB
0.1−j0.321+0.029+j0.07 = 0.129−j0.251 = 0.282∠−62.79
E2 = 2.5 × 35.46∠45
0.556∠ − 50.6 6∠0
E2 = 88.65∠45
−0.277∠ − 56.3 5∠0
VB = =
∆
2.78∠ − 50.6 + 1.662∠ − 56.3 Q 11) In the network shown in Figure 1.269 find the
= value of E2 such that current through the 2+j3Ω is
0.282∠ − 62.79
1.764 − j2.148 + 0.922 − j1.38 zero.
=
0.282∠ − 62.79 5Ω A 2+j3 Ω B 4Ω
2.686 − j3.528 4.43∠ − 52.7
= =
0.282∠ − 62.79 0.282∠ − 62.79
= 15.87∠10.01 30∠0o V j5 Ω 6Ω Vx
VB 15.87∠10.01
I= = = 2.64∠10.01A
6 6
value of E2 such that current through the 8+j8Ω is Figure 1.269
zero. Solution:
10 Ω A 8+j8 Ω B 15 Ω It is given that current through the 2+j3Ω is zero
this is possible only when VA = VB i., VA − VB = 0
Applying KCL to node VA
50∠0o V j10 Ω 10 Ω VA − 30 VA
E2 + = 0
5 j5

1 1 30
+ VA − = 0
5 j5 5
Figure 1.268 [0.2 − j0.2] VA = 6
Solution: 0.282∠ − 45VA = 6
It is given that current through the 8+j8Ω is zero VA = 21.27∠45
this is possible only when VA = VB i., VA − VB = 0
Applying KCL to node VB simultaneous equations are

VB − Vx VB
+ = 0 0.355∠ − 22V1 + 0.133∠90V2 = −5∠90
4 6

1 1 Vx 0.133∠90V1 + 0.212∠ − 38.7V2 = 10
+ VB − = 0
6 4 4
0.355∠ − 22 0.133∠90

Vx ∆ =
[0.166 + 0.25] VB − = 0 0.133∠90 0.212∠ − 38.7
4
Vx
= 0.416VB = 0.0752∠ − 60.7 − 0.0177∠180
4
Vx = 1.664VB = 0.0368 − j0.0655 + 0.0177 = 0.0545 − j0.0655
= 0.0852∠ − 50.2

VA = VB 0.355∠ − 22 −5∠90

0.133∠90 10
Vx = 1.664 × 21.27∠45 V2 =
∆
Vx = 35.39∠45
3.55∠ − 22 + 0.665∠180
——————————————- =
∆
3.3 − j1.33 − 0.665 = 2.635 − j1.3
=
∆
2.93∠ − 26.25
=
0.0852∠ − 50.2
= 34.38∠24

value of V2 using nodal analysis
j3 Ω
V1 V2
-j5 Ω
3Ω 6Ω 10∠0o A
5∠90o A
Figure 1.270
V1 V1 − V2 V1 − V2
+ + + 5∠90 = 0
3 −j5 j3
[0.33 + j0.2 − j0.33] V1 − [j0.2 − j0.33] V2 = −5∠90
[0.33 − j0.133] V1 + j0.133V2 = −5∠90
0.355∠ − 22V1 + 0.133∠90V2 = −5∠90
Applying KCL to node V1
V2 V2 − V1 V2 − V1
+ + − 10 = 0
6 −j5 j3
− [j0.2 − j0.33] V1 + [0.166 + j0.2 − j0.33] V2 = 10
j0.133V1 + [0.166 − j0.133] V2 = 10
0.133∠90V1 + 0.212∠ − 38.7V2 = 10
1.11. Supernode Chapter 1. Module 1 Basic Concepts
1.11 Supernode

−5 0.25
• When an voltage source appears be-
−2 −1

tween two nonreference nodes and any 5 + 0.5
V1 = = = −7.33
elements connected in parallel with it ∆ −0.75
then combining these nodes is called a

0.5 −5
supernode.

1 −2 −1 + 5
V2 = = = −5.33
∆ −0.75

the node voltages
the node voltages
10Ω
4Ω
- -3 A
+
22 V
-
+
2A 7A
2Ω 4Ω 3Ω
1Ω
5Ω
Figure 1.271
-8 A
Solution: -25 A
10Ω
Figure 1.273
V1 2V V2
- Solution:
+
2A 7A 4Ω
2Ω 4Ω -3 A
V2 22 V
V1 - V3
+
3 (Ref) 3Ω
Figure 1.272
1Ω
There is voltage source between V1 and V2 and by 5Ω
supernode equation is -8 A
-25 A
V1 V2
+ −2+7 = 0
2 4
0.5V1 + 0.25V2 = −5 Ref
Figure 1.274
V2 − V1 = 2
There is voltage source between V3 and V2 i.e.,
V1 − V2 = −2
V3 − V2 = 22
Simultaneous equations are V3 = V2 + 22
0.5V1 + 0.25V2 = −5 Apply KCL for node V1

V1 − V2 = −2 V1 − V2 V1 − V3
+ − (−8) − (−3) = 0
3 4
where ∆ is 0.5833V1 − 0.333V2 − 0.25V3 = −11

0.5 0.25 0.5833V1 − 0.333V2 − 0.25(V2 + 22) = −11
∆ = = −0.5 − 0.25 = −0.75
1 −1 0.5833V1 − 0.583V2 = −5.5 (1)
There is voltage source between V2 V3 and by 10Ω

applying supernode equation
-4 A
V2 − V1 V2 V3 − V1 V3 V2 20 V
+ + + + (−3) + (−25) = 0 V1 - V3
+
3 1 4 5
2Ω
4Ω
5Ω
−0.5833V1 + 1.333V2 + 0.45V3 = 28
-5 A
−0.5833V1 + 1.333V2 + 0.45(V2 + 22) = 28 -10 A
−0.5833V1 + 1.783V2 = 18.1 (2)
Simultaneous equations are Ref
0.5833V1 − 0.583V2 = −5.5 Figure 1.276

−0.5833V1 + 1.783V2 = 18.1 There is voltage source between V3 and V2 i.e.,
V3 − V2 = 20
where ∆ is V3 = V2 + 20
Apply KCL for node V1

0.5833 −0.583
∆=
= 1.04 − 0.339 = 0.7
−0.583 1.783 V1 − V2 V1 − V3
+ − (−5) − (−4) = 0
2 10
0.6V1 − 0.5V2 − 0.1V3 = −9

−5.5 −0.583 0.6V1 − 0.5V2 − 0.1(V2 + 20) = −9
0.6V1 − 0.6V2 = −7

18.1 1.783 −9.8 + 10.5 (1)
V1 = = = 1V
∆ 0.7 There is voltage source between V2 V3 and by
applying supernode equation
0.5833 −5.5
V2 − V1 V2 V3 − V1 V3
−0.5833 18.1 10.55 − 3.2 + + + + (−4) + (−10) = 0
V1 = = = 10.48V 2 4 10 5
∆ 0.7
−0.6V1 + 0.75V2 + 0.3V3 = 14
V3 = V2 + 22 = 10.48 + 22 = 32.48 −0.6V1 + 0.75V2 + 0.3(V2 + 20) = 14
—————————————————— −0.6V1 + 1.05V2 = 8 (2)
0.6V1 − 0.6V2 = −7
the node voltages
−0.6V1 + 1.05V2 = 8
10Ω
where ∆ is

-4 A 0.6 −0.6
∆ = = 0.63 − 0.36 = 0.27
20 V −0.6 1.05
-
+

−7 −0.6
2Ω
8 1.05

−7.35 + 4.8
V1 = = = −9.44V
∆ 0.27
4Ω
5Ω 0.6 −7

−0.6 8 4.8 − 4.2
-5 A -10 A V2 = = = 2.22V
∆ 0.27
V3 = V2 + 20 = 2.22 + 20 = 22.22V
———————————————————–
Figure 1.275 Q 4) In the circuit shown in Figure ?? determine the

Solution: node voltages V1 , V2 , V3 , V4
V2 V3 − V4 = 0.2vy
V3 − V4 = 0.2(V4 − V1 )
+
Vx V3 − 1.2V4 = −0.2V1 = −0.2 × (−12)
2Ω
-
14 A V3 − 1.2V4 = 2.4
Ω
5
0.5Vx
0.
−1.2V4 = 2.4 − V3
V1 -
+
V3 V4 = 0.833V3 − 2
12 V
Apply KCL for super nodes V3 and V4
2.
5
1Ω
+
Ω
V3 − V2 V4 − V1 V4
-
− 0.5Vx + + = 0
Vy
y
V
2 2.5 1
0.2
+
−.4V1 − .5V2 + .5V3 + 1.4V4 − .5Vx = 0

−.4V1 − .5V2 + .5V3 + 1.4V4 − .5(V2 − V1 ) = 0
V4
Solution: 0.1V1 − V2 + 0.5V3 + 1.4V4 = 0
−V2 + 0.5V3 + 1.4V4 = 1.2
V2 −V2 + 0.5V3 + 1.4(0.833V3 − 2) = 1.2
−V2 + 1.66V3 = 4 (2)
+
Vx Simultaneous equations are
2Ω
-
14 A
Ω
5
0.5Vx 2.5V2 − 0.5V3 = −10

0.
V1 -
−V2 + 1.66V3 = 4
+
V3
12 V
where ∆ is
2.
5

1Ω
+
2.5 −0.5
Ω
-
∆= = 4.15 − 0.5 = 3.65

-

Vy
−1 1.66
y
V
0.2
+

−10 −0.5

V4 4 1.66 −16.6 + 2
V2 = = = −4V
∆ 3.65

2.5 −10
Figure 1.277
−1 4 10 − 10
From the figure 1.277 it is observed that V3 = = = 0V
∆ 3.65
V4 = 0.833V3 − 2 = 0.833 × 0 − 2 = −2
V1 = −12V

V1 = −12, V2 = −4, V3 = 0, V4 = −2
V2 − V1 V2 − V3
+ − 14 = 0 ———————————————————–
0.5 2
−2V1 + [2 + 0.5]V2 − 0.5V3 = 14
−2(−12) + 2.5V2 − 0.5V3 = 14 the node voltages
2.5V2 − 0.5V3 = −10 (1) 1/4Ω
3A
22 V
-
V3 − V4 = 0.2Vy
+
1/3Ω
8A 25 A
1Ω 1/5Ω
Vy = V4 − V1
Figure 1.278
Vx = V2 − V1 Solution:
The circuit is redrawn with node labeling is as 6Ω

1/4Ω 2Vx 12 Ω
-
3A +
22 V +
V2 V3 + 6V
V1 - 12Ω 6Ω
-
+
Vx
-
1/3Ω
8A 25 A I0
1Ω 1/5Ω
Figure 1.280
4 (Ref) Solution:
The circuit is redrawn with node labeling is as
Figure 1.279 shown in Figure 1.281
From the figure it is observed that there is voltage
6Ω
source between node V2 and V3
V3 − V2 = 22 2Vx
- V2 12 Ω
V3 = V2 + 22 V1 + V3
+ + 6V
V1 − V2 V1 − V3 12Ω 6Ω
-
+ +8+3 = 0 Vx
-
1/3 1/4
7V1 − 3V2 − 4V3 = −11 I0
7V1 − 3V2 − 4(V2 + 22) = −11 4 (Ref)
7V1 − 7V2 = 77 (1)
Figure 1.281
From the figure it is observed that there is dependent
V2 − V1 V2 V3 − V1 V3 voltage source between node V1 and V2
+ + + − 25 − 3 = 0
1/3 1 1/4 1/5
−7V1 + 4V2 + 9V3 = 28 V3 = 6V
−7V1 + V2 + 9(V2 + 22) = 28 V2 = Vx
−7V1 + 13V2 = −170 V1 − V2 = 2Vx
Simultaneous equations are V1 = V2 + 2Vx
V1 = Vx + 2Vx = 3Vx
7V1 − 7V2 = 77
−7V1 + 13V2 = −170 There is voltage source between V1 and V2 Apply
KCL for supernode V1 and V2
7 −7
∆= = 91 − 49 = 42 V1 − V3 V1 V2 V2 − V3
−7 13 + + + = 0
6 12 6 12
2V1 − 2V3 + V1 + 2V2 + V2 − V3

77
−7
= 0
−170 13 1001 − 1190 12
V1 = = = −4.5V 6Vx − 2V3 + 3Vx + 2Vx + Vx − V3 = 0
∆ 42

7
12Vx = 3V3
77

−7 −170 3×6
−1190 + 539 Vx = = 1.5V
V2 = = = −15.5V 12
∆ 42
V1 = 3Vx = 3 × 1.5 = 4.5
V3 = V2 + 22 = −15.5 + 22 = 6.5V
V1 4.5
———————————————————– Io = = = 0.375A
12 12
———————————————————–
the node voltages and also current I0 using nodal Q 7) In the circuit shown in Figure 1.282 determine
analysis all the node voltages using nodal analysis
4 4
10 V V 3 15 V V 7
- 2 2
V1 V3 V1
-
V3
+
+
25 A
10 1
8 2
5 5
4A 4A
6A 3A

Solution: Solution:
From the figure it is observed that there is a voltage
From the figure it is observed that there is a voltage
V2 − V1 = −15
−V1 + V2 + 0V3 = −15
V2 − V1 = 10
−V1 + V2 + 0V3 = 10 There is voltage source between V1 and V2 Apply
There is voltage source between V1 and V2 Apply 2V1 + 4V1 + 3 − 4V3 + 1V2 + 7V2 − 7V3 + 25 = 0
KCL for supernode V1 and V2 6V1 + 8V2 − 11V3 = −28
8V1 + 4V1 − 4V3 + 6 + 10V2 + 3V2 − 3V3 = 0
−4V1 − 7V2 + 16V3 − 25 − 4 = 0
12V1 + 13V2 − 7V3 = −6
−4V1 − 7V2 + 16V3 = 29

−V1 + V2 + 0V3 = −15
6V1 + 8V2 − 11V3 = −28
−4V1 − 3V2 + 12V3 − 4 = 0
−4V1 − 7V2 + 16V3 = 29
−4V1 − 3V2 + 12V3 = 4
Solving above simultaneous equations
V1 = 6.17V V2 = −8.82V V3 = −0.504V
———————————————————–
−V1 + V2 + 0V3 = 10
12V1 + 13V2 − 7V3 = −6 Q 9) In the circuit shown in Figure 1.284 determine
all the node voltages using nodal analysis
−4V1 − 3V2 + 12V3 = 4
2V
-+
3 V2 2
V1 V3
V1 = 5.5V V2 = 4.46V V3 = −0.406V
———————————————————– 2 4
1A

all the node voltages using nodal analysis
Figure 1.284 • The resistor of 2Ω in series with inductance of

Solution: 5 Ω is connected from node 1 to common node.
From the figure it is observed that there is a voltage j5 Ω

source between node V1 and V3 5Ω 1 2 5Ω
2Ω
V3 − V1 = 2 50∠00 -j5 Ω 50∠900
−V1 + 0V2 + V3 = 2 j5 Ω
There is voltage source between V1 and V3 Apply 3
Figure 1.285
2V1 + 3V1 − 4V2 + 4V3 + 2V3 − 2V2 = 0
—————————————
5V1 − 5V2 + 6V3 = 0
The node equations of a network are
10∠30◦

1 1 1 1
−3V1 + 5V2 − 2V3 − 1 = 0 + + V1 − V2 = (1.3)
6 j5 −j10 j5 6
−3V1 + 5V2 − 2V3 = 1
1 1 1 1 50∠ − 30
− V1 + + + V2 = −
j5 j5 3 + j8 4 4
(1.4)
−V1 + 0V2 + V3 = 2 Derive the network
5V1 − 5V2 + 6V3 = 0
Solution:
−3V1 + 5V2 − 2V3 = 1
j5 Ω
6Ω V1 V2 4 Ω
3Ω
V1 = − 1.667V V2 = −0.1667V V3 = 0.833V
10∠300 -j10 Ω 5∠ − 300
———————————————————– j8 Ω
The node equations of a network are
3

1 1 1 1 50∠0
+ + V1 − V2 = (1.1) Figure 1.286
5 j5 −j5 j5 5
—————————————
1 1 1 1 50∠90
− V1 + + + V2 = − (1.2) The node equations of a network are
j5 j5 2 + j5 5 5
50∠0◦

Derive the network 1 1 1 1
+ + V1 − V2 = (1.5)
Solution: 5 j2 4 4 5
From equation (1.5) it is observed that
1 1 1 1 50∠90
− V1 + + + V2 = (1.6)
• From node 1 to common node there is 5 Ω 4 4 −j2 2 2
resistor in series with 50∠0 voltage source. Derive the network
• The inductance of 5 Ω is connected between Solution:
node 1 and node 2 5Ω V1 4 Ω V2 2 Ω
• The capacitance of 5 Ω is connected from node
1 to common node
50∠00 j2 Ω -j2 Ω 50∠900
Similarly From equation (1.6) it is observed that
• From node 2 to common node there is 5 Ω
3
resistor in series with 50∠90 voltage source.
• The inductance of 5 Ω is connected between Figure 1.287

node 1 and node 2 ————————————–
Derive the network for the following mesh equations. 5Ω V1 4 Ω V2 2 Ω

5 + j5 −j5 0

I1
 
50∠30◦

50∠00 j2 Ω -j2 Ω 50∠900
 −j5 8 + j8 −6   I2  =  −5∠0 
0 −6 10 I3 −20∠0◦
(1.7) 3
1.12 Question Papers

2020-Aug-CBCS 2-b)Find I1 in the the circuit shown Applying KCL to node 1
in Figure 1.289 using nodal analysis. V1 − 30 V1 V1 + 10
j4 
+ + = 0
10  5 + j15 −j10 8 − j5

1 1 1 30 10
− + V1 = −
I1 5 + j15 j10 8 − j5 5 + j15 8 − j5
+
20 0 V -j2.5 2I1 j2 0.146∠41.194V1 = 2.38∠ − 97
-
2.38∠ − 97
V1 = = 16.3∠ − 138.19
0.146∠41.194
Figure 1.289: 2020-Aug
Solution:
j4  JAN-2018-CBCS 2-b JAN-2015 1d)For the network
10  A B
shown in Figure 1.293 find the value of voltage
source V such that current through the 4Ω is zero.
+ I1 Use node voltage analysis.
20 0 V -j2.5 2I1 j2
-
5Ω A 4Ω B 2Ω
Figure 1.290: 2020-Aug

V j2 Ω -j2 Ω
Applying KCL to node A 50 j
VA − 20∠0 VA VA − VB
+ + = 0
10 −j2.5 j4

1 1 50∠90
+ VB − = 0 Figure 1.293: 2018-CBCS-Question Paper
2 −j2 2
[0.5 + j0.5] VB = 25∠90 Solution:
It is given that current through the 4Ω is zero this
0.707∠45VB = 25∠90
is possible only when VA = VB i., VA − VB = 0
VB = 35.36∠45 Applying KCL to node B
VB − 50∠90 VB
2019-July-CBCS 2-b)Find the voltage across the + = 0
2 −j2
capacitor 10 Ω reactance of the network shown in
1 1 50∠90
Figure 1.292 using nodal analysis. + VB − = 0
2 −j2 2
j15 [0.5 + j0.5] VB = 25∠90
5 8 -j5 0.707∠45VB = 25∠90
+ VB = 35.36∠45
30 V -j10 VC 10V Applying KCL to node A

_ VA − V
+
VA
= 0
5 j2

1 1 V
Figure 1.291: 2019-July + VA − = 0
5 j2 5
Solution: V
[0.2 − j0.5] VA − = 0
j15 5
5 V1 8 -j5 0.538∠ − 68.2VA − 0.2V = 0
+ 0.2V = 0.538∠ − 68.2VA
30 V -j10 VC 10V V = 2.69∠ − 68.2VA

_ VA = VB
V = 2.69∠ − 68.2 × 35.36∠45

Figure 1.292: 2019-July V = 95.11∠ − 23.2
JAN-2017-CBCS Determine all the node voltages

of the network shown in Figure 1.294 using nodal V1 10Ω
analysis.
+
4Ω A 4Ω B 6V -
+ Vx - 2A
V4 4Ω
4Ω V3 V2
+ 0.5Vx 1A
10 V -
+
1Ω 2Ω - 5V
Figure 1.294
Solution:
Vx = VA − VB
Figure 1.296
Applying KCL to node A
VA − 10 VA − VB V2 = 5
+ + 0.5Vx = 0
4 4 Applying KCL to node 3
1 1 VB
+ VA − + 0.5Vx − 2.5 = 0 V3 + 6 − V2 V3
4 4 4 + +2 = 0
0.5VA − 0.25VB + 0.5(VA − VB ) = 2.5 10 1
1 1 V2 6
+ V3 − + +2 = 0
0.5VA − 0.25VB + 0.5VA − 0.5VB = 2.5 10 1 6 10
VA − 0.75VB = 2.5 1.1V3 − 0.16V2 = −2.6
1.1V3 − 0.16 × 5 = −2.6
Applying KCL to node B
1.1V3 − 0.8 = −2.6
VB VB − VA 1.1V3 = −1.8
+ −1 = 0
4 4 V3 = −1.63
−0.25VA + 0.5VB = 1
Applying KCL to node 4
VA − 0.75VB = 2.5 V4 V4 − V2
+ −2 = 0
−0.25VA + 0.5VB = 1 2 4
0.75V4 − 0.25V2 − 2 = 0
Solving the above equations 0.75V4 − 0.25 × 5 − 2 = 0
0.75V4 − 1.25 − 2 = 0
VA = 6.4V VB = 5.2V
0.75V4 = 3.25
V4 = 4.33
JULY-2017-CBCS 2 b)For the circuit shown in
Figure 1.295 determine all the node voltages.
V1 = V3 + 6 = −1.63 + 6 = 4.37
V1 10Ω V2
CBCS JAN 2017) Find i using nodal analysis for the
+ + network shown in Figure 1.297
6V - - 5V
2A 4Ω 2Ω
A
V3 V4 - +
0.5i1
1Ω 2Ω
+ +
3V - 2A
- 4V
4Ω i1
Figure 1.295
Solution: The circuit is redrawn which is as shown
in Figure 1.296 Figure 1.297
Solution: Solving above simultaneous equations

VA − 4
i1 = = 0.5VA − 2 V1 = 1.5V V2 = 0.9V V3 = −0.5V
2
Applying KCL to node A
————————————–
VA − 3 − 0.5i1 VA − 4 DEC-2015 1(c)) Find the current i In the circuit
+2+ = 0
4 2 shown in Figure 1.299 using nodal analysis
1 1 0.5i1
+ VA − + 2 − 0.75 − 2 = 0 j4 Ω
4 2 4 10 Ω A B
0.75VA − 0.125i1 = 0.75
0.75VA − 0.125(0.5VA − 2) = 0.75
+
0.75VA − 0.0625VA + 0.25 = 0.75 20∠00 V i1 -j2.5 Ω 2i1 j2
-
0.6875VA = 0.5
VA = 0.727v
Figure 1.299
i1 = 0.5VA − 2 = 0.5 × 0.727 − 2 = −1.636A
Solution:
————————————- VA
i1 = = j0.4VA
−j2.5
JULY 2016) In the circuit shown in Figure 1.298
determine all the node voltages using nodal analysis Applying KCL to node A
2V VA − 20 VA VA − VB
+ + = 0
- 10 −j2.5 j4
+

1 1 1 VB
+ − VA − −2 = 0
V1 3 V2 2 10 j4 j2.5 j4
V3 [0.1 − j0.25 + j0.4] VA + j0.25VB = 2
[0.1 + j0.15] VA + j0.25VB = 2
2 1A 4 0.18∠56.3VA + 0.25∠90VB = 2
Figure 1.298 Applying KCL to node B

Solution:
VB − VA VB
From the figure it is observed that there is a voltage + − 2i1 = 0
j4 j2
1 1

j0.25VA + + VB − 2(j0.4VA ) = 0
j4 j2
V1 − V3 = 2 −j0.55VA + [−j0.25 − j0.5] VB = 0
V1 + 0V2 − V3 = 2 −j0.55VA − j0.75VB = 0
There is voltage source between V1 and V3 Apply 0.55∠ − 90VA + 0.75∠ − 90VB = 0
2V1 + 3V1 − 3V2 + 4V3 + 2V3 − 2V2 = 0
5V1 − 5V2 + 6V3 = 0
0.18∠56.3VA + 0.25∠90VB = 2
0.55∠ − 90VA + 0.75∠ − 90VB = 0
−3V1 + 5V2 − 2V3 − 1 = 0
−3V1 + 5V2 − 2V3 = 1

0.18∠56.3 0.25∠90
∆ =
0.55∠ − 90 0.75∠ − 90
V1 + 0V2 − V3 = 2
0.135∠ − 33.7 − 0.1375∠0 = 0.112 − j0.75 − 0.1375
5V1 − 5V2 + 6V3 = 0
−3V1 + 5V2 − 2V3 = 1 −0.0255 − j0.75 = 0.75∠ − 92

2 0.25∠90

0 0.75∠ − 90 1.5∠ − 90
VA = = = 2∠2V V2 − V1 V2 V2 − V3 V4 − V1 V4 − V3
∆ 0.75∠ − 92 + + + + = 0
2 2 1 2 2
−V1 + 2V2 − 1.5V3 + V4 = 0
0.18∠56.3 2

0.55∠ − 90 0 +12 + 2V2 − 1.5 × 24 + V2 + 24 =
−1.1∠ − 90
VB = = = −1.46∠2V 2V2 = 0
∆ 0.75∠ − 92
V2 = 0
The current I0 is also zero

VA
i1 = = j0.4VA = 0.4∠90 × 2∠2 = 0.8∠92 JULY-2014 1(a)) The node equations of a network
−j2.5
are
1 1 1 1 50∠0
2014-JUNE 1(c)) For the network shown in the + + V1 − V2 = (1.1)
5 j2 4 4 5
Figure ?? find the current in I0 using nodal analysis.

1 1 1 1 50∠90
− V1 + − + V2 = (1.2)
4 4 j2 2 2
2Ω 24 V +- 2Ω Derive the network

Solution: From equation (1.1) it is observed that
2Ω I 0 1Ω • From node 1 to common node there is 5 Ω

12 V -+ 2Ω +
- 24 V
• 4 Ω restor is connected between node 1 and
node 2
Figure 1.300
• The inductor of 2 Ω is connected from node 1
Solution: to common node
From the figure it is observed that V1 =
−12V V3 = 24. Also it is observed that there is Similarly From equation (1.2) it is observed that
a voltage source between V2 and V4 this forms the
supernode between node 2 and node 4. • From node 2 to common node there is 2 Ω
V4 Super Node
• 4 Ω restor is connected between node 1 and
node 2
2Ω 24 V +- 2Ω • The capacitor of 2 Ω is connected from node 1

to common node
V1 V3
2Ω V2 1Ω
5Ω 4Ω 2Ω
12 V -+ 2Ω +- 24 V 1 2
50∠00 j2 Ω -j2 Ω
Figure 1.301 50∠900
V4 − V2 = 24
Figure 1.302
V4 = V2 + 24 ————————————————————–
JULY-2013 1-b For the network shown in Figure
By KCL for the supernode 1.293 determine the node voltages by nodal analysis.
5Ω 1 4Ω 2 2Ω ————————————————————
2012-JUNE 1(c)) Find the power dissipated in the
10Ω resistor as shown in the Figure 1.304 by node
voltage method.
5∠00 j2 Ω -j2 Ω
5∠900 12 A
3 V1 V2 V3
Figure 1.303: 2018-CBCS-Question Paper 4Ω 2Ω 4Ω

Solution: 15 A 8Ω 8Ω 10Ω 4Ω +
- 32 V
V1 − 5∠0 V1 V1 − V2 Figure 1.304
+ + = 0
5 j2 4
Solution:
1 1 1 5∠0 V2
+ + V1 − − = 0 Applying the KCL for Node V1
5 j2 4 5 4
[0.45 − j0.5] V1 − 0.25V2 = 1
V1 V1 V1 − V2
0.672∠ − 48V1 − 0.25V2 = 1 + + − 15 + 12 = 0
8 8 4
0.5V1 − 0.25V2 − 0V3 = 3

V2 − 5∠90 V2 V2 − V1 Applying the KCL for Node V2
+ + = 0
2 −j2 4
V1

1 1 1

5∠90 V2 V2 − V1 V2 − V3
− + + + V2 − = 0 + + = 0
4 4 −j2 2 2 10 4 2
−0.25V1 + 0.85V2 − 0.5V3 = 0
−0.25V1 + [0.75 + j0.5] V2 = 2.5∠90
−0.25V1 + 0.9∠33.69V2 = 2.5∠90
Applying the KCL for Node V3
Simultaneous equations are V3 − V2 V3 V3 − 32
+ + − 12 = 0
0.672∠ − 48V1 − 0.25V2 = 1 2 4 4
0V1 − 0.5V2 + 1V3 = 20
−0.25V1 + 0.9∠33.69V2 = 2.5∠90

0.672∠ − 48 −0.25
∆ =
−0.25 0.9∠33.69 Simultaneous equations are
0.6∠ − 14.31 − 0.0625∠0 = 0.581 − j0.148 − 0.0625
0.518 − j0.148 = 0.538∠ − 16 0.5V1 − 0.25V2 − 0V3 = 3

1 −0.25 −0.25V1 + 0.85V2 − 0.5V3 = 0
2.5∠90 0.9∠33.69 0.9∠33.69 + 0.625∠90
V1 = = 0V1 − 0.5V2 + 1V3 = 20
∆ 0.538∠ − 16
0.75 + j0.5 + j0.625 1.35∠56.3 V1 = 18.1 V2 = 24.21 V3 = 32.1

V1 = = = 2.5∠72.3
0.538∠ − 16 0.538∠ − 16 Power dissipated in the 10Ω resistor is

0.672∠ − 48 1

−0.25 2.5∠90
V32 32.12
1.68∠42 + 0.25 P10 = = = 58.6W
V2 = = 10 10
∆ 0.538∠ − 16
2011 DEC 1(b)) Find the currents in all the resistors
1.24 + j1.12 + 0.25 1.86∠36.9 in the circuit shown in Figure 1.305 by node voltage
V2 = = = 3.45∠52.9 method.
0.538∠ − 16 0.538∠ − 16
(15 + 2)
1Ω = = 8.5V
2

2 10
1Ω 4Ω

−1 2
V3 =
∆
2A 8 V +- (4 + 10)
4Ω = = 7V
2
V1 − V2 8.5V − 8
i12 = = = 0.5A
Figure 1.305 1 1
Solution: V1 − V3 8.5V − 7
i13 = = = 1.5A
1Ω 1 1
V2 − V3 8V − 7
i23 = = = 0.25A
1 4
V1 1Ω V2 4Ω
V3 V3 7
i3 =
= = 1.75A
4 4
2A 8 V +- 4Ω JULY-2001 For the network shown in Figure 1.307
determine the current I by nodal analysis.
5Ω 1 4Ω 2 2Ω
Figure 1.306
From the figure it is observed that V2 = 8
Applying the KCL for Node V1 -j2 Ω
50∠00 +
_ j2 Ω +
_ 50∠90
0
I
V1 − V2 V1 − V3
+ −5 = 0
1 1
2V1 − V2 − V3 = 2 3
2V1 − 8 − V3 = 2
2V1 − V3 = 10 Figure 1.307: 2018-CBCS-Question Paper
Solution:
Applying the KCL for Node V3
1 1 1 50∠0 V2
+ + V1 − − = 0
V3 − V1 V3 − V2 V3 5 j2 4 5 4
+ + = 0
1 4 4 [0.45 − j0.5] V1 − 0.25V2 = 10
−V1 − 0.25V2 + 1.5V3 = 0 0.672∠ − 48V1 − 0.25V2 = 10
−V1 − 0.25 × 8 + 1.5V3 = 0
−V1 + 1.5V3 = 2
V2 − 50∠90 V2 V2 − V1
Simultaneous equations are + + = 0
2 −j2 4

V1 1 1 1 5∠90
2V1 − V3 = 10 − + + + V2 − = 0
4 4 −j2 2 2
−V1 + 1.5V3 = 2
−0.25V1 + [0.75 + j0.5] V2 = 25∠90
−0.25V1 + 0.9∠33.69V2 = 25∠90

2 −1
∆= = (3 − 1) = 2

−1 1.5
∆=2 0.672∠ − 48V1 − 0.25V2 = 1
−0.25V1 + 0.9∠33.69V2 = 2.5∠90
10 −1

2 1.5

0.672∠ − 48 −0.25
V1 = ∆ =
∆ −0.25 0.9∠33.69
0.6∠ − 14.31 − 0.0625∠0 = 0.581 − j0.148 − 0.0625

0.518 − j0.148 = 0.538∠ − 16

1 −0.25
2.5∠90 0.9∠33.69 0.9∠33.69 + 0.625∠90
V1 = =
∆ 0.538∠ − 16
0.75 + j0.5 + j0.625 1.35∠56.3

V1 = = = 2.5∠72.3
0.538∠ − 16 0.538∠ − 16

0.672∠ − 48 1

−0.25 2.5∠90 1.68∠42 + 0.25
V2 = =
∆ 0.538∠ − 16
1.24 + j1.12 + 0.25 1.86∠36.9

V2 = = = 3.45∠52.9
0.538∠ − 16 0.538∠ − 16
For the network shown in Figure ?? determine the Voltage between node 2 and 3
current Node Voltages by nodal analysis.
5
V2 − V3 = 10
2A
V2 10 V
V1 V3
Apply supermesh for node 2 and 3
-
+
5
1 V2 − V1 V2 V3 − V1 V3
2 + + + − 2 + 3.5 = 0
5 1 5 2
6A 3.5 A
[−0.2 − 0.2] V1 + V2 [1 + 0.2] + V3 [0.2 + 0.5] = −1.5
−0.4V1 + 1.2V2 + 0.7V3 = −1.5
Figure 1.308: supernode10-1 Simultaneous equations are

Solution:
0.4V1 − 0.2V2 − 0.2V3 = 4
Applying KCL to node 1 0V1 + V2 − V3 = 10
−0.4V1 + 1.2V2 + 0.7V3 = −1.5
V1 − V3 V1 − V2
+ +2−6 = 0
5 5
[0.2 + 0.2] V1 − 0.2V2 − 0.2V3 = 4
0.4V1 − 0.2V2 − 0.2V3 = 4 V1 = 10V V2 = 5V V3 = −5V
Important: All the diagrams are redrawn and solutions are prepared. While
preparing this study material most of the concepts are taken from some text books
or it may be Internet. This material is just for class room teaching to make
better understanding of the concepts on Network analysis: Not for any commercial
purpose

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Chapter 1

Module 1 Basic Concepts

If n number of resistors are connected in parallel then

Sign convention for the application of Kirchoff ’s law

Branch Current Rule Using Method 1

Figure 1.5 Current I2 is

Find the power dissipated in the 3 Ω resistor

1.2 Source Transformation Technique

4 and 8 Ω are in parallel

Q 3) For the circuit shown in Figure determine the

Figure 1.19 V = I × R = 3 × 4 = 12V

Figure 1.23 5Ω 10V 10Ω

Figure 1.25 Figure 1.29

The other method to find current in the above circuit I

Resistance in other baranch Figure 1.35

3Ω Q 6) In the circuit shown in Figure 1.257 determine

+ v0 100Ω + 3V Replace the current source and parallel resistor 4 Ω

Figure 1.40 V = I × R = 30 × 4 = 120V

Figure 1.41 Figure 1.45

100 mA v0 100Ω 100Ω 110 mA R = 2 + 4 = 6Ω

100 mA current source with 100 Ω resistor and 240 V +- 6Ω

210 mA v0 100 V 180

Figure 1.43 12Ω 24Ω

4.8Ω Figure 1.51

- 2Ω Replace the voltage source of 10 V by current source

2Ω 10 V I in parallel with resistor of 200 Ω

Replace the parallel resistors of 200 and 80 Ω by a 50Ω

30Ω 50Ω Figure 1.60

1.3 Question Papers

The equivalent current source is

JULY-2016 Using source transformation find the

Figure 1.79: JULY-2017-Question Paper

−1.51 − 102I − 29.5I = 0

21.375 V Figure 1.89

Figure 1.86: JULY-2016-Question Paper 10 40

1.4 Source Shifting & Transformation Techniques

Figure 1.95 Figure 1.99

Figure 1.101 Figure 1.104

Figure 1.109 Figure 1.114

Aug 2020 (2018 scheme EE) 2 c) Apply source

Figure 1.113: 2018-DEC Figure 1.118: 2018-DEC

1.5 Star Delta Transformation

Rxy Ryz Rzx (Rxy + Ryz + Rzx )

2Rxy Rzx Figure 1.125

In the circuit 5 Ω and 3.6 Ω are in series, 6 Ω

7.2 Ω 27.6 Ω are in parallel, which is in series

7.2 × 27.6 Figure 1.129

Figure 1.128 Q 3) For the circuit shown in Figure shown in Figure

As shown in the Figure 1.129, by looking by any Figure 1.130

parallel Figure 1.140

Figure 1.137 Q 5) 2018-Dec (2010 scheme) Find the resistance of

Q 4) Determine the equivalent resistance between

18.667 Ω Figure 1.143

In the circuit 1.143 at NPB 37.5 Ω 3.875 Ω and

Figure 1.149 Figure 1.152

Figure 1.151 Figure 1.154

The total current flowing in the network is Figure 1.155

Simultaneous equations are Figure 1.156