gives it. The current through this can
Network analysis
Network is the interconnection of
change suddenly but the voltage can
network components.
not (any sudden change in the voltage
Circuit is the network with at least
across it requires infinite current). The
one closed path.
voltage across it remains constant
Network components: There are two
when the current through it is zero. In
types of network components, active
this
V (t ) 
and passive components.
I (t )  C
dV (t )
.
dt
either only dissipate energy or store
and
then
supply
+
first
energy.
and
V
-
Passive components are those, which
1 t
I ( )d
C 
I
Resistors, capacitors and inductors are
Inductor: It is the device that
passive elements.
opposes the change in the current
Resistor: It is device that opposes the
through it. It is a linear passive and
flow of current through it. It is a linear
bilateral device that stores energy and
passive and bilateral device that
gives it. The voltage across this can
dissipates power. The current through
change suddenly but the current can
and the voltage across this can change
not (any sudden change in the current
in any fashion. In this V=IR and
through it requires infinite voltage).
V
.
R
The
through
it
remains
constant when the voltage across it is
+
I
Capacitor: It is the device that
opposes the change in the voltage
zero.
I (t ) 
In
this
1 t
V ( )d .
L 
across it. It is a linear passive and
+
bilateral device that stores energy and
V (t )  L
I
V
-
V
current
-
I
dI (t )
dt
and
polarity of the potential difference
Practical
Independent
voltage
source:
+
 In the case of passive elements, the
between their terminals depends on
V
IR
R
the direction of the current flow. The
V
I
terminal through which the current
E
IDEAL
E
PRACTICAL
I
-
enters becomes positive and the
terminal through which the current
and
comes out becomes negative.
Active components are those which
can
continuously
supply
energy.
Voltage and current sources are the
active elements.
Voltage source:
Voltage source is the device that
creates the potential difference (emf)
between its terminals. Voltage source
can be independent, in which the emf
is independent of any other quantity
in the network in which it is present,
or dependent in which the emf
depends on some other quantity in the
network.
Any given voltage source can be
practical or ideal.
This is the voltage source with emf E
internal
resistance
R .
Its
terminal voltage depends on the
magnitude and the direction of the
current through it. (It experiences the
loading effect). For this source,
V  E  IR . The emf of a practical
voltage source is its open circuit
voltage.
Ideal independent voltage source:
This is the voltage source with zero
internal resistance. For this source, the
polarity and the magnitude of the
potential difference is independent of
the magnitude and the direction of
current through it. (If such a voltage
source is connected between two
points in a network, the magnitude
and the polarity of the potential
difference between those two points
gets fixed).
If a current of I A is entering ideal
source current depends on some
voltage source of emf V volt through
current or voltage in the network.
the positive terminal, the source
Any given current source can be
dissipates VI Watts of power. If a
practical or ideal.
current of I A is leaving the voltage
Practical
source of voltage V volt through the
source:
Independent
+
IL
positive terminal, the source supplies
current
V
R
IL
VI watts of power.
IS
R
IDEAL
Dependent Voltage source is the
depending on either a current (CCVS)
or potential difference between two
PRACTICAL
-
voltage source with its emf E
IS
V
V
This is the current source with source
current I S and internal resistance R  .
Its load current I L depends on the
points in the network (VCVS).
magnitude and the direction of the
+
voltage across it. (It experiences the
R
I
V
E= kix
or kvx
loading effect). For this source,
IL  IS 
-
+
-
Even this can be practical as shown or
ideal with R=0.
supplies current to the circuit.
current
practical current source is its short
circuit current.
Current source is the source which
The
V
. The source current of a
R
source
can
Ideal independent current source:
This is the current source with infinite
be
independent, in which the source
current is independent of any other
quantity in the network in which it is
present, or dependent in which the
internal resistance. For this source, the
polarity and the magnitude of the load
current
is
independent
of
the
magnitude and the direction of the
voltage across it. (If such a current
source is connected in a loop of a
Resistive DC networks: These are
network, the magnitude and the
the networks that contain resistors and
polarity of the current through the
DC voltage and current sources.
loop get fixed).
DC Resistive Network analysis is
Practical dependent current source
one in which one or more currents or
is the current source with its source
voltages in the given network are
current I S depending on either a
determined.
current through some branch (CCCS)
In the first step of the analysis, a set
or the potential difference between
of linearly independent equations are
two points in the network (VCCS).
formed with the quantities to be
determined, using Kirchhoff’s voltage
+
IL
and current laws knowing the V-I
IS= kix
or kvx
R
relations of different components and
V
then they are solved to determine the
required quantities.
-
With R   , it is the ideal dependent
KVL: It states that the sum of the
current source
voltages around a closed path is equal
With a potential difference between
to zero. (While applying KVL, we
the terminals of an ideal current
should never pass through a current
source of current I s is V volts, it
source).
supplies a power of VI watts if the
KCL: It states that the sum of the
terminal through which the current is
currents meeting at a node is zero.
coming out is at higher potential,
 To
otherwise it dissipates that power.
determine
the
potential
difference between two points in a
circuit, assign the polarities of the
voltages across each component
(may be given in the problem
itself), start at the -ve terminal,
the voltage across the series
pass through the network along the
resistor.
shortest path that does not contain
Linear network Any given network
a current source and note the total
is linear, if it contains the passive
change in the voltage up to the +ve
components
(R,L,C,M)
terminal. If the total change in the
independent
and/or
potential is positive the polarity is
dependent,
correct,
voltage and/or current sources.
otherwise
change
the
ideal
and/or
and
linearly
practical
polarity and assign the magnitude
Active network: It is the network
of the potential difference.
with at least one active element.
 In a given circuit if there is a
Passive network: It is the network
resistor in parallel with a voltage
with only passive elements.
source,
be
Series combination of resistors:
neglected in the analysis of the
Resistors connected in such a way
remaining part of the network, if
that the current through each resistor
that circuit does not contain any
is same irrespective of their values.
dependent source that depends on
If
the current through the parallel
connected in series, their equivalent
the
resistor
can
resistor.
Ri 1  i  N
are the resistors
N
 In a given circuit if there is a
resistance is Re   Ri
i 1
resistor in series with a current
Parallel combination of resistors:
source,
be
Resistors connected in such a way
neglected in the analysis of the
that the voltage across each resistor is
remaining part of the network, if
same irrespective of their values.
circuit
the
does
resistor
not
can
contain
any
dependent source that depends on
Ri 1  i  N
the resistors
If V is the voltage across the series
connected in parallel, their equivalent
combination of conductances G1 and
If
resistance is Re 
are
1
N
1
R
i 1
Current
G2 ,the voltage across G1 , V1  V
division
i
,
in
and
parallel
the
voltage
G2
G1  G2
across
G2 ,
G1
.
G1  G2
branches: If a current of I A enters a
V2  V
parallel combination of resistors R1
Mesh or loop analysis is the analysis
and
R1 ,
which is basically done for a network
R2
, and the current through
R1  R2
that contains less number of loops
R2 ,
I1  I
R2 , I 2  I
the current through
R1
.
R1  R2
through different branches. In this we
assign loop currents and determine
If a current of I A enters a parallel
combination of conductance G1 and
G2 ,
I1  I
the
current
than nodes, to determine the currents
through
G1 ,
G1
, and the current through
G1  G2
G2
G2 , I 2  I
.
G1  G2
them by solving the equations formed
using KVL. (Never pass through a
current source while writing the loop
equation, if there is a current source
common to two loops, use the concept
of super-mesh).
Voltage division in series circuit: If
Nodal analysis is the analysis which
V is the voltage across the series
is basically done for a network that
combination of resistors R1 and R2 ,
contains less number of nodes than
R1
the voltage across R1 , V1  V
,
R1  R2
and
V2  V
the
R2
.
R1  R2
voltage
across
R2 ,
loops. In this, considering one node as
the reference (datum) node, voltages
are assigned to different nodes and
determine
them
by
solving
the
equations formed using KCL. ( If
there a voltage source between two
5. Find V in the following circuits.
nodes, use the concept of super-node).
R
Problems
1. Find the current through 5
4R
V
resistor if r=0 and r= 1
2A
5Ω
5Ω
5V
rΩ
3Ω
2Ω
2Ω
V?
4Ω
2A
2. Find the current through each
resistor if r= 1 .
1Ω
5V
rΩ
1Ω
6. Find the current through each
1Ω
1Ω
1Ω
1Ω
3. Find the current through 5
resistor if r= 1 .
1Ω
5V
rΩ
1Ω
1Ω
5V
rΩ
1Ω
resistor if r=0 and r= 1
5Ω
7. Find I x .
5V
rΩ
2Ω
2Ω
Ix
3Ω
4. Find the current through each
5V
resistor.
20 Ω
5Ω
2Ω
3Ω
8. Find R, if the power dissipated in
2V
3V
5 resistor is 20W.
5Ω
12. Find V.
6V
20 Ω
50 V
R
2Ω
16V
N1
N2
4Ω
Vx
9. Find V, if i=2A.
3Ω
2Ω
V
13. Find the currents i1 and i2 .
5Ω
10 Ω
10 Ω
i
200Ω
10. Find I.
2Ω
300Ω
4V
DC
10 Ω
400Ω
20 Ω
2V
2. 2 V
DC
DC
1V
i
i1
20 Ω
10 Ω
i2
14. Find I x in the following circuits.
1Ω
5Ω
3Ω
11. Determine the currents through all
the branches
2Ω
2Ω
2Ω
4Ω
6Ω
10V
DC
4Ω
10V
ix
2Ω
20Ω
10Ω
2Ω
5A
25 Ω
2A
20Ω
1A
10V
3Ω
2Ω
Ix
10Ω
18. Find the currents in all the
branches.
30Ω
40Ω
L
Ix
2V
DC
6Ω
L
7A
L
30Ω
40Ω
12Ω
8A
4Ω
15. If the currents through all the
3Ω
resistors are zero, find V1 , V2 and V3 .
4Ω
20Ω
5A
12A
DC
V1
5Ω
6Ω
3Ω
2Ω
7Ω
12Ω
V2DC
7Ω
19. Find Rx
V3
DC
+
16. Find the power dissipated in each
component
9V
-
Rx
2KΩ
10mA
1KΩ
3KΩ
3Ω
20. Find V
2A
10V
6R
3R
DC
+
8V
2A
4V
DC
-4A
6R
V
1V
-
10V
-3A
21. Find i
17. Find Vx and I x
2Ω
4Ω
5A
10 V
+
60V
10Ω
2Ω
IX
VX
-
3Ω
2A
i
22. Determine the power supplied by
25. Find the power absorbed by each
the 2A current source
component.
2A
DC
7A
6Ω
12Ω
4Ω
2Ω
1V
DC
3V
8A
4A
DC
4Ω
1Ω
4V
3Ω
26. Find v1 .
23. Find all node voltages
3Ω
10Ω
DC
2Ω
4Ω
6V
1A
DC
3A
1Ω
5V
2A
1Ω
+
V1
-
27. Find R and G if the powers
2Ω
supplied by 5A and 40 V sources are
100W and 500 W respectively.
9V
DC
R
DC
1mA
4.7kΩ
-110 V
2.2kΩ
6A
5A
G
DC
40 V
3.3kΩ
3mA
2.2kΩ
10kΩ
28. Find v1 .
5Ω
24. Find V1 and V2 .
+ v1 -
10A
3Ω
+
DC
v1 -
240V
6Ω
30Ω
10Ω
4A
8Ω
1Ω
12Ω
+
v2 DC
60V
29. Find I and power absorbed by
each component
+
2
33. Find i2 and i3 .
-
30 Ω
va
120 V
15 Ω
va
+
4V
i2
6V
DC
DC
i
2 mA
5 kΩ
i3
+
-
1000 i3
5 kΩ
0.5 i2
-
+ vao -
+
30 Ω
2
vao
120 V
34. Determine the currents through all
15 Ω
the branches.
i
4.7 kΩ
ix
9V
30. Find Vx .
+ va -
DC
4.7 kΩ
4.7 kΩ
4.7 kΩ
+ vx 4Ω
2Ω
50 V
0.1
DC
0.1vx
va
+
4.7 kΩ
0.1
ix -
100 V
35. Find i1 , i2 and i3 .
31. Find the currents through all the
i2
12 Ω
0.1 vx
branches and the power dissipated in
5Ω
1Ω
i1
each component.
+ vx 6V
ix
12 Ω
1.5 V
2V
i3
2 ix
24 mA
3Ω
2 kΩ
6 kΩ
36. Find i
1Ω
2.5 i
-
2Ω
+
32. Find ix
3Ω
20 V
ia
5Ω
2Ω
4Ω
ix
3Ω
4A
60 V
+
-
6ia
4Ω
5A
i
i
+ va 3Ω
40. Find Vx , I in , I s and the power
2.5 Ω
supplied by the dependent source.
0.5 va
DC
10 V
1.5 V
2Ω
iin
4Ω
6Ω
1Ω
4Ω
3Ω
3 VDC
DC
+ vx +
2Ω
2 kΩ
+
2V
2Ω
Is
6A
-
-
10 kΩ
i
5 kΩ
5 mA
41. Determine all the current and
voltages in the circuit.
20 kΩ
I3
I2
37. Find I A , I B and I C .
DC
60V
+
i
-
0.4
4 vx
8V
I4
5 I2
+
V1 20Ω
-
+
V2
-
0.25V1
+
V4
5Ω
+
V5
I5
iA
iB
iC
42. Find ix and i y .
+
vx
-
5.6 A
18 kΩ
9Ω
0.1vx
2A
Iy
5A
10ms
0.8Ix
38. Find the currents I1 , I 2 , I 3 and I 4
i1
i2
25 Ω
+ v1 -
40ms
Ix
100 Ω
10 Ω
0.2v1
43. Find ix .
25 A
i4
i3
20 Ω
25 Ω
39. Find Vy if I z  3 A .
2A
2Ω
+
5V
2Ω
vx
-
3
vx
ix
+
vy
-
10 Ω
1Ω
iz
1.5 ix
5Ω
2A
48. Find vx .
I1
5KΩ
20KΩ
20Ω
2mA
3I1
Ix
4A
50Ω
44. Find ix and v8 .
8Ω
10A
25Ω
-
49. Find v1 and i2 .
7A
3Ω
2ix
40Ω
+
Vx
100Ω
5A
+
V8
2A -
3A
9Ω
-
i2
ix
v3
+
50Ω
45Ω
20Ω
30Ω
45. Find i.
+
- 5 i2
0.02v3 +
0.02v1
-
100V
1Ω
i
50. Find V if v1 =0.
5Ω
4A
20
46. Find v1 and i1 .
6A
+
- 0.1v1
Ω
Ω
5V1
2Ω
10
5A +
V1
-
V
96V
+
1Ω
+ v1 3Ω
6Ω
40Ω
-
2Ω
6Ω
2A
v1
i1
4A
51. Find va .
+
0.8va
10Ω
40Ω
20Ω
10A
-
47. Find v p .
50Ω
+
Vp
-
40Ω
2.5A
2Ω
2A
5A
5A
200
Ω
8A
5Ω
+
Va
-
2.5Ω
two points (with polarities) due to
52. Find v1 .
each source acting alone. Similarly,
50Ω
20Ω
30Ω
+
V1
-
5A
the current through any branch, other
than the branch that contains ideal
0.01v1
+
- 0.4v1
ideal current source, in a circuit that
contains many independent sources in
a linear network, is the sum of the
53. Find i1
currents (with polarities) through that
+
branch due to each source acting
-
0.5 i1
30Ω
2A
3V
DC
alone.
DC
4V
i1
6V
two
emf Vth  Voc the open circuit voltage
vy
2Ω
4Ω
Any
represented as a voltage source with
vx
DC
theorem:
terminal, linear active network can be
54. Find k if Vy  0 .
1Ω
Thevinin's
between the terminals, and a series
3Ω
2A
kvx
+
-
resistor Rth equal to the resistance
between
the
terminals,
with
all
independent sources replaced with
Superposition principle (Applicable
only to linear networks): The potential
their
internal
network
resistances.
contains
the
If
the
dependent
difference between any two points,
sources Rth is found out by calculating
other than the terminals of an ideal
the short circuit current I sc and using
voltage source, in a circuit that
the formula Rth 
contains many independent sources in
a linear network, is the sum of the
potential differences between those
Vth
.
I sc
Norton's theorem: Any two terminal,
linear network can be represented as a
current source with source current
, and the maximum power delivered is
I n  I sc
E2
.
4 Rl
the
short
circuit
current
between the terminals, and a shunt
resistor Rth equal to the resistance
between
the
terminals,
with
all
independent sources replaced with
their internal resistances. Here also if
the network contains the dependent
sources, Rth is determined using Vth
V
and Rth  oc .
I sc
In any network, let the voltage across
and current through an element,
(which is neither a coupled coil nor
the element of a dependent source) be
V and I respectively, keeping all the
voltages and the currents in the
network same that element can be
replaced by a voltage source of
+
Active
Network
Substitution theorem:
VOC
Active
Network
ISC
voltage V or a current source with
current I.
-
Reciprocity theorem:
RTH
This is applicable to linear time
VTH=VOC
IN=ISC
RTH
invariant passive network. Let V be
the voltage source in branch ‘a’ and I
Maximum power transfer theorem:
be the current in branch ‘b’, if we
Consider a practical voltage source
keep the same voltage in branch ‘b’
with emf (could be the Thevinin’s
the same current will flow in branch
volt.age of a network) E and a source
‘a’.
resistance (could be the Thevinin’s
resistance of a network) Rs , the power
V1
LTI
Passive
Network
I1
I2
LTI
Passive
Network
delivered by the source (network) is
maximum if the load resistance Rl  Rs
If the networks are same
VI V2
 .
I1 I 2
V2
Problems:
+
10Ω
1. In the network shown if P1 is the
50V
DC
100V
DC
vc
power dissipated in R when V2 is
25Ω
15Ω
zero and P2 is the power dissipated
-
10Ω
power dissipated in R when both
+
in R when V1 is zero, find the
10Ω
DC
10Ω
14V
sources are on.
10Ω
-
R
+
2Ω
Resistive
network
v1
5Ω
v2
15A
3Ω
2. In the network shown, with ia and
-
vb on and vc  0 ix  20 A , with ia
10 Ω
and vc on and vb  0 ix  5 A and
2Ω
4Ω
L
L
+
2V
when all sources are on
ix  12 A
L
L
.
3Ω
6Ω
Find ix due to each source acting
alone
20V
60V
10V
2KΩ
+
2KΩ
ix
2KΩ
ia
Resistive
network
DC
vc
+ix
vb
DC
3. Find Thevinin’s equivalents of the
following networks
5A
5ix
2KΩ
+
250Ω
-
+ ix
5V
20Ω
250Ω
+
19Ω
5Vx
+
5ix
100mA
7.5KΩ
Vx
-
-
5. Check
whether
or
not
the
following networks are reciprocal.
Ie
αIe
Re
+
Rc
2Ω
3Ω
Rb
VS
I
-
6Ω
10V
+
I
1Ω
5V
4. Find Norton’s equivalents of the
following networks
+
15Ω
12Ω
2V
0.1A
50Ω
0.1v1
5V
DC
+
DC
+
200Ω
v1
100Ω
-
Ix
2Ω
-
2Ix
3Ω
I