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1
Basic Circuit Elements

1
...
In fact circuit theory has revolutionised the process of analysis
in all branches of engineering be it Civil, Mechanical, Electronices or Computer engineering
...
Circuit theory is a simplified approximation of a more exact field theory
...
Of course
an electrical engineer is more interested in voltage and current from which charge, power
energy fields etc
...

The following approximations are made while writing circuit equations:
(i) The distributed capacitance effects or displacement currents from one conductor
to another have been ignored
...

(iii) The radiation effects are ignored
...

The two basic quantities to define the functions of circuits are charge and energy
...
An atom has positive charges (protons) in
its nucleus and an equal number of electrons (negative charges) surround the nucleus
making the atom neutral
...
The basic unit of charge is the
charge on an electron
...
An electron has a charge of
1
...

When a charge is transferred from one point in the circuit to another point it constitutes what is known as electric current
...
Its unit is ampere
...

Mathematically,
dq

...
1)
dt
If charge q is expressed in coulomb and time in second, 1 amp flow of current through
a section is equivalent to approx
...
24 × 1018 electrons per second through the
section
...

However, its form can be converted i
...
energy can be converted from one form of energy
to another e
...
Electro-mechanical energy conversion, Electro-chemical energy conversion,
MHD energy conversion, photoelectric energy conversion etc
...
e
...
The
movement of charges contribute to current and the amount of work done per unit charge
is the potential difference between the two points
...

If a differential charge dq is given a differential energy dw, the rise in potential of the
charge

i=

v =

dw
dq

dq
dt
dw dq
×
= p
v × i =
dq
dt
which gives rate of change of energy with time and is equal to power
...
(1
...
(1
...
(1
...
2 CIRCUIT ELEMENTS
Basically there are three circuit elements in electric circuits
...
We shall study these elements in terms of the physical phenomenon,
occurring in these elements, the field phenomenon associated with them and finally its
relevance in circuit theory
...
These collisions being inelastic, these electrons lose energy
...
The time rate of flow
of electrons through the section of the material constitutes the current
...
It has been observed
experimentally that if J is the current density in A/m2 σ is the conductivity of the material
then
J =σ E

...
5)
Consider a wire of length l and cross-section A, the total current through the wire
I = JA

...
6)
and voltage drop across the wire
v = E
...
(1
...
l
l
=
=

...
8)
I
J
...
The unit is ohm after the name of the
discoverer who derived this relation experimentally
...
The field was
found to be magnetic in nature by Oersted when he brought compass needle in its vicinity
which experienced force and got deflected
...
The direction of force is found to be experimentally perpendicular to the current carrying
conductor
...
1
...


Fig
...
1 Magnetic field lines surrounding a current carrying conductor
...
With the thumb pointing in the direction of the
current, the fingers of the right hand encircling the wire, point in the direction of the
magnetic field
...
A magnetic compass could be used to measure
approximate direction of these lines
...
(1
...
(1
...
The flux could
be changed by either changing (i) the relative position between the two circuits continuously
or by changing the (ii) current in the circuit
...


v =k


...
11)

z
z


...
12)

Therefore,

Ψ = v dt

Also

q =

i dt

in electrostatic field
...
(1
...
Since flux linkages are
related to the magnetic field B which in turn depends on current i, we can write
Ψ = NΦ = N

z ez

dB ds i

j


...
14)

where the quantity in the bracket is defined as inductance of the circuit
...
e
...
However, if i produces flux
linkages Ψ2 in another circuit, the parameter is termed as mutual inductance, denoted by
M and is expressed as
Ψ2 = M 2 i

...
15)
We know that

v = d Ψ/dt


...
16)

and

Ψ =L i


...
17)

d Ψ/dt = L

Therefore,

di
=v
dt


...
18)

if L is constant
...
In fact, the
inductance variation is shown as in Fig
...
2
...
1
...


Now taking equation (1
...
(1
...

The differential equations describing the circuit in Fig
...
3
...
(1
...
1
...
c
...


Even though under steady state condition the current through the inductor is theoretically infinite but at t = 0+ the current is zero and the increase in current is gradual with time
...
Again this field is not
visible but its presence can be felt by bringing another charge in the vicinity of this charge
...
Also, if there are two metallic parallel plates separated by some
distance d as shown in Fig
...
4 and if one of the plates is charged with q+ charges, the other
plate is charged with q– charges by induction and an electric field is set up between the
plates and the two plates experience a force between each other
...
Now D =
A where D is the field flux density and
A is the area of plates
...


Fig
...
4

Therefore E =
the plate is q
...


q
V
=
where V is the voltage across the plates when the charge on
d
∈A
q
∈A
=C =
V
d


...
19)

6

Network Analysis

Here C is known as capacitance between the parallel plates separated by a distance d
...
If q is measured in
coulombs and V in volts, C is measured in Farads
...
(1
...
But from the integral equation for charge q, it is
found that for finite value of i, the charge can’t change at t = 0+ and it remains q0
...

Let us take up a few examples to illustrate the application of some of the relations
obtained in the previous discussions
...
1: The current wave form in a series circuit of 500 Ω resistance and a
500 mH inductor is given in Fig
...
1 (a)
...


Fig
...
1
...


The voltage across the inductor

di
From 0 to 2 m sec, the change in current di is zero hence
= 0
...

From 2 to 3 m sec over 1 m sec, the change in current is 20 mA

i
...


=

20 × 10 −3
= 20 A/sec
1 × 10 −3

di
= 500 × 10–3 × 20
dt
= 10 volt
...
Hence voltage across
inductor is zero
...

L

Hence voltage

Example 1
...
Fig
...
1
...
1 H
...
E1
...


Solution: In an inductor current i is given
1
v dt
i =
L
From 0 to 2 m sec v = 0
...
1 2

z

4
2

= 100 mA

8

Network Analysis

From 4 to 6 m sec, v = 0 since there is no change in the current it continues to be same
i
...
100 mA
From 6 to 10 m sec
...
1

10

zb

g

−5 dt = − 200 mA

6

It decreases linearly between 6 to 10 m sec
...
is
–200 + 100 = –100 mA
From 10 to 12 m sec
...
Hence current increases linearly by
an amount
1
0
...

Let us next consider application of non-sinuoidal voltages and currents in a capacitive
circuit
...
3: Consider the application of a voltage wave form shown in Fig
...
3 (a)
to a 0
...

Determine the current wave form
...


(b) Current wave form
...
E
...
3

Solution: Since current in a capacitor is given by
i =C

dv
dt

0
From 0 to 2 m sec dv = 0 hence = 0 Hence the current in this period is zero
...
Hence

v = kt, k =

10
= 10 × 103 v/sec
1 × 10 −3

dv
= 0
...
2 × 10–6 × 10 × 103

i =C

= 2 mA

or

10 V/m sec

Basic Circuit Elements

Beyond 3 m sec since v is constant

9

dv
=0
dt

Hence
i =0
The variation of current is shown in Fig
...
3(b)

1
...
1

SINGLE-PHASE CIRCUIT

Let us consider an inductive circuit and let the instantaneous voltage be
v = Vm sin ωt


...
21)

Then the current will be i = Im sin (ωt – φ) where φ is the angle by which the current
lags the voltage (Fig
...
5)
The instantaneous power is given by
p = vi = Vm sin ωt
...
(1
...
In this case the current
flows in the direction of voltage drop
...
If the circuit is purely reactive the
voltage and current will be 90º out of phase and hence the power will have equal positive
and negative half cycles and the average value will be zero
...
22) the power
pulsates around the average power at double the supply frequency
...
22) can be rewritten as
p = VI cos φ (1 – cos 2 ωt) – VI sin φ sin 2 ωt

...
23)
=

Fig
...
5

b

g

Voltage, current and power in single phase circuit
...
1
...
1
...


(i) The component P marked I pulsates around the same average power VI cos φ but
never goes negative as the factor (1 – cos 2ωt) can at the most become zero but
it will never go negative
...

(ii) The component marked II contains the term sin φ which is negative for capacitive
circuit and is positive for inductive circuit
...
This component is known as reactive power as it travels back
and forth on the line without doing any useful work
...
23) is rewritten as
p = P (1 – cos 2 ωt) – Q sin 2 ωt


...
24)

Both P and Q have the same dimensions of watts but to emphasise the fact that Q
represents a nonactive power, it is measured in terms of voltamperes reactive i
...
V Ar
...
This means a capacitor is a
generator of positive reactive V Ar, a concept which is usually adopted by power system
engineers
...
1
...


Fig
...
7

V-I relations in a capacitor
...
If Q is the total
reactive power requirement of the load and Q′ is the reactive power that the capacitor can
generate, the net reactive power to be transmitted over the line will be (Q – Q′)
...
The
phase modifier controls the flow of reactive power by suitable excitation and hence the

Basic Circuit Elements

11

voltage is controlled
...

It is interesting to consider the case when a capacitor and an inductor of the same
reactive power requirement are connected in parallel (Fig 1
...


Fig
...
8

Power flow in L-C circuit
...
The line power will, therefore, be zero
...
In one half cycle at a
particular moment the capacitor is fully charged and the coil has no energy stored
...

Consider the case when the circuit is purely resistive when φ = 0
P = V I (1 – cos 2 ωt)
= V I – V I cos 2 ωt


...
25)

which means, the average value of power in a purely resistive circuit consists of two
components, one a constant value VI and superimposed over this is a sinusoidally varying
with double the supply frequency and its average value is zero
...
The wave form is shown in Fig
...
9
...
1
...


Next we consider a purely inductive circuit when φ = 90º
The power equation becomes
p = V I cos φ (1 – cos 2 ωt) – VI sin φ sin 2 ωt
= –V I sin 2 ωt

...
26)
This means the average power is zero
...
1
...

A positive loop shows that power is drawn from the source and negative loop shows that
power is fed back to the source
...
Thus power or energy
is taken from the source for a quarter cycle and then it is returned back to the source over

12

Network Analysis

the next quarter cycle
...
This concept implies that the 90º voltage current phase relationship must be for either an inductor or a capacitor since these are the two basic electrical
elements that are capable of storing energy
...


Fig
...
10

An inductor is a device capable of storing energy in the form of moving charge or in
the form of a magnetic field
...

The following example illustrates the relationship between the reactive power and the
electric field energy stored by the capacitor
...
1
...


Fig
...
11

Relationship between electric field energy and reactive power
...
1
...
(1
...
e
...
(1
...
(1
...
(1
...
(1
...
The energy
stored by the capacitor,

∴

Reactive power =

W =
Now

v =

=

∴

2

1
Cv2
2
1
C

z

idt =


...
32)
1
C

VmωC
2

2

2

R ω C +1

⋅

cos (

Vm cos ( ωt + φ )
R2ω2C 2 + 1

V 2 cos2 (wt + f) V 2 cos
1
C◊ m
=
W =
2
R 2 w 2C 2 + 1
R2w


...
33)


...
34)

dW
V2
⋅ 2 cos ( ωt + φ ) ⋅ sin (
= 2 2 2
dt
R ω C +1

=

V 2 ωC
⋅ sin 2 ( ωt + φ )
R ω2C 2 + 1
2

= Q sin 2 (ωt + φ)

...
35)
From this it is clear that the rate of change of electric field energy is a harmonically
varying quantity with a frequency double the supply frequency and has a peak value equal
to Q
...

1
...
2

The 3-phase Circuit

Assuming that the system is balanced which means that the three-phase voltages and
currents are balanced
...
(1
...

p = Vaia + Vbib + Vcic
= Vm sin ωt Im sin (ωt – φ) + Vm sin (ωt – 120º) Im sin
(ωt – 120 – φ) + VmIm sin (ωt + 120º) sin (ωt + 120º – φ)
= V I[2 sin ωt sin (ωt – φ) + 2 sin (ωt – 120º) sin (ωt – 120º – φ)
+ 2 sin (ωt + 120º) sin (ωt + 120º – φ)]

14

Network Analysis

= V I[cos φ – cos (2 ωt –φ) + cos φ – cos (2 ωt – 240º – φ)
+ cos φ – cos (2 ωt + 240º – φ)
= 3 V I cos φ

...
37)
This shows that the total instantaneous 3-phase power is constant and is equal to three
times the real power per phase i
...
p = 3P, where P is the power per phase
...
This does not mean that the reactive power is of
no importance in a 3-phase system
...
Similarly, even though the sum of reactive
power instantaneously in 3-phase system is zero but in each phase it does exist and is equal
to V I sin φ and, therefore, for 3-φ the reactive power is equal to Q3φ = 3V I sin φ = 3Q, where
Q is the reactive power in each phase
...
Nevertheless the
reactive power in a 3-phase system is expressed as Q3φ
...

1
...
3

Complex Power

Consider a single phase network and let
V = |V|ejα and I = |I|ejβ

...
38)
where α and β are the angles that V and I subtend with respect to some reference axis
...
e
...
(1
...
(1
...
(1
...
The magnitude of S = P 2 + Q 2 is termed
as the apparent power and its units are volt-amperes and the larger units are kVA or MVA
...
It is to be noted that Q is positive when (α – β) is positive
i
...
when V leads I i
...
the load is inductive and Q is –ve when V lags I i
...
the load is
capacitive
...
e
...
Therefore, to
obtain proper sign for reactive power it is necessary to find out V I* rather than V*I which
would reverse the sign for Q as
V *I = |V|e–jα |I|ejβ = |V| |I|e–j(α – β)
= |V| |I| cos (α – β) – j |V| |I| sin (α – β)
= |V| |I| cos φ – j |V| |I| sin φ

Basic Circuit Elements

= P – jQ
P = Re [V

*

15


...
42)
I] and Q = – Im [V

*

I]


...
43)

1
...
The first one is based on the kind
of elements in the network e
...
time-invariant or nonlinear or time variant etc
...
e
...

Here we shall study classification based on the second approach
...
3
...
e
...
Then the network is classified as linear if for excitation v1 (t) + v2 (t) the
response is c1 (t) + c2 (t)
...

1
...
2

Passivity

Some networks have the property of absorbing/dissipating or storing energy
...
However, this energy
returned is never more than the energy stored
...

Let network N be connected to a source S
...

If v(t) is the voltage of the source and i(t) is the resulting current into the network N, the
average power delivered into the network is p(t) = v(t) i(t) and the energy delivered w(t) over
time t is given by
t

w(t) =

z

v(t ) i (t ) dt


...
44)

−∞

This energy will always be non-negative for a passive network i
...
over time t there
will always be some energy delivered to the network or at least there will not be any energy
when the network would have returned all the energy back to the source or it would have
stored some energy
...
Mathematically for a passive network
...
(1
...
(1
...
That is for
an active circuit
t

z

v( t ) i( t ) dt < 0


...
47)

−∞

1
...
3

Lumped

Many devices in electric system are distributed in space e
...
transmission lines, the windings of transformers or that of the generators, are distributed in a way
...
However, if we are interested in steady
state and terminal quantities, it is sufficient to assume the parameters to be lumped rather
than distributed
...
e
...
Otherwise, we assume the parameters to be lumped
...

1
...
4 Bilateral
The elements to be considered for electric network are assumed to be bilateral which means
the voltage and current relations are same irrespective of direction of flow of current e
...

resistance, inductance, etc
...
g
...

1
...
5

Time Invariance

A network is said to be time-invariant when there is some response to a certain excitation
irrespective of time of application of excitation
...
Here
the values of the parameters are assumed to be constant at all times and do not change with
time
...
3
...
Consider network N with terminals 1 and 2 as shown in
Fig
...
12
...
1
...


Say if v1 is excitation and i2 response as in Fig
...
12 (a) and if v2 is excitation then i1
is the response as in Fig
...
12 (b) The network is reciprocal if for v1 = v2, i1 = i2
...
The network which does not satisfy the condition

Basic Circuit Elements

17

is a non-reciprocal network
...
Since L and C can store energy in the form of electro-magnetic and
electro-static fields respectively their representation needs special mention
...
(1
...
The capacitor which has initial voltage
v(o) can be represented as shown in Fig
...
13(a)
...
1
...


Similarly if an inductor initially has a current i(o) the current at anytime t is given by
1
v( t ) dt + i ( o )

...
49)
L
Therefore, i(o) could be considered as a d
...
current source in parallel with an initially
relaxed (initial condition zero) inductor as shown in Fig 1
...


i(t) =

1
...
An ideal transformer shown in Fig
...
14 is described by the
following equations:

(a)

(b)
Fig
...
14

The ideal transformer
...
1
...
1
...
(1
...
(1
...
The v-i relationships indicate idealized relations expressing Faraday’s law
(v1 = n v2) and Ampere’s law (i2 = –n i1)
...

Also it can be seen easily that the total energy input to the ideal transformer is zero
...
(1
...
It can be summarized that an ideal transformer is
(i) an abstract concept;
(ii) characterized by a single parameter n, the turns ratio;
(iii) the voltage and current are idealized relations derived out of Faraday’s law and
Ampere’s law respectively;
(iv) One where a resistance R connected to secondary side is reflected as n2R when
referred to primary side; and
(v) a passive device
...
4
...
1
...


Fig
...
15 The perfect transformer
...
(1
...
(1
...
The total energy delivered to the transformer is
t

=

z

(v1i1 + v2i2 ) dt

−∞

Basic Circuit Elements
t

=

z FGH L

1

−∞
i1

=

z

IJ
K

di1
di
+ M 2 i1 dt +
dt
dt

L1 i1 di1 +

i1 , i2

z

19

t

z FGH

L

−∞

M ( i2 di1 + i1 di2

0

0

1
1
2
2
L1 i1 + M i1 i2 + L2 i2

...
54)
2
2
For perfect transformer to be passive, the energy stored should be non-negative i
...


=

2
2
L1 i1 + 2 M i1 i2 + L2 i2 ≥ 0


...
55)

To prove that it is non-negative, we prove that the minimum of this is non-negative
and hence the expression will be non-negative for all other values
...

The above expression reduces to
f (x) = L1 i12 – 2 Mx i12 + L2 x2 i12
df ( x )
2
2
= – 2M i1 + 2 x L2 i1 = 0
dx

or

x =

M
L2

Substituting value of x in f (x) we have
f (x) =
or
or
or

Let

FL
GH

1

−2

M2
M2
+
L2
L2

I ≥0
JK

L1L2 – M 2 ≥ 0
M 2 ≤ L1L2
M2
≤1
L1 L2

K2 =

M2
L1 L2

or

K=

M
L1 L2

Since this inequality is true, the inequality (1
...

Here K is known as co-efficient of coupling and its maximum value is unity when this
transformer is known as a perfect or perfectly coupled transformer
...
To establish the difference we know
v
that in case of ideal transformer 1 = n
v2
In case of perfect transformer
v1
L di dt + M di2 dt
= 1 1
v2
L2 di2 dt + M di1 dt

20

Network Analysis

=

L1 di1 dt +

L1L2 di2 dt

L2 di2 dt +

L1 L2 di1 dt

Multiplying and dividing the RHS by L2 we have
L1

v1
=
v2

=

L2
L1

L2 L1 ⋅ L1 di1 dt +
L2
di
⋅ L2 2 +
L1
dt

L2
L1

L1 L
L1 L2 d

L1
L2


...
58)

This shows for perfect transformer to be the same as ideal transformer n =

L1 L2
...
Fig
...
16
shows relation between a perfect and an ideal transformer
...
16

1
...


THE GYRATOR

A gyrator is another two-port terminal device shown in Fig
...
17

(a)

(b)
Fig
...
17

A Gyrator
...
1
...
(1
...
(1
...
1
...
The direction of arrow associated with r gives
the direction of gyration
...
It can be shown that gyrator is not a
reciprocal device, it is anti-reciprocal
...
(1
...
Suppose the gyrator is terminated through a resistance
R, it’s equivalent on the input side is given by
v1 = –ri2 = – r

FG −v IJ = r r i
HRK R
2

1

= ( r2 G ) i1


...
62)

Thus, the equivalent resistance at the input terminals equals r2 times the conductance
terminating at the output terminals
...

A very interesting case is found when the gyrator is terminated through an inductor
or a capacitor
...
The equivalent of this when
referred to input side is obtained as follows
...
1
...


FG
H

v1 = − ri2 = ( − r ) −C dv2
dt
2
=rC

di1
di
= Leq 1
dt
dt

IJ = rC d (r
K dt

...
63)

22

Network Analysis

This shows that through inversion a r2 times capacitor is equivalent to an inductor and
it can be shown that an inductor is inverted as a capacitor
...

These can be simulated through the use of a gyrator terminated through a suitable capacitor
...
1
...
1
...


V1 = –I2 r1 and V2 = I1 r2
The arrow here indicates that the device has a forward transmission path and this has
no backward transmission
...
However if r1 ≠
r2 it is an active device as power can be made negative as shown here
...
(1
...
The input impedance of the gyrator is
Zin =

=

1
...
(1
...
(1
...
1
...

It can be seen from the first set of equations that when i1 is in the reference direction
i2 is also in the reference direction and hence the current is said to be inverted in going
through the negative converter
...
Therefore, the
first set of equations characterise a current negative converter
...


(a)
Fig
...
20

(b)

Negative converters (a) Current negative converter (b) Voltage negative converter
...
Suppose these are terminated through an inductor L
...
(1
...

Thus at the input terminals, the equivalent inductance is – k2L i
...
negative of the k2L
...

Similar conclusions can be drawn by terminating the device with R or C
...
7

INDEPENDENT SOURCES

So far we have considered passive elements when interconnected, form a network
...
Mainly there are two types of sources (i) Voltage
source (ii) Current source
...

1
...
1

Ideal Voltage Source

An ideal voltage source is a two terminal device whose terminal voltage is independent of
the current drawn by the network connected to its terminals
...
This means an ideal voltage source should have zero
internal resistance
...
Fig
...
21 shows the
symbols and reference conventions for such a source and the voltage current characteristic
...
1
...
c
...


24

Network Analysis

When the voltage generated does not change with time, it is represented by a battery
as shown in Fig
...
(i)
However, in actual practice, there is no voltage source which does not have internal
resistance, however small it could be and, therefore, an actual voltage source is always
associated with an internal resistance as shown in Fig
...
22
...
1
...


Therefore, the terminal voltage decreases as the current drawn by the external network increases and the voltage current relation is as shown in Fig
...
22(b)
...
7
...
e
...
Here again it makes no sense to consider open circuiting of the
terminals of a current source as this again imposes two conflicting requirements at the terminals
...
Fig
...
23 shows the symbol and reference convention for the current source and also
the v-i characteristic of such a source
...
1
...


Certain devices may be represented by the model shown in Fig
...
24 wherein a resistor
is connected across the current source
...
1
...


(a)

(b)

Fig
...
24 (i) Current source (ii) v-i charisteristic
...


Basic Circuit Elements

1
...
These devices have two pairs of terminals, one pair corresponds to the controlling quantity and the other the controlled quantity
...

There are controlled sources where the voltage/current is proportional to derivative of
some other voltage or current
...


1
...

(ii) Shifting the positions of the sources in the network
...
e
...

Suppose there are two voltage sources in series as shown in Fig
...
25, the equivalent
will be one source with voltage (v1 + v2)
...
We know that two generators can’t be

(a)
Fig
...
25

(b)

(a) Two voltage sources in series, (b) Two current sources in parallel
...

Fig
...
26 (a) shows a resistance R connected across a voltage source v and Fig
...
26
(b) shows a resistance R in series with a current source when an external network is
connected across the terminals of these source circuits
...
1
...


(a)
Fig
...
26

(a) R connected across a voltage source,

(b)
(b) R connected in series with a current source
...
To this, the basic criterion
is that the terminal conditions (v and i) of the two networks, the source and the external
network must remain same before and after the transformation
...
Refer to Fig 1
...

I

v ( t)

R

v 1 (t)

Fig
...
27

E xt
...


v = v1 + I R


...
68)

v
v
− 1

...
69)
R R
Equation (1
...
1
...
69) which has been derived
from equation (1
...
1
...
1
...

Similarly if the series element is an inductor or a capacitor with the voltage source Fig
...
28
(a) and (b), the equivalent current source is given by the circuit as shown in Fig
...
28
...
1
...


Similarly current sources can be transformed into voltage sources

Basic Circuit Elements

Fig
...
29

Current to voltage source transformation
...
1
...
1
...
(1
...
(1
...


is = i +

1
L

z

v1 dt


...
72)

z

or

L is = Li + v1 dt
Differentiating the equation gives
L

di
dis
+ v1
=L
dt
dt

dis
di

...
73)
−L
dt
dt
Similarly an equivalent voltage source can be obtained when a current source is connected across a capacitor
...
However for more than one passive elements when considered, above method can’t be used, as the determination of the source equivalent involves the
solutions of the differential equation
...
Here, before transformation could be
carried out, it is necessary to shift the source within the network
...
31
...
1
...


It is to be noted that whereas current distribution remains unaffected, the voltage distribution in the network is changed since node ‘o’ is at the same potential as ‘d’ after the shift
...
1
...


Fig
...
32

Shifting of current sources
...

While drawing graph from a network, the source transformation makes following
suggestions:
(i) Elements in parallel with voltage source or in series with a current source can be
eliminated from the graph
...

While preparing a network before analysis following thumb rules are to be followed for
convenience
...
However, if it is to be analysed based on state variable approach, the network may
contain both kinds of sources
...
10

DOT CONVENTION FOR COUPLED CIRCUITS

The significance of dot convention is explained with the help of Fig
...
33
...
1
...


terminal
...
The following experimental procedure can be used to establish dotted ends
of the transformer windings
...
c
...
Connect a MC voltmeter
across the secondary winding
...

If there are more than two windings, similar procedure can be followed for identifying
relationship between each pair of windings
...

Suppose we are required to write loop equations for a mutually coupled circuit shown
in Fig 1
...

r1

v( t)

i1

M

i2

r2

Fig
...
34 Coupled circuit
...
The voltage induced in primary will be determined by the direction of i2 with
respect to its dot in secondary loop
...
Therefore, for current
i1 it will be rise in voltage and the equation will be

di1
di
- M 2 = v(t)
dt
dt
and for the secondary loop, similarly
i1r1 + L1

i2 r2 + L2

di2
di
−M 1 = 0
dt
dt

30

Network Analysis

The equivalent two loops are as shown in Fig
...
35
...
1
...
1
...


Consider Fig
...
36 which is a magnetically coupled network

Fig
...
36

Magnetically coupled network
...
1
...


Fig
...
37

Equivalent of Fig
...
36
...

Voltage induced in coil 1
Due to coil 2, the current through coil 2 is (i2 – i3) entering the circular dot, therefore the
voltage induced in coil 1 will be with positive polarity at the dot
...

Voltage induced in coil 2
Due to coil 1, the current (i1 – i2) enters the dot, therefore positive polarity at the dot of
coil 2
...

Voltage induced in coil 3
Due to coil 1 the current entering the dot is (i1 – i2) hence the positive polarity at the dotted
terminal of coil 3
...
Hence the equivalent circuit given as follows:

Fig
...
38

Equivalent of Fig
...
36 with voltage sources
...

When the two coils are to be interconnected it is important to know whether the
mutual inductance M is aiding or opposing
...
Refer to Fig
...
39(a) where the mutual inductance is
aiding their self inductances whereas in (b) it is opposing
...
1
...
39(b) then,
+ L3

z

32

Network Analysis

Fig
...
39

(a) Two coils aid and (b) Oppose
...
1
...


LA = L1 + L2 + 2M
LB = L1 + L2 – 2M

LA − LB
4
This method provides a very convenient way of determining the value of mutual
inductance between the coils
...
40 are both zero
...

Therefore, the linkages are zero and the value of M is zero
...


M =

Example 1
...

Since the flux through the two coils opposes each other, the mutual inductance is
substractive from each coil and hence

Fig
...
1
...
73H
15
L1 + L2

Example 1
...
E1
...

LA = 25 + 10 – 10 = 25H
LB = 30 + 10 – 15 = 25H
Therefore,

LC = 35 – 15 – 10 = 10H

1
1
1
2+2+5
1
+
+
=
=
Leq
25 25 10
50
50
= 5
...

A given network will always have a unique solution in terms of its nodal voltages and
loop currents
...
It would be seen that sometimes it is more convenient to use nodal voltages
as independent variables, sometimes loop currents as independent variables and yet sometimes a suitable combination of both from the point of view of minimum independent
variables required to describe the netowrk and hence the number of equations to seek
solution of the network
...
Therefore, from solution point of view, whichever is
smaller of the two should be used for analysis of a given network
...
nodal vollages) as we may not be interested in the other
kind of variables (viz
...


Example 1
...
E1
...
E1
...
E1
...
7: Two gyrators are connected in cascade
...


Fig
...
7

v1 = –ir1
v = –i1r1 = i2 r2
v2 = –ir2
To eliminate v and i
v2 =

v1
r
r = 2 v1
r1 2 r1

−r1
i1
r2
The relations show that two gyrators connected in cascade behave as an ideal transr
former with ratio as 2
...
1
...

1
...
A circuit has voltage and current given as
v = 100 sin

FG ωt − π IJ
H 6K

and i = 10 sin

FG ωt + π IJ
H 6K

Determine the power loss
...
3
...
Is the circuit inductive or capacitive
...
4
...
Determine and
V2
(ii) P = I2 R
...
5
...
Determine the energy stored by the capacitors
...
6
...
Determine the equation for the current in the
circuit
...
7
...
P1
...


compare the power in the circuit using the formulae (i) P =

Fig
...
7

1
...
A gyrator is terminated through a series R, L circuit as shown in Fig
...

Determine an equivalent two element network shown in Fig
...
8(b)

(a)

(b)
Fig
...
8

1
...
Three coils are interconnected with winding sense shown by dots
...
Determine Leq between
1 1′

36

Network Analysis

Fig
...
9

1
...
Two coils are wound on the same core and have negligible resistance as shown in the
Fig
...
10
...


Fig
...
10



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