network analysis & synthesis unit-ii norton’s theorem, …€¦ · this theorem is very basic...

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B y : D E E P A K P A N D E Y e m a i l : d . p a n d e y 2 9 0 6 @ g m a i l . c o m

Page 1

NETWORK ANALYSIS & SYNTHESIS

UNIT-II

2.1 SUPERPOSITION THEOREM

This theorem is very basic theorem of network analysis and

very useful to solve a network where two or more sources are

present and connected in any manner.

2.1.1 Statement: In any linear and bilateral network, if a

number of sources are acting simultaneously, the resultant

response (current or voltage) in any branch is the algebraic

sum of the responses that would be produced in it when each

source acts alone replacing all other independent sources by

their internal resistances (or impedances).

2.1.2 Steps Involving to solve a network using Superposition

theorem:

Step-1: Take only one independent source and deactivate the

other independent sources. Obtain the branch current/voltage. Step-2: Repeat, Step-1 for each of the independent sources.

Step-3: To determine the net current/voltage in the branch, add

algebraically (for ac network, this addition should be phasor

addition) all the current or voltages obtained in Step-1 and Step-

2 for each branch. The net response in each branch is then

obtained.

*Note:

It is to ne noted that this theorem is applicable to determine the response where response and excitation have linear

relation between each other. So power in any branch cannot

be determine using Superposition theorem, because power

in any branch has quadratic relation with voltage (=V2/R) or current (=i2R).

For networks containing dependent sources, during the operation, the dependent sources should be present in

the network i.e. they should not be deactivated during

the process.

Example-1: Determine the current I by using superposition

theorem of network shown in fig.1(a).

Solution: Let us first assume that the 20V source is in operative

mode while the current source of 4A is deactivated mode. In

such condition, the circuit can be shown by fig.1(b).

In fig.1(b), 20 + 4V0 -10I’ + V0 = 0

Or 4 + V0 - 2I’ = 0 (i)

& V0 = - 4I’ (ii)

Using (ii) in (i), we get, 4 + (-4I’) -2I’ = 0

Or A.I ' 6606

4 . (Direction is same as I)

Now remove 20V source and set current source in operative

mode (fig.1(c)).

Applying KCL, we get;

I’’ + I4Ω + (-4) = 0

Or

04410

4 000

VVV

V0 = -80/15 = -5.33V

∴

A..V

"I 665210

3355

10

5 0

. (Direction is

opposite to the direction of I).

∴ I = I’ + I” = 0.66A + (-2.665A) = -2.005 A (direction is same as shown in fig. 1(a)).

2.2 THEVENIN’S THEOREM

Thevenin theorem is a network reduction method which

reduces a complex network in a simpler form.

Complex

Network of

independent &

dependent

sources

ZL

.

.

X

Y

+

-

IZL

.

X

Y

+

-

I+

-

ZTH

VTH

(a) A complex network (b) Thevenin’s Equivalent

.

Fig.2

2.2.1 Statement: It states that a linear active network

constituted of independent and/or dependent voltage and

current sources and linear passive elements can be replaced at

any pair of terminal by an equivalent source VTH in series with

an equivalent resistance RTH (or impedance ZTH). The value of

VTH (called the Thevenin voltage) is equal to the potential

difference between the selected terminals when they are open-

circuited, and RTH (or ZTH) is the Thevenin equivalent

resistance (or Thevenin equivalent impedance) looking into the

network at the selected with the independent active sources set

Network Theorems (Applications to dependent & independent sources): Superposition theorem, Thevenin’s theorem,

Norton’s theorem, Maximum power transfer theorem, Reciprocity theorem, Millman’s theorem, Compensation theorem,

Tellegen’s theorem.

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to zero, leaving their internal resistance (or impedance) in the

circuit. If the network contains a dependent source, the control

variable for the dependent source must be contained within the

same network.

2.2.2 Steps Involving to solve a network using Thevenin’s

theorem:

Step-1: Remove the load resistance (or impedance) and find the

open circuit voltage across the open-circuited load terminals.

Step-2: Step-2 involves the determination of equivalent

Thevenin’s resistance (or impedance), which may be understood

in following methods:

(a) For networks containing independent sources only: If all the sources are independent, then RTH (or ZTH) can be

obtained by deactivating the constant sources (for voltage

source, short-circuit it by internal resistance and for current

source delete the source by open circuit) and find the

internal resistance of the source side looking through the

open-circuited load terminals. This is RTH (or ZTH).

(b) For networks containing independent and dependant sources both: There are two methods to explain:

Method-1: Find open-circuit voltage (VOC) across the

chosen terminals, then short the load terminals to calculate

short-circuit current (ISC) through the shorted terminals. The

internal resistance/ impedance (Thevenin’s equivalent resistance or impedance) of the network is then obtained as:

SC

OCTH

I

VZ .

Method-2: Remove the load resistance and apply a d.c.

driving voltage Vdc at open circuited load terminals.

Keeping the other independent sources deactivated during

this time. A d.c driving current Idc will flow in the circuit

from the load terminals due to Vdc. The internal resistance of

the network is then obtained as: dc

dcTH

I

VZ .

Step-3: Now connect VTH and ZTH in series across the load

terminals as shown in fig.(b). Then load current can be

calculated as: LTH

TH

ZZ

VI

.

Example-2: In given fig.3(a), obtain the current Ix by

Thevenin’s theorem.

Solution: Here we need to determine the value of current

between the points x & y.

Calculation of VTh:

Remove the load resistance (2Ω), then the circuit is reduced

as shown in fig.3(b).

Applying KVL in fig.3(b),

10 – VOC – 3×1 = 0 ⇒ VOC = 7V = VTh

Calculation of RTh:

1. Short-circuit the load terminals x-y (fig.3(c)). 2. In fig.3(c), from KVL;

10 – 1(Isc + 3) – 2Isc = 0 ⟹ A.I sc 3323

7 .

3. Here 3332

7

.I

VR

sc

ocTh Ω.

So with the help of Thevenin’s theorem, fig.3(a) may be

reduced as shown in fig.3(d);

x

y

Ix+

-VTh=7V

Fig.3(d) Thevenin’s Equivalent

2Ω

RTh = 3Ω

From fig.3(d), A.I x 4123

7

2.3 NORTON’S THEOREM

Norton’s theorem is also a network reduction method to

solve network and converse of Thevenin’s theorem. It is also

called the ‘dual’ of Thevenin’s theorem.

Complex

Network of

independent &

dependent

sources

ZL

.

X

Y

+

-

I ZL

.

X

Y

+

-

IZNIN

(b) Norton’s Equivalent

..

(a) A complex network

Fig.4

2.3.1 Statement: It states that a linear active network

constituted of independent and/or dependent voltage and

current sources and linear passive elements can be replaced at

any pair of terminal by an equivalent current source IN,

parallel with an equivalent resistance RN (or impedance ZN).

The value of IN (called the Norton equivalent current) is equal

to the current between the selected terminals when they are

short-circuited, and RN (or ZN) is the Norton equivalent

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resistance (or Norton equivalent impedance) looking into the

network at the selected with the independent active sources set

to zero, leaving their internal resistance (or impedance) in the

circuit. If the network contains a dependent source, the control

variable for the dependent source must be contained within the

same network.

2.3.2 Steps Involving to solve a network using Norton

theorem:

Step-1: Short the load terminals and find the short circuit current

flowing through the shorted load terminals. This will give the

value of Norton equivalent current (IN).

Step-2: Step-2 is exactly same as for the Thevenin’s theorem, in which RTH (or ZTH) is similar to RN (or ZN).

Step-3: Now connect IN and ZN in parallel across the load

terminals as shown in fig.(b). Then load current can be

calculated as: NLN

NI.

ZZ

ZI

.

Example-3: Obtain Norton’s equivalent of the network shown

in fig.5(a).

10 0∟

0Aj15Ω 2Ω

-j5Ω 3Ω

A

B

Fig.5(a) Solution: Calculation of Isc: Short-circuit the terminals AB and find the

short-circuit current between these terminals (fig.5(b)).

0

0

0

6718116

90150

1532

15010

..

j

jIsc

⇒ Nsc IA..I 0

418499

10

0

∟

0 A

j15Ω 2Ω

-j5Ω 3Ω

A

BFig.5(b)

Isc

j15Ω 2Ω

-j5Ω 3Ω

A

BFig.5(c)

O.C

ZTh

Calculation of RN: Deactivate all the sources (here only one

current source, open-circuit it). Refer fig.5(c).

ZTh = (2+3+j15) ∥ (-j5) = 5155

5155

jj

jj

0

00

4631811

9056718115

..

.. 08811111 ..

Norton’s equivalent of network shown in fig.5(a), may be

drawn in fig.5(d).

ZN = 11.11 -81.8 IN = 9.49 18.4

Fig.5(d) Norton’s Equivalent

∟0A

∟0

Ω

2.4 MAXIMUM POWER TRANSFER THEOREM:

Maximum power transfer theorem is used to find the value

of load resistance (in ac, impedance) for which there would b

maximum amount of power transfer from source to load.

2.4.1 Statement: It states that, in a linear network having

energy source and impedances, maximum power is transferred

from source to load when the load impedance is the complex

conjugate to the Thevenin equivalent impedance of the

network as seen from the load terminals.

2.4.2 Explanation: Let assume a network shown in fig. in which

VS and ZS are the Thevenin equivalent voltage and impedance of

the network respectively. If I be the current through the load,

then;

LSLS

S

LS

S

XXjRR

V

ZZ

VI

So, Power (real power) PL = I2RL

L

LSLS

S RXXRR

V

22

2

(1)

In first step, we need to examine about the condition when

maximum power will flow from source to load.

In this condition; 0L

L

dX

dP

From eq.(1)

22

2

LSLS

LS

LL

L

XXRR

RV

dX

d

dX

dP

2222 2

LSLS

LSLS

XXRR

XXRV

(2)

Now, 0L

L

dX

dP, we get from eq.(2);

On substituting the value of XL as –XL in eq(1),

LSLS

LSL

LS

SL

R.RRR

R.VR.

RR

VP

222

22

XL = -XS

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=

THLL

TH

TH

RRR

R

V

22

2

(3)

From eqn. (3)

For RL = 0 (short circuit) ⟹ PL = 0; & for RL = ∞ (open circuit) ⇒ IL = 0, hence, again PL = 0. So for proper value of power, RL should be finite. It is clear from eqn. (3), for maximum value of PL; its

denominator should be minimum as:-

dPL/dRL = 0 & d2D/dR2L should be a positive value.

Where D = SLL

S RRR

R2

2

.

Now:

SL

L

S

LL

RRR

R

dR

d

dR

dD2

2

= 0

= 012

2

L

S

R

R

So,

Or in the other words,

i.e., *SL ZZ . This means that the load impedance is complex

conjugate of the source impedance.

Now the maximum power transferred from the source to the

load, S

S

SS

SSmax

R

V

RR

RVP

4

2

2

2

Example-4: Determine the amount of maximum power

delivered to the network in example-2.

Solution: In example-2, VTh = 7V and RTh = 3Ω. According to

maximum power transfer theorem for maximum power delivered from source to load, RL should be equal to RTh. So in this

condition, the value of load resistance should be equal to 3Ω

(instead of 2Ω).

Maximum power delivered =34

7

4

22

Th

Th

R

V= 4.083 W.

2.5 RECIPROCITY THEOREM

2.5.1 Statement: In any branch of a network, the current (I)

due to a single source of voltage (V) elsewhere in the network

is equal to the current through the branch in which the source

was originally placed when the source is placed in the branch

in which the current (I) was originally obtained.

In simple words, the location of the voltage source and the

current through the resistance (or impedance) may be

interchanged without any change in current.

*Note:

It is to be noted that the polarity of the voltage source should identical with the direction of branch current in each

position.

This theorem is applicable only to single source networks. The network should not have any time varying element.

2.5.2 Steps for solving a network using Reciprocity theorem:

Step-1: First we need to select the branches, between which this

theorem is to be established.

Step-2: Now the current in branch is obtained by using any of

the conventional method.

Step-3: Then interchange the voltage source between the

concerned branches.

Step-4: Calculate the current in branch where the voltage source

was existing earlier.

Step-5: It may be observed that currents obtained in Step-2 and Step-4 are same.

Example-5: Verify Reciprocity theorem by finding the current

through the branch CD in the network shown in fig.6(a).

(2 - j2) Ω (1 + j2) Ω

(2 +

j2)Ω

-j10

Ω

100 0 V∟ 0

Fig.6(a)

I

A B C

DEF

Solution: Applying nodal analysis at B,

V..Vj

Vjjj

BB082172984

22

100

43

1

10

1

22

1

Now current .A...

..

j

VI B 0

0

0

3135891613535

82172984

43

Now, interchange the position of voltage source and position of

current I (fig.6(b)).

(2 - j2) Ω (1 + j2) Ω

(2+

j2)Ω

-j10Ω

100 0 V∟ 0

Fig.6(b)

I

A B C

DEF

In fig.6(b), nodal equation at B is

V..Vj

Vjjj

BB031806247

43

100

43

1

10

1

22

1

So current .A...

..

j

VI B 0

0

0

3135891645812

31806247

22

Hence the theorem is verified.

RL = RS

(RL + jXL) = (RS - jXS)

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2.6 MILLMAN’s THEOREM

This theorem is utilized for a network containing number of parallel voltage sources. These all parallel sources can be

reduced to one equivalent one.

2.6.1 Statement: A network containing number of voltage

sources (V1, V2, V3, …, Vn) having internal resistances (R1, R2,

R3, .. Rn) respectively, then the arrangement can be replaced by

a single equivalent source V in series with an equivalent series

resistance R (in a.c impedance Z) as shown below:

DC DC DC DC

V1 V2 Vn

R1 R2 Rn

RL RLV

R

Fig.7(a)

In reduced network, as per Millman’s theorem;

n

nn

G......GG

GV.......GVGVV

21

2211 &

nG....GGGR

21

11.

2.6.2 Explanation: Assuming a network is shown in fig.7(b),

applying KCL at point x, we get

DC DC DC

V3 Vn

R3 R4 Rn

DC DC

V1

R1 R2

I

V2RL

x

y

I1 I2 I3 I4

In

Fig.7(b)

I = I1 + I2 + I3 +……..+In

Or,

nn

n

n

GV.......GVGVGV

R

V.....

R

V

R

V

R

V

332211

3

3

2

2

1

1

The equivalent resistance (for ac, impedance) be the parallel

combination of all the resistance, so;

nR

........RRRR

11111

321

Or G = G1 + G2 + G3 +…….+ Gn Thus equivalent voltage at x;

n

n

G.....GGG

I.......III

G

IV

321

321

Or, n

nn

G......GG

GV.......GVGVV

21

2211

And equivalent resistance,

nG....GGG

R

21

11

2.6.3 Steps Involving to solve a network using Millman’s

theorem:

Step-1: Obtain the conductance (G1, G2,…...) of each voltage

source (V1, V2,…….) and find equivalent conductance G¸

removing the load.

Step-2: Apply Millman’s theorem to find V by:

n

nn

G......GG

GV.......GVGVV

21

2211

Step-3: Determine the equivalent series resistance (R) with the

equivalent source (V), by

nG....GGG

R

21

11

Step-4: The current through the load is then given by:

LL

RR

VI

, RL being the load resistance.

*Note: In ac, replace G by Y and R by Z (similarly RL by ZL).

Example-6: Using Millman’s theorem, find the current through

RL in the network shown in fig.8(a).

Solution: From Milliman’s theorem, network may be replaced

by:

(1) a single voltage source V, where

-

+

-

+

-

+2V 4V 8V

5Ω 10Ω 15Ω

10Ω

Fig.8(a)

x

y

15

1

10

1

5

1

15

18

10

14

5

12

321

332211

GGG

GVGVGVV

= - 0.266V.

(2) A resistance R, where

15

1

10

1

5

1

11

321

GGG

R

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= 2.72Ω

Now the circuit can be

reduced to fig.8(b),

From fig.8(b),

72210

2660

.

.I

= - 0.021A

2.7 COMPENSATION THEOREM

Compensation theorem is a convenient means to determine

changes in voltages and currents in a circuit when resistance (or

impedance) is changed in one of the branches of the circuit.

ACVTH

ZTH

ZL

IL

.

+

+

ACVTH

ZTH

ZL+ΔZIL

.

+

+,

. .

(a) Thevenin equivalent

of circuit supplying a

load ZL

(b) Thevenin equivalent

of circuit supplying a

load ZL+ΔZ

AC

ZTH

ZL+ΔZΔIL

.

.VC (= ILΔZL)

(c) Compensation theorem circuit

Fig.9

2.7.1 Statement: In a linear time-invariant network when the

resistance (impedance) in a branch varies by ΔR (or ΔZ), the

change in voltage is equal to that produced by an opposing

voltage source of magnitude VC = IΔR (or ΔZ), where I is the

current flowing through the branch prior to the variation in R

(or Z) and the current can be obtained by assuming that this

source VC has been connected in series with R+ΔR (or Z+ΔZ)

when all other sources in the network are replaced by their

internal resistance (or impedance).

2.7.2 Explanation: Consider the Thevenin equivalent circuit of

network is shown in fig.9(a), which is connected across ZL.

From fig.9(a), load current is given by LTH

THL

ZZ

VI

From fig.9(b), load current due to change in load impedance is

given by LLTH

TH'L

ZZZ

VI

.

The change of current being termed as ΔIL, we find as

LTH

TH

LLTH

THL

'LL

ZZ

V

ZZZ

VIII

=

LLTH

L

LTH

TH

ZZZ

Z

ZZ

V

=

LLTH

LL

ZZZ

ZI

Form the fig.9(c), it can be seen that

LLTH

CL

ZZZ

VI

Hence it is clear that VC = IΔZL termed as compensating

voltage.

Example-7: In the network shown in fig.10(a), find the change

in current flowing through the impedance Z, when its value

changed from (4 – j3)Ω to (6 + j5)Ω using compensation theorem.

10 30 V∟ 0

IΔI

VC=IΔZ

(a) (b)Fig.10

Z Z+ΔZ

Solution: From fig.10(a), current in the network:

A.j

I 00

1383234

3010

Now, change in impedance;

ΔZ = [(6+j5) – (4-j3)] = (2+j8)Ω

So from Compensation theorem;

VC = IΔZ = V..j. 00 831425168213832

In fig.10(b), 56

831425160

j

..

ZZ

VI

C

093102112 .. A.

2.8 TELLEGEN’s THEOREM

This theorem is one of the most general theorem which is applicable to any network made up of lumped two terminal

elements.

2.8.1 Statement: Tellegen’s theorem states that any instant of

time, the sum of instantaneous power delivered to an electric

circuit is zero.

Mathematically, for all instant of time t, Tellegen’s

theorem may be expressed as:

n

kkk )t(i)t(v 0 , for all values of t.

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2.8.2 Explanation: For a complex, the power supplied by active

circuit is given by

m

kkkmm )t(i)t(v)t(i)t(v...)t(i)t(v)t(i)t(v

12211

The power dissipated or stored in the passive network is

expressed as follows:

m

kkkmm )t(i)t(v)t(i)t(v...)t(i)t(v)t(i)t(v

12211

Since the rates of energy supplied and dissipated (or stored) are

equal, it is clear that;

m

kkk

m

kkk )t(i)t(v)t(i)t(v

11

or 01 1

m

k

m

kkkkk )t(i)t(v)t(i)t(v

2.8.3 Steps Involving to solve a network using Tellegen’s

theorem: Step-1: Find the branch voltage drop and corresponding branch

currents using conventional analysis.

Step-2: Then this theorem can be justified by summing all

products of branch voltage and currents.

If the set of voltages and currents is taken corresponding to

two different instants of time, t1 and t2, Tellegen’s theorem is

applicable, from where we get;

m

kkk

m

kkk )t(i)t(v)t(i)t(v

112

121 0

Example-8: Verify Tellegen’s theorem for the network shown

in fig.11.

I1

I2 I3

16Ω

j3Ω -j4Ω 20 0 V∟ 0

Fig.11

Solution: From the circuit shown in fig.11, we can write

00

1 93616080

43

4316

020..j.

jj

jjI

By current division rule;

422343

49361 02 .j.

jj

j.I

A

V1 = (0.8 – j0.6)×16 = 12.8 – j9.6 V

814243

39361 03 .j.

jj

j.I

A

V2 = V3 = 7.2 + j9.6 V

& 0020V V

Now,

4

1kkk iV

6080208142442233608016 .j..j.j.j.j.j.

0

Hence, Tellegen’s theorem is proved.

EXERCISE

Q.1. Find the current through the capacitor and voltage across

4Ω resistance of the a.c. network shown in fig.A by using

superposition theorem.

Q.2. Obtain VAB using Millman’s theorem in the circuit given in

fig.B and verify the result using superposition theorem.

20Ω

10Ω

Fig.B

50Ω

+

-

+

-

P

Q

15

0V

10

0V

40Ω 30Ω

Q.3. (i) What should be the value of ZL for maximum power

transfer from source to load in network shown in Fig.C?

(ii) What is the value of maximum power transfer from source to

load in network shown in Fig.C?

ZB

ZB44

0 -

45

V∟

0

ZA

ZA

ZL

Where:

ZA =

1F

1H

1F 1HZB =

Fig.C

Q.4. Obtain the Norton’s equivalent of network shown in fig.D.

1Ω 1Ω

+ - + -+

-

a

b

1A I1 I2

10I1 10I2

Fig.D

network analysis & synthesis unit-ii norton’s theorem, …€¦ · this theorem is very basic...

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