EE2092!3!2011 Network Theorems

8/13/2019 EE2092!3!2011 Network Theorems

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Network Theorems Professor J R Lucas 1 May 2011

3 Network Theorems J . R. Lucas

Fundamental laws that govern electric circuits are

Ohms Law and Kirchoffs Laws.

Ohms Law

v(t) i(t)

v(t) =R .i(t)

Can be extended to cover inductances and capacitors

under alternating current conditions under transient conditions.

R

v(t)

i(t)


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Generalised Ohms Law

Stated as

v(t) = Z(p) . i(t)

where p = d/dt = differential operator

Z(p)Impedance function of the circuit

The differential equation governs behaviour of circuit.

For a resistor, Z(p) = R

For an inductor Z(p) = L p

For a capacitor, Z(p) = pC1

Z(p)

v(t)

i(t)


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With alternating current,

p = jso that governing circuit behaviour is written as

V = Z(j).I

For a resistor, Z(j) = R

For an inductor Z(j) = jL

For a capacitor, Z(j) = Cj1

Cannot analyse electric circuits using Ohms Law only.

Need Kirchoffs current law and Kirchoffs voltage law.


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Kirchoffs Current LawBased on the principle of conservation of charge.

algebraic sum of charges within a system cannot change.00

dt

dqq

Total rate of change of charge must add up to zero.

0IIdt

dq Kirchoffs current law

algebraic sum of currents meeting at a point is zero.

at a node, Ir= 0,

where Ircurrents in branches meeting at nodeSometimes stated as sum of the currents entering a node is

equal to sum of currents leaving the node.


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Theorem applicable not only to node, but to closed system.

i1+ i2i3+ i4i5= 0 iaib+ icidie= 0

or i1+ i

2+ i

4= i

3+ i

5

i1

i2

i3

i5i4

iaid

ib

ie

ic


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Kirchoffs Voltage LawBased on principle of conservation of energy.

Requires that the total work done in taking a unit positive

charge around a closed path and ending up at the originalpoint is zero.

Algebraic sum of potential differences taken round a closedloop is zero.

Vr= 0where Vrbranch voltages

va+ vb+ vc+ vdve= 0

Sometimes stated as sum of

emfs taken around a closed

loop is equal to sum of voltage

drops around loop.

va

vd

vb

ve

vcloop


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Using only Ohms Law and Kirchoffs laws, would be tedious.

Network theorems are formulated to simplify calculations.

Although followings examples are with resistive circuits with d.c., for

simplicity, the laws are applicable to a.c. as well.

Example 1Using Ohms Law and Kirchoffs Laws to solve current I in 160

I = I1 I2

100 = 20 I1 + 160 (I1I2) 10 = 18 I116 I270 = 20 I2160 (I1I2)

7 = 16 I118 I2I1= 1 A, I2= 0.5 A

I = 0.5A.

I1 I2

IE1= 100V

20 20

160

E2= 70 V


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Superposition Theorem

If a network comprises of more than one source,

resulting currents and voltages in network determined

by taking each source independently, and

superposing the results.

r(t) = r1(t) + r2(t)

Excitationsk1e1(t)andk2e2(t)response k1 r1(t) + k2r2(t)

Linear

Passive

Bilateral

Network

e1(t)

e2(t)

r(t)

Linear

Passive

Bilateral

Network

e1(t)

r1(t)Linear

Passive

Bilateral

Network

e2(t)

r2(t)


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Example 2

Solve earlier problem using Superposition theorem.

Solution

E1= 100 V

20 20

160 E2= 70 V

+E1= 100 V

20 20

160 i1

20 20

160 E2= 70 V

i2


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for circuit 1,

source current =

i1=2.647 = 0.294 A

Similarly for circuit 2,

source current=

i2 = 1.853 = 0.206 A

unknown current i = i1+ i2= 0.294+ 0.206= 0.500 A (same answer as before)

A647.2778.37

100

180

2016020

100

20//16020

100

180

20

A853.1778.3770

180

2016020

7020//16020

70

180

20


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Thevenins Theorem(or Helmholtzs Theorem)

A linear, active, bilateral network across one port

replaced by Thevenins voltage source andequivalent series impedance (Thevenins impedance).

Comparison under open circuit conditions

EThevenin = Voc, and ZThevenin = Zin

Linear

ActiveBilateral

Network

ZThevenin

ETheveninZin Voc


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Example 3

Consider earlier example to illustrate Thevenins Theorem.

Solution

To calculate the current in 160 ,find the Thevenins equivalent

circuit across the terminals afterdisconnecting (open circuiting)

the 160 resistor.

E1= 100 V

20 20

160 E2= 70 V

A

B

E1= 100 V

20 20

160

E2= 70 V


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Under open circuit conditions,

Current flowing = (10070)/40 = 0.75 A

Voc,AB = 1000.75 20 = 85 V

ETh= 85 VInput impedance across AB (with sources shorted)

= 20//20 = 10 .

ZTh= 10 .Thevenins equivalent circuit isredrawn with branch AB

reintroduced.

i= (same result as earlier)

ETh= 85 V

ZTh=10

160

A

B

i

A5.016010

85


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Nortons Theorem

Dual of Thevenins theorem

states that any linear, active, bilateral network,considered across one of its ports, can be replaced byNortons current source and an equivalent shunt

admittance (Nortons Admittance).

Inorton= Isc, and Ynorton = Yin

YNortonINorton

Linear

Active

Bilateral

Network

Yin Isc


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Example 4

Consider same example to illustrate Nortons Theorem.

Solution

To calculate the current in the

160 , find the Nortonsequivalent circuit across the

terminals after short-circuiting

the 160 resistor.

E1= 100 V

20 20

160 E2= 70 V

E1= 100 V

160 E2= 70 V

A

B

Isc

20 20


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Short circuit current Iscis given by

Isc = 100/20 + 70/20 = 8.5 A

Inorton= 8.5 ANortons admittance = 1/20 + 1/20 = 0.1 S

Nortons equivalent circuit is

and the current in the unknown resistor is

A5.000625.01.0

00625.05.8

(same result as before).

8.5 A0.1 S 160

A

B

or

0.00625 S


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Reciprocality Theorem

In a linear passive bilateral network an excitation and

the corresponding response may be interchanged.In a two port network,

if e(t) at port (1) produces response r(t) at a port (2),

then if same e(t) is applied instead to port (2), sameresponse r(t) would occur at port (1).

Linear

Passive

BilateralNetwork

Port 1 Port 2e(t)

r(t)Linear

Passive

BilateralNetwork

Port 1 Port 2e(t)

r(t)


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Example 5

Consider earlier example, but

with one source and current in

the 160 branch.Verify reciprocality theorem.

Solution

E1= 100 V

20 20

160

20 20

160 E1= 100 V

I1

I2

E1= 100 V

20 20

160


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Compensation Theorem

After circuit is analysed, it is realised that only a small

change is required to a component to get desired result.

Compensation theorem allows us to compensate properlyfor such changes (ZZ+Z) without sacrificing accuracy.

+

Linear

ActiveBilateral

NetworkZ

IV

Linear

PassiveBilateral

NetworkZ+

II . V

Linear

Active

Bilateral

Network

Z+

I +I LinearActive

Bilateral

Network

Z

I +IV+V

I .


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Voltage drop across the modified branch

V+V = (Z+Z)( I+I) = Z . I + Z . I + (Z +Z) . I

from the original network, V = Z . I

V = Z . I + (Z +Z) . I

I already known from the earlier analysis,

change Z also known,

I .Z is a known fixed value of voltage, and may thusbe represented by a source of emf I.Z .Using superposition theorem,

original sources give rise to the original current I,while change corresponding to the emf I.Z mustproduce the remaining changes in the network.


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Example 6

From example 4, the current in 160 resistor is 0.5 A.It is required to change R by a quantity R such that thecurrent in the 160 resistor is 0.600 A.I = 0.5

I = 0.60.5 = 0.1 A

E1= 100 V

20 20

160

IE2= 70 V


23/46


20 20

(160+R)

I.R

I 20//20160

5.0)(

R

RI

i.e. R

R

170

5.0)(1.0

17 + 0.1 R = (-) 0.5 Ri.e. R = ()17/0.6 = () 28.333

Required R = 16028.333 = 131.67 Checking with Thevenins theorem

From Example 2, Thevenins circuit with

the 160 replaced by 131.67 .Ai 6.0

667.13110

85

(same result)

ETh= 85 V

ZTh=10

131.67

A

B

i


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Maximum Power Transfer Theorem

A car battery is rated at 12 V but can haveopen circuit voltage ~12.6 to13.8 V.

Similarly, if we take 9 pen-torch batteries,they too will have a terminal voltage of

91.5 = 13.5 V.

You would be aware, that if your car battery is dead, you

cannot use 9 pen-torch batteries and start your car.

Why is that ?

Because pen-torch batteries, although having same opencircuit voltage does not have necessary power (or current

capacity) and hence required current could not be given.


25/46


Or if stated in different words,

Pen torch batteries have a high internal resistance(say 90.5)

car battery have a low internal resistance (say 0.05)so that with pen torch batteries voltage would drop

without giving necessary current.

For example, for a 12V, 5W bulb (current 0.4 A),

voltage of a car battery would drop by 0.02 V where as

that of the 9 torch batteries would drop by 1.8 V.

Both operate at near 12 V and would be satisfactory.


26/46


If instead, the starting current of a car (say 40A) is

supplied, the drop would be 2 V for the car battery,which is acceptable.

1513.5

10

5

00 5 10 15 20 25 30 35 40 i A

V

Car battery

9 pen torchbatteries

v= Er i


27/46


Even the short-circuit current that can be supplied by

the 9 pen torch batteries is just 3A and hence in no way

can the 40 A current be supplied.

This means that a given battery (or any other energy

supply, such as the mains) can only give a limited

amount of power to a load.

Maximum power transfer theorem defines this power,and tells us the condition at which this occurs.

For example, if we consider the above battery,

maximum voltage would be given when the current is

zero, and maximum current would be given when the

load is short-circuit (load voltage is zero).


28/46


Under both these conditions, there is no power

delivered to the load.

Thus obviously in between these two extremes must be

the point at which maximum power is delivered.

The Maximum Power Transfer theorem states that for

maximum active power to be delivered to the load, load

impedance must correspond to the conjugate of thesource impedance (or in the case of direct quantities, be

equal to the source impedance).

Let us analyse this, by first starting with the basic case

of a resistive load being supplied from a source with

only an internal resistance (this is the same as for d.c.)


29/46


Resistive Load supplied from a source with only an

internal resistance

Consider a source with an internal

emf of E and an internal

resistance of r and a load of

resistance R.

current I =

Load P = I

2

. R =

Only R is a variable.

r

RE

I

V

P

R

rRE

RrR

E

2


30/46


To obtain maximum power transfer to the load,

0)(21)()(

2

4

2

rRRrRrR

E

dR

dP

[Note: It is the maximum, rather than maximum or minimum,because from physical considerations there must a maximum

power in the range. Second derivative is not required to seewhether it is maximum or minimum].

(R+r)22R (R+r) = 0or R + r2R = 0

i.e. R = r for maximum power transfer.


31/46


value of maximum power = r

Er

rr

EP

4

22

max

load voltage at maximum power = 2.. E

rrr

E

RrR

E


32/46


33/46


Since there are two variables R and X, for maximum

00

X

Pand

R

P

0)(21)()(

)()(

22

222

2

RrRXxRr

XxRr

E

and )(2

)()( 222

2XxR

XxRr

E

The second equation gives x+ X = 0 or X = - x

Substituting in the first equation gives

(r+R)2R . 2(r+R) = 0

i b i


34/46


Since R cannot be negative,

r + R 0.

r + R

2R = 0i.e. R = r

Z = R + j X = rj x = z*

Therefore for maximum power transfer, the load

impedance must be equal to the conjugate of the source

impedance.


35/46


Load of f ixed power factor suppl ied from a source

with an internal impedance

Given power factorf

)()( XxjRr

E

jXRjxr

EI

R and X are no longer independent

power factor f =

or X = = k . R

2222 ).()()()( RkxRr

E

XxRr

EI

r + jx

R +jXE

I

V

22 XR

R

R

f

112

2


36/46


Load Power P = I2. R =R

RkxRr

E

22

2

).()(

for maximum power 0dRdP

kRkxRrRRkxRr

RkxRr

E)..(2)(21).()(

).()(

22

222

2

= 0

(r+R)2+ (x+k.R)2

2R(r+R)

2k.R(x+k.R) = 0r2+2 r.R + R2+ x2+ 2k.x.R + k2.R22 R.r2 R22k.R.x2 k2.R2= 0

i.e. r2R

2+ x

2k

2.R

2= 0

i.e. R

2

+ k

2

R

2

= r

2

+ x

2

. i.e. R

2

+ X

2

= r

2

+ x

2

i.e. Z = z

Mill Th


37/46


Millmanns Theorem

For a number of admittances

Y1, Y2, Y3 .Yp. Yq,

Yn are connected togetherat a common point S.

If the voltages of the free

ends of the admittances withrespect to a common

reference N are known to be

V1N, V2N, V3N .VpN.

VqN, VnN, then Millmannstheorem gives the voltage of the common point S with

respect to the reference N as follows.

S Y1

Y2

Y3Yp

Yn

n

1

2

3

p

Nreference

Yq

q


38/46

A t i f th Mill th i th


39/46


An extension of the Millmann theorem is the

Equivalent generator theorem

For a system of voltage sources operating in parallel

n

kk

n

kkk

eq

Y

YE

E

1

1

, n

k

keq YY1

E1

Z1

E2

Z2

Ek

Zk

En

Zn

Ee

Ze

ZL

IL

ZL

IL

E l 6


40/46


Example 6

Using Millmanns theorem

find the current I.

Solution

equivalent generator has

VEeq 85

20

1

20

1

7020

1100

20

1

10

20

1

20

1

1eqZ

Hence current I =5.0

16010

85

A (same answer)

E1= 100 V

20 20

160

I

E2= 70 V


41/46

F d l t k f Mill th


42/46


For nodal network, from Millmanns theorem

, so that

Iq = Yq(VqN

VSN) = Yq.VqNYq.

n

pp

n

ppNp

SN

Y

VY

V

1

1

n

pp

n

ppNp

Y

VY

1

1

n

p

p

pN

n

p

qNpq

n

p

p

n

p

pNp

n

p

pqNq

q

Y

V-VYY

Y

VY-YVY

I

1

1

1

11

)(

F d fi it ti i bl k i i t i l


43/46


For a definite summation, variablep or kis immaterial

n

p

pqn

k

k

qp

n

k

k

pN

n

p

qNpq

q VVY

YY

Y

V-VYY

I1

11

1

)(

)(

For the mesh network,Comparing equations,

statement ofRosens theorem

n

k

k

qp

pqY

YY

Y

1

n

ppqpqq VVYI

1)(

Star Delta Transformation


44/46


Star-Delta TransformationNodal-Mesh Transformation when n = 3

CSBSAS

ASCSCA

CSBSAS

CSBSBC

CSBSAS

BSASAB

YYY

YYY

YYY

YYY

YYY

YYY

.,

.

,.

YAB

YBC

YCA

A

BC

YAS

YBSYCS

A

BC

Delta Star Transformation


45/46


Delta-Star Transformation

In 3 node case, equal branches in both nodal network (known

as star) and mesh network (known as delta).

Thusdelta connectednetwork equivalentstar

Z

Z Z

Z Z ZZ

Z Z

Z Z ZZ

Z Z

Z Z ZAS

AB CA

AB BC CA

BSAB BC

AB BC CA

CSCA BC

AB BC CA

.

,

.

,

.

ZAS

ZBSZCS

A

BC

ZAB

ZBC

ZCA

A

BC

Proof:


46/46

Proof:Compare the two networks

ZCA//(ZAB+ ZBC) = ZCS+ ZAS

i.e. ASCSBCABCA

BCABCA ZZZZZ

ZZZ

)(

similarly

BSAS

CABCAB

CABCAB ZZ

ZZZ

ZZZ

)(

and CSBSABCABC

ABCABC ZZZZZ

ZZZ

)(

elimination of variables gives the desired results.

CABCAB

BCAB

CABCAB

CAABBSAS

ZZZ

ZZ

ZZZ

ZZZZ

..

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EE2092!3!2011 Network Theorems