EE2092!2!2011 Coupled Circuits Dependant Sources - Copy

8/13/2019 EE2092!2!2011 Coupled Circuits Dependant Sources - Copy

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Coupled Circuits Professor J R Lucas 1 November 2010

Coupled CircuitsJ. R. Lucas

ResonanceResonance in sound.

At resonance

maximum vibration (string) and hear the greatest sound (water column)

Same idea is present in electrical engineering.

Resonance basically occurs when a quantity,such as voltage or current, becomes a maximum.

Maximum one quantity minimum with another


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Resonance, Mutual Inductance & coupled circuits Professor J R Lucas 2 May 2011

So that resonance could occur at a minimum value of a

quantity as well.

For example, for a given source voltage,

currentwouldbemaximumwhen

impedanceof the circuit is aminimum.

In an a.c. circuit,

would have a minimum value when X is zero

(Possible because inductive and capacitive reactances

have opposite signs).

Z

VI

22 XRjXRZ


3/100


When X is zero, the circuit is purely resistive and the

power factor of the circuit becomes unity.

Thus resonance is also defined in terms of the powerfactor of a circuit becoming unity.

Three main methods of defining resonance condition:

(a) current through a circuit for a given source voltagebecomes amaximum:

voltage across a circuit for a given source currentbecomes a minimum,

admittance of the circuit becomes a maximum, or impedance of the circuit becomes a minimum.


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(b) voltage across a circuit for a given source currentbecomes amaximum:

current through a circuit for a given source voltagebecomes a minimum,

impedance of the circuit becomes a maximum, or admittance of the circuit becomes a minimum.

and

(c)power factorof the circuit becomesUnity:

when the impedance of the circuit is purely real,

when the admittance of the circuit is purely real, or when the voltage and the current are in phase.


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Condition (a) occurs in series circuits, and is usuallyreferred to asSeries Resonance.

Condition (b) occurs in parallel circuits, and is usuallyreferred to asParallel Resonance.

While series resonance and parallel resonance are exclusive

conditions, unity power factor condition could correspondto either series resonance or parallel resonance.

In complicated circuits, the unity power factor condition

could also give displaced answers from the other twoconditions.


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When does oscillations occur ?What determines natural frequency of oscillations ?

Consider the simple pendulum.Why does it oscillate ?

Because of stored energy in the system

potential and kinetic energy.

During oscillations, energy transfers

between potential energy and kinetic energy.

Natural frequency of oscillations, depends

on the value of gravity, length and so on.Friction in the medium would reduce the energy and cause the

pendulum to slow down and finally stop.


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In an electric circuit,

Energy is stored in the

electrostatic fieldin thecapacitance electromagnetic fieldin theinductance.

Oscillations occur when the energy gets transferred

between these two forms.

Resistance would cause energy losses

results decreasing magnitude of oscillations.


8/100Resonance, Mutual Inductance & coupled circuits Professor J R Lucas 8 May 2011

Series Resonance

Occurs in a circuit where the energy storage elements

are connected in series.

At angular frequency of ,Z = R + jL + = R + j

Consider value of when |I| becomes a maximum for agiven V |Z| becoming a minimum, or |V| becominga minimum for a given I.

R LCj

1

)1

(C

L

Cj

1


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magnitude of impedance = | Z |, | Z |2= R

2+

phase angle of Z, =

Condition for maximum or minimum | Z | can either be

obtained by

differentiation of | Z | or differentiation of even | Z |2 or inspection from physical considerations.

Since | Z |2consists of the sum of two square terms, the

minimum value for any of components would be zero.

2)1

(C

L

R

CL /1tan 1


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Since only the second term is dependant on , theminimum value could occur when second term is zero.

i.e. for minimum value of | Z |2 ,= 0 or

or

This would also correspond to the minimum value of Z.If the current were considered,

| I | = ,

phase angle of I=

)1

( CL

CL

1

LCo

1

22 )1()1(C

LR

E

CLjR

E


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Q R 0

high Q

low R

low Qhigh R

| I |

/2

/2

o

o

high Q

low R

low Q

hi h R

Q R 0lead

lag

phase

angle


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Now consider the frequency at which the power factor

of the circuit becomes unity.

CapacitiveReactance

XL

XC

XL+XC

InductiveReactance


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Occurs when imaginary-part(impedance) 0or imaginary-part(current) 0.

= 0, or at = o. (same as before).i.e. j

Shape of curve is dependant on value of rof the circuit.

Asr 0,perfect resonance occurs, I at o

Asr 0, angle of current with respect to voltage tendsto either /2 or /2, changing at o.

Series resistancerdefines quality of the resonance.

If r is low,Plossis low and hence the quality is high.At resonance, low r means r


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Quali ty of a Circui t

In simplest terms, we could defined quality of the

circuit in terms of either ratio of Lo to ror1/Co to r.Quality = at resonance, for a series circuit

[For a parallel circuit, quality would actually increase

when value of resistance R rather R 0, andhence inverse ratio would define the quality]

A means of measuring quality of a resonant circuit has

been defined,but which would appear to cause confusing results if the

series and parallel circuits are considered.

rCr

L

o

o

1


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Quality Factor Q

Necessary to have a unique definition

In terms of quantities which do not vary with circuit.Quality of resonanceis arelationshipbetween

maximum energy storedin energy storing elements(L or C)and

energy dissipationin resistive elements in the circuit.

For consistency with original definitions, Qfactor has a

multiplying constant 2, in addition to ratio of energies.

Q =cycleperndissipatioenergy

energystoredtotal2


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Determination of total stored energy

Consider the simple pendulum.

maximum kinetic energy occurs at thelowest point

maximum potential energy occurs at thehighest point

total stored energy does not change iffriction were absent.

It is easier to calculate either the

maximum kinetic energy only at the lowest point, or the maximum potential energy only at the highest point,rather than calculate both at any other point.


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Total stored energy in Electrical Circuit

maximum energy stored in the electromagnetic fieldoccurs in the inductor when current is a maximum,

maximum energy stored in the electrostatic fieldoccurs in the capacitor when voltage is a maximum.

Total energy does not change unless dissipated by the

resistive elements in the circuit.For calculation purposes, energy stored at either the

peak current or the peak voltage is considered.

For a series circuit, it is easier to consider the currentthrough the inductance rather than the voltage across

the capacitor.


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Verification of Energy based and Impedance based formulae

Necessary to verify whether energy based formula for Q

is compatible with the impedance based formula for Q.

For the series circuit, for a current of i(t) = Imsin tMaximum energy stored in inductor = L Im

2

Energy dissipated per cycle in series resistor = r Irms2T,

where T = 2Q-factor for series circuit would be given by

Q =

which is the expected result.Similar verification can be done for parallel circuit.

r

L

Ir

LI

TrI

LI

m

m

rms

m

/2)2/(

.22

2

21

2

2

21


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Relationship between Q-factor and bandwidth

In a resonance curve, Pmax

occurs at Imax(or Vmax) at

resonance frequency (o).SinceP I2(or V2), halfpower occurs when I (or V)

magnitude decreases by afactor of 2.Points (and ) on the

resonance curve where half power occurs are known ashalf power points.

Im

2

mI

0 o


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Frequency range between the half power points is

defined as the bandwidth of the resonance curve.

Fractional deviation of frequencyfrom resonance

=For a series resonant circuit, we have the condition

| I | = ,

and RCR

LQ

o

o

1

,LC

o

1

o

o

22

)

1

()

1

( CLR

E

CLjR

E

22

1.

o

oo

o

L

RL

E


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so that | I |

2

1

2

2

1

1.

o

o

o

Q

L

E

2

1

2

2

1

o

ooo

Q

L

E


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near resonance, if the Q factor is high (as usually is),

o and =,

| I | =

| I | is a maximum when = 0, giving |Imax| = ,

at the half power points,.

i.e. i= = =

2

o

o

o

21

222

1

2

2

414

1

QR

E

QL

E

o

R

E

max2

1max2

1 , IIPP

2

maxIRE

21

21

22

2141 QR

E


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i.e. 1 + 4 Q2

2= 2, giving 4 Q

2

2= 1

= + , =

Bandwidth= ( ) = ( )

= =

Q-factor = bandwidthfrequencyresonance

[This same definition can be shown to betrue for parallel resonance as well]

Parallel Resonance

R

L

Cj

1

Q2

1

Q2

1

Q

o


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Parallel resonance occurs in a circuit where

the different energy storage elements are

connected in parallel.

At an angular frequency of ,

Y = R1

+ Lj1

+ Cj =)

1(

1

LCj

R

magnitude of admittance = | Y |, | Y |2=

2

2 )

1(

1

LC

R

parallel resonance is seen to occur when imaginary part is zero.

i.e. =Also corresponds to unity power factor.

L1 C


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In practical circuits, series resonance and parallel

resonance can occur in different parts of same circuit.

The unity power factor resonance may correspond to

one of these.

Example 1

Find the types of resonance

and resonance frequenciesof the circuit shown.

Solution

Z = R + jL1+ = R + jL1+

R L1

C

L2

CjLj

CjLj

1

1

2

2

CL

Lj

2

2

2

1


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Consider each type of resonance condition in turn

(a) when the power factor is unity

equivalent impedance is purely real.

Therefore L1+ = 0

i.e. L1+ L2- 2

L1L2C = 0, or ,where Leq=

It is seen that the equivalent inductance is L1// L1.

Thus unity power factor resonance frequency

corresponds to = CLeq1

CL

L

2

2

2

1

CLCLL

LL

eq

1

21

212

21

21

LL

LL


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(b) when the impedance of the circuit is a minimum

Examination shows that this also corresponds to a

minimum value of impedance or series resonance.

This need not be the case for all examples.

(c) when the impedance of the circuit is a maximum

| Z |2 = R

2 +

Z will have a maximum value of at (1 - L2 C) = 0

resonant frequency for parallel resonance =

2

2

2

21

1

CL

LL

CL2

1


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Example 2

Find the unity power factor resonance

frequency of circuit.

Also determine the parallel resonance frequency

if R = 20 , L = 10 mH, and C = 4 F.

Solution

Z = =

for unity power factor resonance, impedance Z is purely real.

R

C

L

CjLjR

CjLjR

1

1)(

CRjLC

LjR

)1( 2


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CRL

LC

R

)1( 2 , i.e. C R2 = L (1 - 2L C)

2= CLCRL2

2

, or = CLCRL2

2

Unity power factor resonance

=

Now consider the magnitude of the impedance

| Z |2 =

for maximum value of | Z |, | Z |2 must be a maximum.

Hzsrad 3.729/6.458210410100

4001041010

66

63

22222

222

)1( RCLC

LR


30/100


= [(1-2LC)2+ 2C2R2].2L2i.e.

(R

2

+

2

L

2

)[2(1-

2

LC).(2LC) + 2C2

R

2

] = 0i.e. (1-2LC)2L2+ 2C2L2R2+ 2R2LC(1-2LC)

+ 22L2LC(1-2LC)(R2+ 2L2)C2R2=0

4

L

4

C

2

+

2

[-2L

3

C + C

2

L

2

R

2

- 2R

2

L

2

C

2

+2 L

3

CL2C

2R

2] +L

2+ 2R

2LC - C

2R

4= 0

substituting values for the components

-0.1610-1841.2810-122 + 1.294410-4= 04810620.80901015= 0

d

Zd 2||


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This quadratic equation in can be solved as follows2 = 4106(161012+ 0.80901015)

= 4106

28.723106

= 26.723106

rad/s = 791.4 Hz

It can be seen that this resonance frequency is close but

not the same as the earlier result of 729.3 Hz.


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Loci Diagrams for RL and RC circuits

(a) Series RL circuit

For a constant voltage V at constant

frequency applied,

If resistance R of the circuit varies,VR+ VL= V= a constant.

V = (R + j L) .I = (R + j X) .I

If V is taken as reference, thecurrent I would lag the voltage by

a phase angle .

R L

V,

I

IVR

VL

P


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VR is in phase with I and VL would be quadrature

leading I (or current I lagging the voltage VLby 90o).

Since VRand VLmust be mutually perpendicular, when

R varies, point P must move along a semi-circle.

This semi-circle is the locus of point P as R is varied.

Consider the variation of I as R varies

and X is kept fixed.V = VL+ VR.

VLis first drawn perpendicular to I

(such that current lags voltage by 90o

).The locus of the point P will again be

a semi-circle.

I

VRVL

P


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Since X is fixed, there will be a definite proportion

between length of phasor I and length of phase VL.

Also, I is always lagging the voltage VLby 90o.

Thus the locus of I must also be a semi-circle lagging

the semi-circle for voltage VLby 90oas shown.

(b) Series RL circuit with inductance L having a finite

resistance rIn practical inductances winding

resistance r may not be negligible.

VX would have two componentscorresponding to inductive and

resistive components respectively.

R LI r

V, VX


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The locus of the node between

the resistor and the inductor is

no longer a semi-circle.

However, for total resistanceR + r the locus of that point

with the pure inductance part remains a semi-circle.

If X is variable, then ratio of VRto Vrremains aconstant and value of the internal resistance of the coil

can be determined.

IVR

VL

PVr

VX


36/100


(c) Series RC circuit

Series RC circuit analysed in a similar

manner to series RL circuit.

However current would be leadinginstead of lagging the voltage.

Unlike in the practical inductor, the

practical capacitor does not have

significant losses.

Thus the practical locus diagram is

same as the theoretical diagram.

(d)Parallel RL and RC CircuitsLoci diagrams of parallel RL and RC circuits is obtained in

a manner similar to obtaining that for the series circuits.

IVR

VCP


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Mutual Inductance

Mutual coupling between coils exist, when one coil is in

the magnetic field created by another coil.

When a varying current ip(t) flows in the primary

winding, then a varying flux pis produced in the samecoil and produces a back emf ep(t).

Lp Ls

p m

ep esMi

pis


38/100


Part of the flux produced mcan link with a second coil.Since this flux will also be varying, an induced emf es

will be produced in the second coil.

ep= Np td

d p

, es= Ns td

d m

, m= k . p

kcoefficient of coupling

mutual flux m primary flux pk < 1, k 1 (slightly less but very close to unity).

In the linear region of the magnetisation characteristic,

flux produced p current ipproducing the flux.Also, mutual flux m current ip.


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Thus, induced emf es rate of change of current ip.The constant of proportionality is defined as the Mutual

inductance Msp.

i.e. td

idM

td

ide

p

sp

p

s

, where p

ps

p

ms

spi

kN

i

NM

Thus the (coefficient of) Mutual inductance is definedas the flux linkage produced in a secondary winding per

unit current in the primary winding.

The sign associated with the mutual inductance can be

positive or negative dependent on the relative directionsof the two windings.

Di i f l i d d i d il ill d d


40/100


Direction of voltage induced in second coil will depend

on the relative direction of winding of the two coils.

Flux p in a coil is related to thecurrent ip producing the fluxthrough the self inductance of coil

Lpand may be expressed in terms

of the dimensions of the magneticpath (length l and cross-sectionA) as follows.

i.e. Lp= p

pp

p

pp

p

pp

i

AHN

i

ABN

i

N

, but Np ip= Hpl

Lp= l

AN

li

AiNN p

p

ppp 2

l

Np Ns

Th l i d b i il l d i d i


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The mutual inductance may be similarly derived in

terms of the dimensions.

Msp = p

ppsp

p

pssp

p

pssp

i

AHNk

i

ABNk

i

Nk

= l

ANNk

li

AiNNk spsp

p

ppssp

similarly, Mps = l

ANNk psps

Coupling between the primary and the secondary, for allpractical purposes, will be identical to the coupling

between the secondary and the primary, so that kps= ksp.

Th i b h f ll i l M i


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Thus it can be seen that for all practical purposes, Mpsis

identical to Msp, and would usually be denoted by a

single quantity M and a single coefficient k.

M = ,

Lp = , Ls=

giving M2= k

2LpLs or M = k spLL

l

ANkN ps

lANp

2

lANs

2

E t d i i f t ll l d il


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Energy stored in a pair of mutually coupled coils

Inductor stores energy in electromagnetic field =2

21 LI .

When mutual coupling is present, total energy stored bytwo coils is different from addition of terms.

Difference is the effective energy

stored in the mutual inductance.Total energy stored

= vpipdt + vsisdt

= dtidtidM

dtidLdti

dtidM

dtidL s

ps

sp

sp

p ).().(

= LpipdipM ipdis + LsisdisM isdip

Vp VsLp Ls

isip M

2

21 LI

Thi b d t i


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This may be grouped to give

Total energy stored

= Lpipdip + LsisdisM ipdisM isdip= spsspp iiMiLiL

2

212

21

Therefore the effective energy stored in the mutual

inductance corresponds to M ipis.

E i l t i d t f 2 t ll l d il i i


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Equivalent inductance of 2 mutually coupled coils in series

Coils can either be connected in series so that

fluxes aid each other, or fluxes oppose each other.Since each current is i,

Total energy stored = iiMiLiL .2

2212

121 or

= iiMiLiL .2

2212

121

L1 L2 L1 L2i i

If i l i l t i d t L i id d


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If a single equivalent inductor Leqis considered,

total energy stored =2

21 iLeq .

Thus equating the energies2

21 iLeq = iiMiLiL .

2

2212

121 or iiMiLiL .

2

2212

121

i.e. Leq = L1+ L2+ 2 M or L1+ L22 MThe effective inductance can either increase or decrease

due to mutual coupling dependant on whether the coils

are wound in the same direction or not.

E l 3


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Example 3

Consider a couple of coils connected in series as shown.

Let each coil have N turns,

then total series connected

coil has 2N turns.

If the dimensions of the common magnetic circuit on

which these are wound have areaAand length l

L1= L2= l

AN 2

,

and total coil L = lAN

lAN

22

4)2( .

L1+ L2is obviously not equal to this total.

L2L1

I1 I2

Wh t t ?


48/100


What went wrong ?

If the two coils are wound on the same magnetic circuit,

very closely, then there would be mutual coupling with

the coefficient of coupling almost unity.

Then, M = k spLL 1 lAN

l

AN

l

AN 222

Thus the total inductance would be

L1+ L2+ 2 M = l

AN 24

as expected.

Magnetic circ it Anal sis


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Magnetic circuit Analysis

When a coil is wound round a magnetic core,

the core becomes magnetised

one side becomes a north pole, and other side becomes a south pole.There are different methods of remembering which side

is which.

One of the simplest methods to remember is

When we look at a coil from one side


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When we look at a coil from one side,

if current direction is anti-clockwise, nearer side is anorth pole (also seen from arrow direction of N);

if current direction is clockwise, nearer side is asouth pole (also seen from arrow direction of S).

Consider coil A and coil B,

we can easily visualise that they are both wound in thesame direction.

A B

left hand side right hand side

i e if we examine the coil from the left hand side when


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i.e. if we examine the coil from the left hand side when

a current is entering the left hand side end of the coil,

each coil would produce a north pole at the left hand

side and a south pole at the right hand side.

With a slightly differently drawn diagram it would be

less obvious.

Further if the magnetic circuit was not a straight line,confusion could tend to enter the decision.

A

BA B

left hand side right hand side

A method would be to open out the bent path to make it


52/100


A method would be to open out the bent path to make it

a straight line and then compare directions.

Each time we analyse a magnetic circuit, we would

need to look at the relative directions of the two coils.

However, once wound, the relative directions of the

coils would not change, independent of how we look at

the coils.Thus we can use a simpler method to know the relative

directions of the coils. For this purpose we use dotsto

denote similar ends of different coils.

Dot Notation


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Dot Notation

Relative directions of coils is important in determining

direction of induced voltage with mutual coupling.

Consider the following magnetic circuits, and the

corresponding electrical circuit with polarities defined

by the dotted ends.

Note that there are two possible ways of drawing the

dotted ends, but both give the same relationship to each

other.

A B

A B

A B

The dots do not indicate that one end is a north pole or a


54/100


The dots do not indicate that one end is a north pole or a

south pole, as this would depend on which direction the

current passes in the coil.

Dots indicate similar ends, in that if a changing fluxpasses in magnetic core, then induced voltages would

either all correspond to direction of dot or all to

opposite direction.Thus they indicate similar ends of windings only.

Consider the same core as before, but with the coil B

wound in the opposite direction.

A B

A B

A B

P Q R S

P Q R S

The position of the dots again indicate the different


55/100


The position of the dots again indicate the different

winding directions.

Once the dots are drawn to indicate similar ends, unlike

in the magnetic circuit, there is no real necessity tophysically place the winding diagram in the same

physical position. Thus the following diagrams would

be identical.

Thus it is seen that once the dots have been marked to

identify similar ends, the physical positions have no

meaning and we can draw them where it is convenient.

P Q R S P Q

R

S

Let us now see how once the dots have been marked in


56/100


Let us now see how, once the dots have been marked, in

an electrical circuit the correct directions of the induced

voltage scan be obtained.

If a single coil is considered, the induced voltage would

always be opposite to the direction of current flow in

that winding.

The voltages V1and V2have been drawn as such in the

above diagrams.

A B

V1 V2

I1 I2A B

V1 V2

I1 I2

Let us now see what would happen due to each


57/100


Let us now see what would happen due to each

individual current.

Since each winding is wound in the same direction, the

current I1would induce voltages in A and B in the samesense.

Also if the current I2 is marked as shown, then it too

would induce voltages in A and B in the same sense,and also in the same sense as due to the current I1.

The fluxes and the corresponding induced emfs are thus

additive as the coils are wound in the same direction.

This is also shown by the non magnetic equivalent

circuit.

In other words the voltage drop due to the self


58/100


In other words, the voltage drop due to the self

inductance term and the voltage drop due to the mutual

inductance term have the same sign.

Consider what would happen if direction of current I2ischanged without changing any physical considerations.

Obviously, the direction of the voltage V2 due to thiscurrent I2on its own winding will change in direction.

A B

V1 V2

I1 I2A B

V1 V2

I1 I2

However the induced voltage due to the current I1 will


59/100


However, the induced voltage due to the current I1will

have the same direction for both coils, as the physical

directions of the two coils are still the same as before.

Thus it is seen that the voltage drop due to the self

inductance term and the voltage drop due to the mutual

inductance term have opposite signs.

Let us see what would happen if the direction of one of

the coils were changed but with the currents entering

each winding at the left hand end of the winding.

A B

V1 V2

I1

I2

A B

V1 V2

I1 I2

If the direction of one of the windings is changed the


60/100


If the direction of one of the windings is changed, the

mutual induced voltage will change in direction relative

to the self inductance term.

Thus again the voltage drop due to the self inductanceterm and the voltage drop due to the mutual inductance

term have opposite signs.

The final possibility is if both the direction of thewinding and the current direction are changed.

A B

V1 V2

I1

I2

A B

V1 V2

I1 I2

It will easily be seen that the voltage drop due to the self


61/100


It will easily be seen that the voltage drop due to the self

inductance terms and that due to the mutual inductance

term must have the same sign.

The above derivations were done primarily based on the

magnetic circuits appearing in the above 4 cases.

Let us now see what has happens with the electricalcircuits.

The summary of the 4 cases are shown.

V V


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mutual term has

same sign self inductance term

mutual term has

opposite sign to self inductance

term

mutual term hasopposite sign to self inductance

term

mutual term has

same sign self inductance term

A B

V1 V2

I1 I2

A B

V1 V2

I1 I2

A B

V1 V2

I1 I2

A B

V1 V2

I1 I2


63/100

If both currents either enter at dotted end or both


64/100


If both currents either enter at dotted end, or bothcurrents leave at dotted end, the sign associated with

mutual inductance is positive, and mutual inductance

term in voltage drop would have the same sign as theself inductance term.

If one current enters at a dotted end and the other

current leaves at dotted end, sign associated withmutual inductance is negative, and the mutual

inductance term in the voltage drop would have the

opposite sign to the self inductance term.

Example 4


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Example 4

For circuit shown, write down voltage drop across AB.

VAB = R. i1e(t) + (L p i1+ M p i2)

where p = d/dtVBA = R. i1+ e(t) (L p i1+ M p i2)

BL1R

L2

A

e(t)

M

i1

i2

Since both currents enter at respective dotted ends, mutual


66/100


Since both currents enter at respective dotted ends, mutual

inductance term and self inductance term have same sign.

True even if voltage is measured in opposite direction.

Usually advisable to treat the self inductance terms and

the mutual inductance terms together within parenthesis

to avoid wrong negation.

With steady state a.c. analysis, the input quantities aresinusoidal with frequency , so the differential wouldgive a multiplication of and a phase shift of

Thus p = j would give the required equations.The impedance due to mutual inductance for a.c. wouldin general be j

Non-coupled Equivalent circuit of simple coupled circuits


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Non coupled Equivalent circuit of simple coupled circuits

When mutual coupling is present,

voltage drops in a particular branch depends on

currents in that branch currents in other branch with which coupling existsSeries and parallel equivalents of branch circuits containing

mutual coupling cannot be easily implemented.

can sometimes be avoided by obtaining a non-coupled equivalent circuit.

(a) coupled coil s being on two arms of a T-junction


68/100


(a) coupled coil s being on two arms of a T junction

Consider pair of mutually

coupled coils on two arms

of a T-junction.

By marking currents i1and i2

flowing in them, and a current (i1i2) flowing in the

common branch, Kirchoffs current law has been applied.Applying Kirchoffs voltage law between PR, and RQ,

VPR= td

idM

td

idL 211

= L1p i1M p i2or VPR= jL1i1jM i2 with sinusoidal current

L1 L2Mi1 i2

i1i2P QR

Current i1 leaves dotted end of L1, while the current i2


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Cu e t 1 eaves dotted e d o 1, w e t e cu e t 2enters dotted end of L2,

sign associated with the mutual inductance is negative,

voltage drop term due to mutual inductance has oppositesign to that due to self inductance term

and VRQ= L2p i2M p i1

If a non-coupled equivalentcircuit is to be obtained,

voltage drops in a branch

should only correspond tocurrents in it own branch.

P Q

R

LA LB

Lm

i1 i2

i1i2

Re-write the 2 equations as follows to achieve this.


70/100


Re write the 2 equations as follows to achieve this.

VPR = L1p i1- M p i2M p i1+ M p i1,

i.e. VPR = (L1M) p i1+ M p (i1i2)

similarly

VRQ = (L2M) p i2+ M p (i1i2)

These two equations would be satisfied with

LA= L1M, LB= L2M, and Lm= M

This transformation will be valid, independent of what

directions marked for currents in the diagrams.

[Compare with: Expression for parallel equivalent of tworesistors is independent of original markings of the current

directions when proving.]

L L2M L1- M L2 - M


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Note that L M appears when the two coils are

opposing each other.

In this case the common branch has an added +Mappearing on it.

In like manner, if the two dots were at the further ends,

the equations would be unchanged and the equivalentcircuit would also be unchanged.

L1 L2M

L1 M

M

L2 M

M L1- M L2 - M


72/100


If position of one of the dots is changed, then the two

coils would be aiding each other, and the terms can be

shown to correspond to L+M with a common term ofM.

L1 L2

M

1

M

2

L1 L2

M

L1+ M

M

L2 + M

Example 5


73/100


Example 5

Write down the non-coupled equivalent circuit for the

coupled circuit shown.

L1

L2

R1 R2

CEM

P

Solution L1R1 R2P


74/100


To get correct sign in non-

coupled equivalent circuit,

consider imaginary currentto flow through coil 1 and coil 2, forgetting other elements.

It is seen that they are opposing.[You will notice that the two coils do not exactly meet at a

common node, but that the node P would have that property ifthe positions of R2and L2were interchanged.]

Since they are series elements, the voltage drop equation

would not change even if the order were changed.[However, intermediate point of connection between theseelements would have a different voltage from earlier, as the

drops are occurring in a different order.]

L2CEM

P

The non-coupled equivalent circuit may thus be drawn


75/100


p q y

This circuit no longer has mutual coupled elements, butthe elements have taken into account the affects of

mutual inductance.

Thus the problem may be solved as for any alternatingcurrent problems.

L1-M

L2-M

R1 R2

C

EM

Transformer as a pair of mutually coupled coils


76/100


Transformer as a pair of mutually coupled coils

Transformer is a pair of mutually coupled coils.

From Kirchoffs voltage law

Vp= Lpp ipM p is

and Vs=(Lsp isM p ip)

For near ideal transformers

|Vp| : |Vs| a|ip| : |is| 1/a

Thus

Vpand aVsare comparable quantities

ipand is/a are comparable quantities.

Vp VsLp Ls

isip M

a:1

Using these comparable variablesip is/a is


77/100


g p

Vp = Lpp ipaM p is/a

and aVs=(a2Lsp is/aaM p ip)

If ipand is/aare unequal,

Leakage current = ipis/a

re-formulated in terms of this leakage current as

Vp = Lpp ipaM p is/a + aM p ipaM p ip

= (LpaM) p ip + aM p .(ipis/a)

and aVs =(a2Lspis/aaM pip)+aM pis/aaM pis/a

=a2

(Ls aM

) pis/a +aMp.(ipis/a)

ip-is/aVp aVs Vs

a:1

Note that the equations are expressed either in terms of


78/100


q p

ip and ipis/a or in terms of is/a and ipis/a.

Allows a non-coupled equivalent circuit to be formed

May also be drawn with ideal transformer added so that

secondary quantities would remain unmodified on that side.

Also, M = k spLL

, Lp/Ls= a

2

so that M = k 2a

LL

p

p

Lp- aM

aM

a2(LsM/a)

ip is/a

ip-is/a

Vp aVs

lp

Lm

a2lsip is/a

ip-is/a

is

VpaVs Vs

a:1

giving aM = k Lp, LpaM = (1k) Lp,


79/100


g g p, p ( ) p,

a2(LsM/a) = a

2(1k) Ls

Since k is the coefficient of coupling,

(1k) corresponds to the leakage.

(1k)Lp and (1k)Lscorrespond to leakage inductances

lpand lsof primary and secondary windings.

Shunt inductance aMcorresponds to the magnetisationinductance Lmof the transformer.

The transformer equivalent circuit is normally drawn

with these variables rather than the self and mutualinductances.

Practical Transformer


80/100


In addition has

resistances rpand rsin primary and secondary windings

has losses (eddy current loss and hysteresis) in the transformer corebased on magnetic flux in core (or on corresponding voltage acrosswindings).

rp, rsare included in primary and secondary sides (rsdrawn asa2rs).

Core loss is represented by a shunt resistance Rc across the

magnetisation inductance.

lp

Lm

a2lsip is/a

ip-is/a

is

VpaVs Vs

a:1

rp a2rs

Rc

Dependent Sources


81/100


p

Passive Circuit Elements

Most basic elementsR, L and C.

Do not generate any electricity.

Either consume energy

in Resistive elementsconvert from electrical form to non-electrical formproduce heat, light etcor store energy

in Capacitive or Inductive elementsin electrostatic and electromagnetic fields.


82/100


Active Circuit Elements (Sources)

Circuit component capable of producing energy.Categorised into

voltage sources and current sources.

independent sources (generated voltage or current doesnot depend on any other circuit voltage or current) dependent sources (generated voltage or current

depends on another circuit voltage or current).


83/100


Independent source

Terminal voltage (or current)depends only on the loading

and

internal source quantity,but not on any other circuit variable.

Symbol forIndependent source

I ndependent Voltage and Current Sources


84/100


v(t) = e(t) for all i(t) v(t) = e(t)Z(p).i(t)

i(t) = I(t) for all v(t) i(t) = I(t)Y(p).v(t)

i(t)

Figure 6(b)Practical current sourceFigure 6(a)Ideal current sourcev(t)Y(p)

I(t)I(t)

i(t)

v t

Figure 5(a)Ideal voltage source

e(t)

i(t)

v(t)

Z(p)

e(t)

i(t)

v(t)

Figure 5(b)Practical voltage source

Dependent Source


85/100


A dependent voltage source (or current source)

would have its terminal voltage (or current)

depend on

another circuit quantity

such as a voltage or current.

Symbol for Dependent source

Four possibilities exist.


86/100


oVoltage dependent (controlled) voltage source

oCurrent dependent (controlled) voltage source

oVoltage dependent (controlled) current source

oCurrent dependent (controlled) current source

+

V +

IoVo

Fi ure 7De endent sources

I

Example 1


87/100


Determine the current I and Vo.

Answer

Kirchoffs voltage law, gives

7.5 = 4 I + 4 Vo+ 5Vo

From Ohms law

1.I =Vo

By substitution,

I = 2.5 A

Vo =2.5 V

+

I

7.5V

Vo

+

4V

+

5V

4

1

+

Figure 8Circuit for example 1

Example 2


88/100


Determine I and Vo.

Answer


6 = 5 (I0.8 Vo)4Vo

From Ohms law

1.I =Vo

By substitution,

I = 1.0 A

Vo=1 V

+

I

6V

Vo

0.8Vo

+

4V

5

1

+


Example 3


89/100


Determine the current I.

Answer


E = R1I + R2(1+) IFrom Ohms law

RoI =VoBy solution of equations

)1(

)1(

21o

21

RR

ERV

RR

EI

o

+

I

E Vo

I

+

RoR2

R1


Operational Amplifier (Op Amp)


90/100


an active circuit elementbehaves as a voltage-controlled voltage source,very versatilecan be used to add, subtract, multiply, divide,

amplify, integrate and differentiate signals.

A practical Op Amp,

available in Integrated circuit (IC) packages

would have inputs and outputs as shown in figure 11.

V+


91/100


Output voltage is limited to the linear range V+to V

.

Ouside these limits, Op Amp is said to be saturated.Null determines the offset.

+

Inverting Input

Non-inverting Input

Output

V

V NullFigure 11Circuit connections of Op Amp


92/100


Vd Voltage difference of input terminals + and

A Gain of Op Amp

Rin Input resistance of Op Amp

RoutOutput resistance of Op Amp

+

RinRout

+

VdAVd

Vout

Figure 12Equivalent Circuit of Op Amp


93/100


Voutlinearly proportional to Vdwith open-loop gain A.

Op Amp has a dependent voltage source AVd.

Parameter Symbol Ideal Typical

Open-loop gain A 105to 108

Input Resistance Rin 106to 1013

Output Resistance Rout 0 10 to 100

Supply Voltage V

+

, V

5 to 24 V


94/100

outdddin VVVVVi


95/100


out

doutoutd

in

R

AVV

R

VV

RRRi

2

21

Figure 14 shows equivalent

circuit of inverting amplifier.

If Rout= 0, and Rin= ,

Vout=A Vd

2

2

1 R

VAAV

R

AVAVdddin

+

R1

Vout

R2

+

A Vd

Rin

Rout

Vd

Figure14Equivalent Circuit

i

VAVVAV outoutoutin


96/100


giving 121 RRR

Thus 212

)1( RRAAR

VV

in

out

If A as for an ideal Op Amp,

Gain = 1

2

R

R

V

V

in

out

,0

AVV outd

Thus Vdis taken as virtual earth, when analyzing.

Non-inverting Amplifier


97/100


Input voltage is applied

directly to non-inverting (+)

input and a small part ofoutput voltage is applied tothe inverting () input from

the R1 R2potential divider.

For an ideal Op Amp,with Rout = 0, Rin =

and A =

Gain = 1

21

RRR

+

V+

V

Figure 15Non-inverting Amplifier

R

R2

Vin

Vout

Summing Amplifier


98/100


With 2 inputs, Summing

amplifier is obtained.

With R1A= R1B

)(1

2

inBinAout VVR

RV

If R1A R1B

)(11

2

B

inB

A

inA

outR

V

R

VRV

+

V+

V

Figure 16Summing Amplifier

R1A

R2

VinA

Vout

R1BVinB

Differential AmplifierR2


99/100


Inputs are connected toboth inverting or non-

inverting terminals

Resultant output basedon difference between

V1Aand V1BFor ideal Op Amp,

Rin= , Rout= 0, and A = .

d

out

VVA

giving0

outdVV

+

V+

V

Figure 17Differential Amplifier

R1A

R2

VinA

Vout

R1BVinB

R3

Vd

IA

IB

VVVVI outinAA

VVVI inBB


100/100

21 RR AA

,31 RR B

B

This gives inBB VRR

R

V31

3

Substitution gives A

inA

A

out

R

V

RRV

R

V

212

11

inA

A

inB

BA

Aout V

R

RV

RR

R

R

RRV

1

2

31

3

1

21

WhenR1A= R1B= R1andR3= R2,

inAinBout VVR

V 2

Documents

EE2092!2!2011 Coupled Circuits Dependant Sources - Copy