bent 1123_chapter 1 - time varying signals

8/6/2019 BENT 1123_Chapter 1 - Time Varying Signals

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Time VaryingSignals

Chapter 1


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Chapter Learning Outcomes

y At the end of this chapter, students should be

able to;

1. identify a sinusoidal waveform and measure itscharacteristics.

2. determine the various voltage and current values of asine wave.

3. differentiate between direct current and alternatingcurrent.

4. apply phasor and trigonometric techniques to analyzealternating signals.

5. use a phasor to represent a sine wave.


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Introduction

y A sinusoid is a signal that has the form of the sine or cosine

function.

y A sinusoidal current is usually referred to as alternating

current (ac).y Such a current reverses at regular time intervals and has

alternately positive and negative values.

y AC circuits are the circuits driven by sinusoidal current or

voltage sources.


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y A constant voltage

source supplies the

same voltage at every

instant.

y A sinusoidal voltage

source supplies avoltage that varies with

time.

Introduction contd


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Sinusoids

y A general expression for the sinusoid,

whereVm

= the amplitude of the sinusoid = the angular frequency in radians/s= the phase

)sin()( J[ ! tVtv m


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The Sinusoidal Wave

y The period, T (s) is the time taken to complete one cycle

y The frequency,f is the number of cycle per second, where

HzT

f1

!


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y is the angular frequency of the sinusoidal function, where

y Vm is the maximum value for the voltage or amplitude of the

sinusoidal voltage.

y is the phase angle of the sinusoidal function. It determinesthe value of the sinusoidal voltage at the time t= 0s.

s/radfT[ 2!


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Example 1

Given a sinusoid, 5 sin(4t 60), calculate its amplitude, phase,angular frequency, period, and frequency.

Solution:

Amplitude = 5, phase = 60o, angular frequency = 4 rad/s,

Period = 0.5 s, frequency = 2 Hz.


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Example 2

For the following sinusoidal voltage, find the value Vat time t= 0s

and t= 0.5s.

V=6 cos(100t+60)

Solution:

Vat t= 0s : 3 V

Vat t= 0.5s : 4.26 V


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Periodic Function

y A periodic function is one that satisfies v(t) = v(t+ nT),for all t

and for all integers n.


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y The starting point ofv2 occurs first in time. Therefore,

y v2 leads v1 by orv1 lags v2 by

y If,

y , v1 and v2 are out of phase

y , v1 and v2 are in phase

y Only two sinusoidal values with the same frequency can be

compared by their amplitude and phase difference.

0{J

0!J


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Trigonometric Identities

y A sinusoid can be expressed in either sine or cosine form.

y When comparing two sinusoids, it is expedient to express both

as either sine or cosine with positive amplitudes.

y With these identities, it is easy to show that

BsinAsinBcosAcosBAcos

BsinAcosBcosAsinBAsin

O!ss!s

tsintcos

tcostsin

tcostcostsintsin

[[

[[

[[ [[

O!rs

s!rs

!rs !rs

90

90

180180


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tsintcos [[ !r 90

Graphical Technique

y A graphical approach may

be used to relate sinusoids

as an alternative to using

trigonometric identities.

Angle:

-ve : clockwise

+ve: counterclockwise


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tsintsin [[ !r 180


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y The magnitude and argument of the resultant sinusoid in cosine form

is readily obtained from the triangle. Thus,

U[[[ ! tcosCtsinBtcos


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y Where

y For example, we may add 3cos tand sin tas shown in

Figure (b) and obtain

A

Btan,BA 122 !! U

r! 153543 .tcostsintcos [[[


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Example 3

Calculate the phase angle between v1 = -10 cos(t+50)and v2 =12 sin(t10). State which sinusoid is leading.

Solution: in order to compare v1 and v2, we must express them in

the same form with positive amplitudes.

METHOD 1: (express them in cosine form)

Using trigonometric identities,

v1 = -10 cos(t+50) v1 = 10 cos(t-130)or v1 = 10 cos(t+230)

v2 = 12 sin(t10) v2 = 12cos(t100)


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Comparing v1 and v2shows the phase difference is 30. We canwrite v2 as

v2 = 12cost130 + 30)orv2 = 12cost+ 260)Hence,

v2 leads v1by 30

METHOD 2: (express them in sine form)

Using trigonometric identities,

v1 = -10 cos(t+50) v1 = 10 sin(t 40) = 10 sin(t1030)

v2 = 12 sin(t10)


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Comparing v1 and v2shows the phase difference is 30.

Hence,

v1 lags v2by 30

METHOD 3: (graphical approach)


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Example 4

Express the following functions in cosine form:

r

r

2010

62

304

tsinc

tsinb

tsina

[

[


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Example 5

a) Express v = 8 cos(7t+ 15)in sine form.b) Convert i = -10 sin(3t 85)to cosine form.


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Example 6

Find the phase angle between

and

Does i1

lead or lag i2

?

r! 2537741 tsini r! 4037752 tcosi


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Example 7

Given

and

Determine the phase angle between the two sinusoids and which

one lags the other.

r! 60201 tsinv [ r! 10602 tcosv [


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y where:

j =

x= real part ofzy= imaginary part ofz

r= the magnitude ofz

= the phase ofz

1


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Relationship between Rectangular and

Polar Form

2 2 1; tany

r x yx

J ! !

c s ; si x r y r J J! !

J J J! ! ! cos sinz x jy r r j r

or

Thus, z may be written as,

Note: addition and subtraction of complex number better perform in rectangular form.Multiplication and division in polar form.


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Basic Properties of Complex Numbers

Addition :

Subtraction :

Multiplication :

Division :

212121 JJ ! rrzz

212

1

2

1 JJ !r

r

z

z

212121 yyjxxzz !

212121 yyjxxzz !


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Reciprocal :

Square Root :

Complex Conjugate :

note:

In general,

J!rz

11

2/J! rz

JJ jrerjyxz !!!

jj

!1

JJJ sinjcose j s!s


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y As we know, the sinusoidal voltage can be represented in sine

or cosine function.

y First, consider the cosine function as in:

y This expression is in time domain.

y In phasor method, we no longer consider in time domain instead

in phasor domain (also known as frequency domain).

( ) c smv t V t [ J!


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y The cosine function will be represented in phasor and complex

number such as:

y For example:

Transform the sinusoid:

v(t) = 12co

s (377t- 60)thus,

( ) c s

c s si

m

m

m m

v t V t

V

Vj

V

[ J

J

J J

!

!

!

39.1066012 jv r! or


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y The phasor representation carries only the amplitude and phase

angle information.

y The frequency term is dropped since we know that the

frequency of the sinusoidal response is the same as the source.y The cosine expression is also dropped since we know that the

response and source are both sinusoidal.


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Sinusoid-phasor Transformation

Time-domain representation Phasor-domain Representation

cos

sin 90

cos I

sin I 90

m m

m m

m m

m m

V t V

V t V

I t

I t

[ J J

[ J J

[ J J

[ J J

r

r

* to get the phasor representation of a sinusoid, we express it in cosine form.


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and

y Thus, the phasor diagram is:

J! mVV II m -=


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Differentiation and Integration

Time Domain Phasor Domain

dtdv Vj[

vdt [jV


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Example 8

Evaluate the following complex numbers:

oo

o

jjb

jja

3010)]43/()403510.[(

*]605)41)(25.[(


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Example 8:Solution

? A ? A

? A

6713515

6713515

334521813

606051813

6054125

.j.

.j.

.j.j

sinjcosj

jj

*

*

*

*o

!

!

!rr!

!

(a) Using polar-rectangular transformation, addition and subtraction,

(b) Using calculator,

272938 .j. !


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Example 9

Transform these sinusoids to phasors:

Vtvbtia

r!

r!

5030sin4)(

)4050cos(6)(


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Example 9: Solution

Refer to sinusoids to phasors transformation table:

a)

b) Since

so

A406I r!

V140t30cos49050t30cos450t30sin4v

90AcosAsin

r!

rr!r!r!

V1404V r!


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Example 10

Given y1 = 20 cos (100t- 30r) and y2 = 40 cos (100t+ 60r). Expressy1 + y2 as a single cosine function.

Solution:

In phasor form

Thus, y1 +y2 =44.72 cos(100t+ 33.4r)

r!!

!

rr!@

r!r!

4.3372.44

64.2432.37

64.34201032.17

60403020

6040&3020

21

21

j

jj

yy

yy


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Example 11

Using phasor approach, determine i(t) for:

r! )tcos()t(i

dt

diidt 2255610


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Circuit Elements in Phasor Domain

y Circuit analysis is much simpler if it is done in phasor domain.

y In order to perform the phasor domain analysis, we need to

transform all circuit elements to its phasor equivalent.

y Transform the voltage-current relationship from time domain to

the frequency domain for each element.


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PhasorRelationships forResistor

y If current through resistor is:

y The voltage across R is V=IR (Ohms Law); in phasor form:

y But; phasor representation of the current is:

y Hence: V=RI

J

J[

!

!

m

m

II

tIi )cos(

J!m

RIV

Jm

II=

Phasor domain

Ohms Law in phasor form


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Time domain Phasor domain


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PhasorRelationships forInductor

y If current through inductor is:

y The voltage across the inductor:

y Which transforms to the phasor

y But

y Hence

JmII=

)tcos(LI)tsin(LIdtdiLv mm r!!! 90J[[J[[

90! J[ mIV

)cos( J[ ! tIi m

IjLIjV m [J[ !!


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The current and voltage are 90o out of

phase (voltage leads current by 90o)


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PhasorRelationships for Capacitor

y Given voltage through capacitor is:

y The current through capacitor is:

y But

y Thus: where

y And

)cos( J[ ! tVvm

dt

dvCi !

)90cos()sin( !! J[[J[[ tCVtCVimm

90! J[ mCVI

901!jCVjCVjI m [J[ !! JmV

)/( CjIV [!


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Voltage Current Relationships

Element Time domain Frequency domain

R

L

C

Riv!

dt

diLv!

dt

dvCv!

RIV!

LIjV [!

Cj

IV

[

!


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Example: Phasor relationship for circuit

elements

a) If voltage v = 10 cos (100t+ 30)is applied to a 50F capacitor,

calculate the current through the capacitor.

b) What is the voltage across a 2F capacitor when the current

through it is i =4sin(106t+ 25) A?

Ans:

a) 50 cos (100t + 120) mA

b) 2 sin (106t - 65) V


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Impedance and Admittance

y Previously:

y In terms of the ratio of the phasor voltage to the phasor current:

y Ohms Law in phasor form for any type of element

Cj

IV,LIjV,RIV

[[ !!!

Cj

I

I

V,Lj

I

V,R

I

V

[[ !!!

ZIVor,I

VZ !!


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Impedances and Admittances ofPassive

Elements

Element Impedance Admittance

R

L

C

RZ!

LjZ [!

Cj

1Z

[!

1Y !

Lj

1Y

[!

CjY [!


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Equivalent circuits at dc and high

frequencies


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Impedance in Rectangular Form

y The impedance may be expressed in rectangular form as

y where R = ReZ is the resistance and X= Im Z is the reactance.

y The impedance is inductive when X is positive or capacitive

when X is negative.

y The impedance may be also be expressed in polar form as

jXRZ !

U! ZZ


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y where

y and

U!! ZjXRZ

R

Xtan,XRZ 122 !! U

UU sinZX,cosZR !!


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Admittance

y It is sometimes convenient to work with the reciprocal of

impedance, known as admittance.

where G = ReY is called the conductance

and B = ImY is called the susceptance.

jBGY,V

1

Z

1Y !!!

jXR

1jBG

!


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Impedance Combinations

y Consider the Nseries-connected

impedances shown in figure.

y The same current I flows through

the impedances.

y Applying KVL around the loop gives

y The equivalent impedance at the input terminals is

or

)Z...ZZ(IV...VVVN

21N

21

+++=+++=

N21eq Z...ZZI

VZ +++== N21eq Z...ZZZ +++=


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Voltage Divider of Series Circuit

y IfN= 2, the current through the

impedances is

y Since V1 = Z1I and V2 = Z2I ,

then:

21 ZZ

V

I +=

VZZ

ZV,V

ZZ

ZV

21

2

2

21

1

1 +=

+=


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Parallel Circuit

y We can obtain the equivalent impedance or admittance of the Nparallel-connected impedances shown in figure.

y The cross voltage each impedance is the same.


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Parallel Circuit

y Applying KCL at the top node

y The equivalent impedance is

y And the equivalent admittance is

N21eqZ

1...

Z

1

Z

1

V

1

Z

1!!

N21eqY....YYY !

)Z

1...

Z

1

Z

1(VI...III

N21

N21!!


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Current Divider ofParallel Circuit

y Eg: When N= 2, as shown in Figure

5.19, the equivalent impedance

becomes

y since V=IZeq=I1Z1=I2Z2 the currents

in the impedances are

21

21

2121eq

eq

ZZ

ZZ

Z/1Z/11

YY1

Y1Z

!

!

!!

IZZ

ZI,I

ZZ

ZI

21

1

2

21

2

1 !

!


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Example 13

Find the input impedance of the circuit in figure below. Assume that

the circuit operates at [=50rad/s.


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Example

Find Zeq in the circuit.

Ans:Zeq = 1 + j0.5


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Example: Admittance

Determine the admittance Y for the circuit in Figure below.

Ans:Y =(250 -j25) mS


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Example 14

Find v(t) and i(t) in the circuit ofFigure below.

;5i


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Example 14: Solution

vs =10 cos4t Vs =

Theimpedanceis

Hencethe current

V010 r

;[ 5.2j51.04j

1

5Cj

1

5Z !v!!

A57.26789.18.0j6.15.2j5

010

Z

VI s r!!

r!!


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Example 14: Solution contd

Hencethe current

Converting I and V to thetimedomain,

V43.6347.41.04j

57.26789.1

Cj

IIZV

Cr!

vr

!!![

V43.63t4cos47.4tv

A57.26t4cos789.1ti

r!

r!


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Example 15

Determine v(t) and i(t).

4i

0.2Hvs = 20sin(10t+ 30) V


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Wye-Delta Transformations


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y Situations often arise in circuit

analysis when the resistors

are neither in parallel nor in

series.y This type of circuit can be

simplified by using three-

terminal equivalent networks.

y These are:

Wye (Y) or tee (T) network

Delta () or pi () network

Wye-Delta Transformations


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Wye (Y) or Tee (T) Network


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Delta () orPi () Network


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Delta to Wye ( - Y) Conversion

y The delta-to-wye and

wye-to-delta

transformations that we

applied to resistive

circuits are also valid forimpedances .

cba

cb

1

ZZZ

ZZZ

!

cba

ca

2

ZZZ

ZZZ

!

cba

ba

3

ZZZ

ZZZ

!


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Wye to Delta (Y - ) Conversion

1

133221

a

Z

ZZZZZZZ

!

2

133221

b

Z

ZZZZZZZ

!

3

133221

c

Z

ZZZZZZZ

!


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Example 16

Calculate vo in the circuit.

Ans:vo(t) = 7.07 cos(10t 60)V


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Example 17

Find current I in the circuit.

Ans:I = A204.4666.3 r


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Example 18

Find I in the circuit.

Ans:I = A8.3364.6 r


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Application: Phase-Shifters

y Employed to correct an undesirable phase shift already present

in a circuit or to produce special desired effects.

y An RC circuit is suitable for this purpose because its capacitor

causes the circuit current to lead the applied voltage.

y RL cir

cuitso

ran

yre

activ

ec

ircu

itscouldal

so

serv

eth

es

ame

purpose.

y Two types:

Leading output

Lagging output


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Phase Shifter: Leading Output

y I leads Vi by some phase angle U,where 0 < U < 90o, depending on thevalues ofR and C.

y the output voltage Vo

across the

resistor is in phase with the current, Voleads (positive phase shift) Vi

C

C1

jXRZ

R

Xtan

!

! U


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y The output is taken across the capacitor

y The current I leads the input voltage Vi by U, but the outputvo(t) across the capacitor lags (negative phase shift) the

input voltage vi(t)

Phase Shifter: Lagging Output


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y As the phase shift U approaches 90o, the output voltage Voapproaches zero.

y For this reason, these simple RCcircuits are used only

when small amounts of phase shift are required.

y

If it is desired to have phase shifts greater than 60

o

, simpleRC networks are cascaded, thereby providing a total phase

shift equal to the sum of the individual phase shifts.

Phase Shifter

R

Xtan C1!U


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Example 19

For the RL circuit shown, calculate the amount of the phase shift

produced at 2 kHz.


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Example 19: Solution

Given f = 2kHz.

The impedance Z,

;T[

;T[

83.62jm5k22jLjZmH5

7.125jm10k22jLjZmH10

!vvv!!p

!vvv!!p

;r!

!

1.6056.69

83.62j100||7.125jZ


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Example 19: Solution (contd)

Using voltage division,

and

)1(V02.423582.0

V3.60j7.184

1.6056.69V

150Z

ZV

i

ii1

r!

r

!

!

)2(V86.57532.0V832.62j100

832.62jV11o

r!

!


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Example 19: Solution (contd)

Sub. (1) into (2),

The output is about 19% of the input in magnitude but leading the

input by 100.

i

io

V1001906.0

V02.423582.086.57532.0V

r!

rr!


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AC Bridges

y An ac bridge circuit is used in

measuring the inductance L of an

inductor or the capacitance Cof acapacitor.

y It is similar in form to theWheatstone bridge for measuring

an unknown resistance and follows

the same principle.

y The bridge is balanced when no

current flows through the meter.

y This means that V1 = V2. Applying

the voltage division principle,

x132

x3

x

21

2

s

x3

x

2s

21

2

1

ZZZZZZ

Z

ZZ

Z

VZZ

ZV,V

ZZ

ZV

!p!

!

!

32

1

x

ZZ

ZZ !


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y In each case, two resistors, R1 and R2, are varied until the ac

meter reads zero. Then the bridge is balanced.

AC Bridges for measuring L and C

s

2

1

x CR

RC !

s

1

2

xL

R

RL !

bent 1123_chapter 1 - time varying signals

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