Download - 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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Time domain Phasor domain
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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Time domain Phasor domain
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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 !