Chapter 11 Alternating Current Circuits Recommended...
Transcript of Chapter 11 Alternating Current Circuits Recommended...
Chapter 11 Alternating Current Circuits
Recommended Problems:
1,3,5,9,11,13,15,17,19,21,23,29,31,33,37,41,43,61,65.
Alternating Current Sources
Generator or dynamo is
a device that operate
according to the
electromagnetic induction
and Faraday's law.
It consists of a loop of wire that can rotate, by an external means,
in a magnetic field. The ends of the loop are connected to slip
rings that rotate with the loop. The slip rings are connected to the
external circuit by two fixed brushes.
Figure 33.1 Schematic diagram of a generator.
It is a device that
produces a current by
converting the
mechanical energy into
electrical energy.
As the loop rotate, the magnetic flux through it changes with time
inducing an emf and so a current in the external circuit.
Suppose that the loop, with area A, rotates with constant angular
speed , and suppose that is the angle between the magnetic
field and the area of the loop. The magnetic flux through the loop
at any time is then
tBABAm coscos
Therefore, the induced emf in the loop is given by
ttABdt
d m sinsin max
This means that the emf varies sinusoidally with
time forming a sinusoidal wave, as shown in the
figure. Because of that a source of AC is
represented in the circuits by the symbol
max
-max
t
I
Io
-Io
t
I
t
(a)
The frequency of commercial
generators in our country and most of
the world is 50 Hz, while it is 60 Hz in
some other countries like the USA and
Canada. Note that the frequency f is
related to the angular speed through
the relation . f 2
Ohm’s law is still hold for AC circuits,
i.e.,
tRR
I
sinmax
tII m sin
The voltage across an element in an
AC circuit is given by
tVV m sin
Im and Vm are called the peak current and the peak voltage,
respectively.
As it is clear from Figure 14.7(b), the average value of an AC is
zero. This doesn't mean, however, that no power is needed or that
no heat is produced in a resistor. Electrons for AC move forth and
back and so produce heat.
It is known that the power dissipated in a resistor is given by
tRIRIP m 222 sin
Since the current is now squared, the power is always positive.
It is easy to show that the average value of 2
1sin 2 t
And so the average value of 2212 is mII
the average power dissipated in a resistor in AC circuits is then
R
VRIP m
m
2
212
21
That is, what is important for calculating the average power is the
mean value of the square of the current or voltage. The square
root of each of these values is known as the rms (root-mean-
square) and is given by
2m
rms
II
2m
rms
VV
The rms values of V and I are sometimes known as the effective
values. They are useful because they can be substituted directly
into the famous power formula to get the average power, that is,
R
VRIVIP rms
rmsrmsrms
22
It is the rms values that are specified or measured. Therefore,
ammeters and voltmeters are designed to read the corresponding
rms values. When we say that the standard voltage in our country
is 220 V, we mean that the rms value of V is 220V.
The peak value of such a voltage is V3112 rmsm VV
Phasor: It is a counterclockwise-rotating vector representing an
AC quantity such that its length represents the maximum value of
the quantity and its projection onto the vertical axis represents the
instantaneous value of the quantity.
I vs t
t
It is clear that the instantaneous value is changing with time as the
phasor rotates while the maximum value is constant.
• Test Your Understanding (1)
Consider the voltage phasor shown. Choose the figure at which
the instantaneous value of the voltage has the largest magnitude.
(a) (b) (c) (d)
Choose the figure at which the instantaneous value of the voltage
has the smallest magnitude.
v
(a)
v
(b)
v
(d)
v
(c)
(a) (b) (c) (d)
Example 33.1 A 900.-W microwave oven is designed to
operate at 220. V. Calculate its resistance and the peak current
when it is operating.
Solution: To find the resistance we have
And for the current we have
8.53
900
22022
P
VR rms
A09.4220
900
rmsrms
V
PI A78.52 rmsm II
Resistors in AC Circuits
Consider the circuit that consists of a
resistor and connected in series with an
AC source.
R
tVv m sin
Let vR, be the voltage at some instant across the resistor. This
voltage must equal to the voltage of the AC source, that is
tVv mR sin
The current passing through the resistor is
tIR
tV
R
vi m
mR
sinsin
As it is clear from the last two equations that
The current i and voltage vR across a resistor in a pure resistive
AC circuits are in phase.
Vm
iR, vR
t
Im
iR
t
iR, vR
vR
Inductors in AC Circuits
Consider the circuit that consists of an
inductor and connected in series with
an AC source.
L
tVv m sin
Let vL, be the voltage at some instant
across the inductor. Again this voltage
must equal to the voltage of the AC
source, that is
tVv mL sin
To find the current passing through the inductor we have
dt
diLvL
)cos()sin(1
tL
Vdtt
L
Vdtv
Li mm
L
Using the identity
)cos(2sin tt
2sin2sin
tItL
Vi m
m
L
mm
X
VI with
LX L and Is called the inductive reactance.
The current i lags behind the voltage vL across an inductor in a
pure inductive AC circuits by 90o .
Vm
iR, vR
t
Im
iR
Im
vR
t
iR, vR
Vm
• Test Your Understanding (2)
Consider the AC circuit shown. The
frequency of the AC source is adjusted
while its voltage amplitude is held
constant. The lightbulb will glow the
brightest at
a) high frequencies b) low frequencies
c) all frequencies. d) It will not glow at all
Capacittors in AC Circuits
Consider the circuit that consists of a
capacitor connected in series with an
AC source.
Let vC, be the voltage at some instant
across the i capacitor. Again this
voltage must equal to the voltage of
the AC source, that is
tVv mC sin
To find the current passing through the capacitor we have
C
tVv m sin
CCvq
)cos( tCVdt
dvC
dt
dqi m
C
Using the identity
)cos(2sin tt
2sin2sin tItVCi mm
C
mm
X
VI with
CXC
1and Is called the capacitive reactance.
The current i leads the voltage vC across a capacitor in a pure
capacitive AC circuits by 90o .
Vm
iR, vR
t
Im
iR
Im
vR
t
iR, vR
Vm
• Test Your Understanding (3)
Consider the AC circuit shown. The
frequency of the AC source is adjusted
while its voltage amplitude is held
constant. The lightbulb will glow the
brightest at
a) high frequencies b) low frequencies
c) all frequencies. d) It will not glow at all
• Test Your Understanding (4)
Consider the AC circuit shown. The
frequency of the AC source is adjusted
while its voltage amplitude is held
constant. The lightbulb will glow the
brightest at
a) high frequencies b) low frequencies
c) all frequencies. d) It will not glow at all
SERIRS RLC AC CIRCUITS L
C R
tVv m sin
Consider the circuit which consists of a
resistor, an inductor, and a capacitor,
connected in series with an AC source. Let
vR, vL, and vC be the voltage at some instant
across each element, respectively.
Since they are connected in series, the voltage across the
combination is CLR vvvv
This voltage must equal to the voltage of the AC source, that is
CLRm vvvtV sin
Note that a voltmeter, when connected across the combination,
does not read (VR)rms + (VL)rms + (VC)rms . What will read then? Let
us see.
To solve this equation we use the phase diagram by letting the
phasor representing I to be along the horizontal axis
where the constant is called the phase
angle between the current and the
applied voltage. Now we have
tRIIRv mR sin
2sin
tXIv LmL
2sin
tXIv CmC
The phase diagram is now
Im
Vm
VR
VL
VC
VR
VL-VC
Vm
Supposing the current passing through each element (which is the
same due their series connection) to be given as
tII m sin
From the diagram we conclude that
22CLRm VVVV
R
CL
V
VV tan
Using the relations RIV mR LmL XIV CmC XIV
ZIXXRIV mCLmm 22
22CL XXRZ
With Z is called the impedance of the circuit with its unit is . The
phase angle is now given as
R
XX CL tan
From the last two equations we conclude that
if the circuit consists only of a resistor with the source, (XL =XC
=0), we have Z=R, and the phase angle is zero, that is I and V are
in phase.
if the circuit consists only of an inductor with the source, (R =XC
=0), we have Z= XL, and the phase angle is 90o, that is I lags
behind V by 90o .
if the circuit consists only of a capacitor with the source, (R =XL
=0), we have Z= XC, and the phase angle is -90o, that is I leads V by 90o .
In general if XL > XC the phase angle is +ve, and I lags behind
V by . In the other hand if XL < XC the phase angle is -ve, and I leadsV by
When XL = XC then phase angle =0. In this case, the
impedance equals the resistance and the current has its peak
value given by RVm
The frequency o at which this occurs is called the resonance
frequency of the circuit. Using the condition XL = XC we get
C
Lo
o
1
LCo
1
• Test Your Understanding (6)
Indicate each part of the figure shown whether
XL = XC , XL >XC , XL < XC .
a) XL < XC b) XL = XC
c) XL >XC
Example 33.4 In a series RLC circuit we have Vm =120 V,
R= 200 , C=4.0 F and f= 60 Hz. Find L such that the voltage
across the capacitor lags the applied voltage by 30o.
Solution
Since the angle is between the
applied voltage and the voltage
across R, and since the angle
between VR and VC is 90o
Im
Vm
VR
VL
VC
30o ooo 609030
663)104)(60(2
11Now
6CXC
200
66373.1tan LCL X
R
XX
Hf
LLX L 83.0120
317
2
317317
Example 33.4 In a series RLC circuit we have Vm =150 V,
R= 425 , C=3.5 F, L=1.25 H, and =377 s-1 .
a) Determine Z.
b) Find the maximum current in the circuit.
c) Find the phase angle.
d) Find the maximum voltage across each element.
Solution
Let us first find the reactances
47125.1377LXL 758105.337711 6CXC
a) The impedance is then
5137584714252222
CL XXRz
b) The maximum current is given by
A292.0513
150
Z
VI mm
c) The phase angle is given by
oCL
R
XX34
.425
758417tan
d) For the maximum voltage across each element we have
Note that since the circuit is more capacitive (XL < XC ), the
current leads the applied voltage by the angle 34o
V124 RIV mR
V138 LmL XIV
V221 CmR XIV
Power in AC Circuits
The instantaneous power delivered by Ac source is
)sin()sin( maxmax tVtIiVP
sin)cos(cossinsinusing ttt
sin)cos()sin(cos)(sin maxmax2
maxmax ttVItVIP
sin)cos()sin(cos)(sin2maxmax avravravr tttVIP
Knowing that 212 )(sin avrt 0)cos()sin( avrtt
cos2
1maxmaxVIPavr cosor rmsrmsavr VIP
The quantity cos is called the power factor given by
max
max
max
cosV
RI
V
VR
RIP rmsavr2
Example 33.5 In a series RLC circuit we have Vm =150 V,
R= 425 , C=3.5 F, L=1.25 H, and =377 s-1 . What is the
power delivered to the circuit.
Solution
From the previous example we have Z=518 .
A206.02Z
VI mrms
WRIP rmsavr 1.18)425()206.0( 22
Resonance in a Series in RLC Circuits
It is known that
22
22
22
CL
rmsrms
rmsrmsavr
XXR
RVRI
z
RVRIP
222222
2 1oCL L
CLXX
222222
22
o
rmsavr
LR
RVP
From the last Equation it is clear that the average power is
maximum when = o (at resonance). At resonance the average
power is given by
R
VP rms
avr
2
The average power versus the
frequency is plotted for two
different values of the frequency.
The sharpness of the curve is
measured by a factor known as
the quality factor defined as
oQ
With is the width of the
resonant power curve at half
maximum and given by
L
R
R
LQ o
• Test Your Understanding (7)
The impedance of a series RLC Circuit is
a) Larger than R
b) Less than R c) Equal to R
d) Impossible to determine.
Example 33.7 In a series RLC circuit we have Vm =20 V,
R= 150 , L=2.0 mH, and =5000 s-1 . What is the value of C for
the current is maximum.
Solution
The maximum current is attained when = o .
FCCLC
o 2102
15000
1
3
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