Therefore, the in-stantaneous current in the resistor is 33.2 where Imaxis the maximum current: From Equations 33.1 and 33.2, we see that the instantaneous voltage across the re-sistor i
Trang 1c h a p t e r
Alternating-Current Circuits
Small “black boxes” like this one are commonly used to supply power to elec- tronic devices such as CD players and tape players Whereas these devices need only about 12V to operate, wall outlets provide an output of 120 V What
do the black boxes do, and how do they
work? (George Semple)
Trang 2n this chapter we describe alternating-current (ac) circuits Every time we turn
on a television set, a stereo, or any of a multitude of other electrical appliances,
we are calling on alternating currents to provide the power to operate them Webegin our study by investigating the characteristics of simple series circuits that con-tain resistors, inductors, and capacitors and that are driven by a sinusoidal voltage
We shall find that the maximum alternating current in each element is proportional
to the maximum alternating voltage across the element We shall also find that whenthe applied voltage is sinusoidal, the current in each element is sinusoidal, too, butnot necessarily in phase with the applied voltage We conclude the chapter with two
sections concerning transformers, power transmission, and RC filters.
AC SOURCES AND PHASORS
An ac circuit consists of circuit elements and a generator that provides the nating current As you recall from Section 31.5, the basic principle of the ac gener-ator is a direct consequence of Faraday’s law of induction When a conductingloop is rotated in a magnetic field at constant angular frequency , a sinusoidalvoltage (emf) is induced in the loop This instantaneous voltage v is
alter-where Vmaxis the maximum output voltage of the ac generator, or the voltageamplitude From Equation 13.6, the angular frequency is
where f is the frequency of the generator (the voltage source) and T is the period.
The generator determines the frequency of the current in any circuit connected tothe generator Because the output voltage of an ac generator varies sinusoidallywith time, the voltage is positive during one half of the cycle and negative duringthe other half Likewise, the current in any circuit driven by an ac generator is analternating current that also varies sinusoidally with time Commercial electric-power plants in the United States use a frequency of 60 Hz, which corresponds to
an angular frequency of 377 rad/s
The primary aim of this chapter can be summarized as follows: If an ac tor is connected to a series circuit containing resistors, inductors, and capacitors,
genera-we want to know the amplitude and time characteristics of the alternating current
To simplify our analysis of circuits containing two or more elements, we use
graph-ical constructions called phasor diagrams In these constructions, alternating
(sinus-oidal) quantities, such as current and voltage, are represented by rotating vectorscalled phasors The length of the phasor represents the amplitude (maximumvalue) of the quantity, and the projection of the phasor onto the vertical axis rep-resents the instantaneous value of the quantity As we shall see, a phasor diagramgreatly simplifies matters when we must combine several sinusoidally varying cur-rents or voltages that have different phases
RESISTORS IN AN AC CIRCUIT
Consider a simple ac circuit consisting of a resistor and an ac generator ,
as shown in Figure 33.1 At any instant, the algebraic sum of the voltages around a
Trang 3closed loop in a circuit must be zero (Kirchhoff’s loop rule) Therefore,
or1
(33.1)
where v R is the instantaneous voltage across the resistor Therefore, the
in-stantaneous current in the resistor is
(33.2)
where Imaxis the maximum current:
From Equations 33.1 and 33.2, we see that the instantaneous voltage across the
re-sistor is
(33.3)
Let us discuss the current-versus-time curve shown in Figure 33.2a At point a,
the current has a maximum value in one direction, arbitrarily called the positive
direction Between points a and b, the current is decreasing in magnitude but is
still in the positive direction At b, the current is momentarily zero; it then begins
to increase in the negative direction between points b and c At c, the current has
reached its maximum value in the negative direction
The current and voltage are in step with each other because they vary
identi-cally with time Because i Rand v Rboth vary as sint and reach their maximum
values at the same time, as shown in Figure 33.2a, they are said to be in phase
Thus we can say that, for a sinusoidal applied voltage, the current in a resistor is
al-ways in phase with the voltage across the resistor
A phasor diagram is used to represent current – voltage phase relationships The
lengths of the arrows correspond to Vmaxand Imax The projections of the phasor
arrows onto the vertical axis give v R and i Rvalues As we showed in Section 13.5,
if the phasor arrow is imagined to rotate steadily with angular speed , its
vertical-axis component oscillates sinusoidally in time In the case of the single-loop
resis-tive circuit of Figure 33.1, the current and voltage phasors lie along the same line,
as in Figure 33.2b, because i Rand v Rare in phase
Note that the average value of the current over one cycle is zero That is,
the current is maintained in the positive direction for the same amount of time
and at the same magnitude as it is maintained in the negative direction However,
the direction of the current has no effect on the behavior of the resistor We can
understand this by realizing that collisions between electrons and the fixed atoms
of the resistor result in an increase in the temperature of the resistor Although
this temperature increase depends on the magnitude of the current, it is
indepen-dent of the direction of the current
We can make this discussion quantitative by recalling that the rate at which
electrical energy is converted to internal energy in a resistor is the power
where i is the instantaneous current in the resistor Because this rate is
propor-tional to the square of the current, it makes no difference whether the current is
direct or alternating — that is, whether the sign associated with the current is
posi-tive or negaposi-tive However, the temperature increase produced by an alternating
Maximum current in a resistor
The current in a resistor is in phase with the voltage
Figure 33.1 A circuit consisting
of a resistor of resistance R
con-nected to an ac generator, designated by the symbol
.
R
∆v R
∆v = ∆Vmax sin ωt
Trang 4current having a maximum value Imaxis not the same as that produced by a direct
current equal to Imax This is because the alternating current is at this maximumvalue for only an instant during each cycle (Fig 33.3a) What is of importance in
an ac circuit is an average value of current, referred to as the rms current As we
learned in Section 21.1, the notation rms stands for root mean square, which in this
case means the square root of the mean (average) value of the square of the rent: Because i2varies as sin2t and because the average value of i2is(see Fig 33.3b), the rms current is2
cur-(33.4)
This equation states that an alternating current whose maximum value is 2.00 Adelivers to a resistor the same power as a direct current that has a value of (0.707)(2.00 A) 1.41 A Thus, we can say that the average power delivered to a resistorthat carries an alternating current is
Irms√i2
2That the square root of the average value of i2 is equal to can be shown as follows: The rent in the circuit varies with time according to the expression sint, so
cur-sin 2t Therefore, we can find the average value of i2 by calculating the average value of sin 2t A
graph of cos 2t versus time is identical to a graph of sin2t versus time, except that the points are
shifted on the time axis Thus, the time average of sin 2t is equal to the time average of cos2t when
taken over one or more complete cycles That is,
(sin 2t)av (cos 2t)av
Using this fact and the trigonometric identity sin 2 cos 2 1, we obtain
When we substitute this result in the expression we obtain
or The factor is valid only for sinusoidally varying currents Other forms, such as sawtooth variations, have different factors.
wave-1/√2
Irms Imax /√2.
I2 max /2,
(i2 )av i2 I2
rms
i2 I2 max sin 2t,
(sin 2t)av 1
(sin 2t)av (cos 2t)av 2(sin 2t)av 1
I2 max
ω
, ∆v R
Figure 33.2 (a) Plots of the instantaneous current i Rand instantaneous voltage v Racross a sistor as functions of time The current is in phase with the voltage, which means that the current
re-is zero when the voltage re-is zero, maximum when the voltage re-is maximum, and minimum when
the voltage is minimum At time t T, one cycle of the time-varying voltage and current has been
completed (b) Phasor diagram for the resistive circuit showing that the current is in phase with the voltage.
Trang 5Alternating voltage also is best discussed in terms of rms voltage, and the
rela-tionship is identical to that for current:
(33.5)
When we speak of measuring a 120-V alternating voltage from an electrical
outlet, we are referring to an rms voltage of 120 V A quick calculation using
Equa-tion 33.5 shows that such an alternating voltage has a maximum value of about
170 V One reason we use rms values when discussing alternating currents and
voltages in this chapter is that ac ammeters and voltmeters are designed to read
rms values Furthermore, with rms values, many of the equations we use have the
same form as their direct-current counterparts
Which of the following statements might be true for a resistor connected to an ac
The voltage output of a generator is given by
Find the rms current in the circuit when this generator is connected to a 100- resistor.
Solution Comparing this expression for voltage output
Thus, the rms voltage is
Figure 33.3 (a) Graph of the current in a resistor as a function of time (b) Graph of the
cur-rent squared in a resistor as a function of time Notice that the gray shaded regions under the
curve and above the dashed line for have the same area as the gray shaded regions above
the curve and below the dashed line for Thus, the average value of i2is I2
max /2.
I2 max /2.
I2 max /2
Trang 6Solving this equation for di, we find that
Integrating this expression3 gives the instantaneous current in the inductor as afunction of time:
(33.7)
When we use the trigonometric identity cos we can expressEquation 33.7 as
(33.8)
Comparing this result with Equation 33.6, we see that the instantaneous current i L
in the inductor and the instantaneous voltage v L across the inductor are out ofphase by (/2) rad 90°
In general, inductors in an ac circuit produce a current that is out of phasewith the ac voltage A plot of voltage and current versus time is provided in Figure
33.5a At point a, the current begins to increase in the positive direction At this
in-stant the rate of change of current is at a maximum, and thus the voltage across
the inductor is also at a maximum As the current increases between points a and
b, di/dt (the slope of the current curve) gradually decreases until it reaches zero at point b As a result, the voltage across the inductor is decreasing during this same time interval, as the curve segment between c and d indicates Immediately after point b, the current begins to decrease, although it still has the same direction it had during the previous quarter cycle (from a to b) As the current decreases to zero (from b to e), a voltage is again induced in the inductor (d to f ), but the po- larity of this voltage is opposite that of the voltage induced between c and d (be-
cause back emfs are always directed to oppose the change in the current) Notethat the voltage reaches its maximum value one quarter of a period before the cur-rent reaches its maximum value Thus, we see that
Figure 33.5 (a) Plots of the
in-stantaneous current i Land
instanta-neous voltage vLacross an
induc-tor as functions of time The
cur-rent lags behind the voltage by 90°.
(b) Phasor diagram for the
induc-tive circuit, showing that the
cur-rent lags behind the voltage by 90°.
for a sinusoidal applied voltage, the current in an inductor always lags behindthe voltage across the inductor by 90° (one-quarter cycle in time)
The current in an inductor lags
the voltage by 90°
Trang 7The phasor diagram for the inductive circuit of Figure 33.4 is shown in Figure 33.5b.
From Equation 33.7 we see that the current in an inductive circuit reaches its
maximum value when cos t 1:
(33.9)
where the quantity X L, called the inductive reactance, is
(33.10)
Equation 33.9 indicates that, for a given applied voltage, the maximum current
de-creases as the inductive reactance inde-creases The expression for the rms current in
an inductor is similar to Equation 33.9, with Imax replaced by Irms and Vmax
re-placed by Vrms
Inductive reactance, like resistance, has units of ohms However, unlike
resis-tance, reactance depends on frequency as well as on the characteristics of the
in-ductor Note that the reactance of an inductor in an ac circuit increases as the
fre-quency of the current increases This is because at higher frequencies, the
instantaneous current must change more rapidly than it does at the lower
frequen-cies; this causes an increase in the maximum induced emf associated with a given
to dim the lights in the theater gradually.
Figure 33.6 shows a circuit consisting of a series combination
of an alternating voltage source, a switch, an inductor, and a
lightbulb The switch is thrown closed, and the circuit is
al-lowed to come to equilibrium so that the lightbulb glows
steadily An iron rod is then inserted into the interior of the
inductor What happens to the brightness of the lightbulb,
and why?
Solution The bulb gets dimmer As the rod is inserted, the
inductance increases because the magnetic field inside the
inductor increases According to Equation 33.10, this
in-crease in L means that the inductive reactance of the
induc-tor also increases The voltage across the inducinduc-tor increases
while the voltage across the lightbulb decreases With less
In a purely inductive ac circuit (see Fig 33.4), L 25.0 mH
and the rms voltage is 150 V Calculate the inductive
reac-tance and rms current in the circuit if the frequency is
60.0 Hz.
Solution Equation 33.10 gives
Maximum current in an inductor
Inductive reactance
Trang 8Using the trigonometric identity
we can express Equation 33.14 in the alternative form
(33.15)
Comparing this expression with Equation 33.12, we see that the current is /2 rad 90° out of phase with the voltage across the capacitor A plot of current andvoltage versus time (Fig 33.8a) shows that the current reaches its maximum valueone quarter of a cycle sooner than the voltage reaches its maximum value
Looking more closely, we see that the segment of the current curve from a to b
indicates that the current starts out at a relatively high value We can understandthis by recognizing that there is no charge on the capacitor at as a conse-quence, nothing in the circuit except the resistance of the wires can hinder theflow of charge at this instant However, the current decreases as the voltage across
the capacitor increases (from c to d on the voltage curve), and the capacitor is charging When the voltage is at point d, the current reverses and begins to in- crease in the opposite direction (from b to e on the current curve) During this time, the voltage across the capacitor decreases from d to f because the plates are
now losing the charge they accumulated earlier During the second half of the
cy-cle, the current is initially at its maximum value in the opposite direction (point e)
and then decreases as the voltage across the capacitor builds up The phasor gram in Figure 33.8b also shows that
Exercise Calculate the inductive reactance and rms current
in the circuit if the frequency is 6.00 kHz.
Figure 33.7 A circuit consisting
of a capacitor of capacitance C
Figure 33.8 (a) Plots of the
in-stantaneous current i Cand
instan-taneous voltage v Cacross a
capac-itor as functions of time The
voltage lags behind the current by
90° (b) Phasor diagram for the
ca-pacitive circuit, showing that the
current leads the voltage by 90°.
Trang 9From Equation 33.14, we see that the current in the circuit reaches its
maxi-mum value when cos
(33.16)
where X Cis called the capacitive reactance:
(33.17)
Note that capacitive reactance also has units of ohms
The rms current is given by an expression similar to Equation 33.16, with Imax
replaced by Irmsand replaced by Vrms
Combining Equations 33.12 and 33.16, we can express the instantaneous
volt-age across the capacitor as
(33.18)
Equations 33.16 and 33.17 indicate that as the frequency of the voltage source
creases, the capacitive reactance decreases and therefore the maximum current
in-creases Again, note that the frequency of the current is determined by the
fre-quency of the voltage source driving the circuit As the frefre-quency approaches zero,
the capacitive reactance approaches infinity, and hence the current approaches
zero This makes sense because the circuit approaches direct-current conditions as
Hence, from a modified Equation 33.16, the rms current is
Exercise If the frequency is doubled, what happens to the capacitive reactance and the current?
Answer X C is halved, and Imax is doubled.
An 8.00- F capacitor is connected to the terminals of a
60.0-Hz ac generator whose rms voltage is 150 V Find the
ca-pacitive reactance and the rms current in the circuit.
Solution Using Equation 33.17 and the fact that
THE RLC SERIES CIRCUIT
Figure 33.9a shows a circuit that contains a resistor, an inductor, and a capacitor
connected in series across an alternating-voltage source As before, we assume that
the applied voltage varies sinusoidally with time It is convenient to assume that
the instantaneous applied voltage is given by
while the current varies as
Trang 10is to determine max Figure 33.9b shows the voltage versus time across eachelement in the circuit and their phase relationships.
To solve this problem, we must analyze the phasor diagram for this circuit.First, we note that because the elements are in series, the current everywhere inthe circuit must be the same at any instant That is, the current at all points in aseries ac circuit has the same amplitude and phase Therefore, as we found inthe preceding sections, the voltage across each element has a different amplitudeand phase, as summarized in Figure 33.10 In particular, the voltage across the re-sistor is in phase with the current, the voltage across the inductor leads the current
by 90°, and the voltage across the capacitor lags behind the current by 90° Usingthese phase relationships, we can express the instantaneous voltages across thethree elements as
(33.19) (33.20)
(33.21)
where V R, V L, and V C are the maximum voltage values across the elements:
At this point, we could proceed by noting that the instantaneous voltage v
across the three elements equals the sum
For the circuit of Figure 33.9a, is the voltage of the ac source equal to (a) the sum of the maximum voltages across the elements, (b) the sum of the instantaneous voltages across the elements, or (c) the sum of the rms voltages across the elements?
Although this analytical approach is correct, it is simpler to obtain the sum by amining the phasor diagram Because the current at any instant is the same in all
ω
∆V L
∆V C
Figure 33.9 (a) A series circuit
consisting of a resistor, an inductor,
and a capacitor connected to an ac
generator (b) Phase relationships
for instantaneous voltages in the
se-ries RLC circuit.
Figure 33.10 Phase relationships between the voltage and current phasors for (a) a resistor, (b) an inductor, and (c) a capacitor connected in series.
Trang 11elements, we can obtain a phasor diagram for the circuit We combine the three
phasor pairs shown in Figure 33.10 to obtain Figure 33.11a, in which a single
pha-sor Imaxis used to represent the current in each element To obtain the vector sum
of the three voltage phasors in Figure 33.11a, we redraw the phasor diagram as in
Figure 33.11b From this diagram, we see that the vector sum of the voltage
ampli-tudes V R, V L, and V Cequals a phasor whose length is the maximum applied
voltage where the phasor makes an angle
Imax Note that the voltage phasors V Land V Care in opposite directions along
the same line, and hence we can construct the difference phasor
which is perpendicular to the phasor V R From either one of the right triangles
in Figure 33.11b, we see that
(33.22)
Therefore, we can express the maximum current as
The impedance Z of the circuit is defined as
(33.23)
where impedance also has units of ohms Therefore, we can write Equation 33.22
in the form
(33.24)
We can regard Equation 33.24 as the ac equivalent of Equation 27.8, which
de-fined resistance in a dc circuit as the ratio of the voltage across a conductor to the
current in that conductor Note that the impedance and therefore the current in
an ac circuit depend upon the resistance, the inductance, the capacitance, and the
frequency (because the reactances are frequency-dependent)
By removing the common factor Imax from each phasor in Figure 33.11a, we
can construct the impedance triangle shown in Figure 33.12 From this phasor
dia-gram we find that the phase angle
(33.25)
Also, from Figure 33.12, we see that When (which occurs
at high frequencies), the phase angle is positive, signifying that the current lags
behind the applied voltage, as in Figure 33.11a When the phase angle
is negative, signifying that the current leads the applied voltage When
the phase angle is zero In this case, the impedance equals the resistance and
the current has its maximum value, given by The frequency at which
this occurs is called the resonance frequency; it is described further in
Section 33.7
Table 33.1 gives impedance values and phase angles for various series circuits
containing different combinations of elements
Figure 33.11 (a) Phasor diagram
for the series RLC circuit shown in
Figure 33.9a The phasor V Ris in
phase with the current phasor Imax, the phasor V L leads Imaxby 90°, and the phasor V C lags Imaxby 90° The total voltage makes
an angle with Imax (b) fied version of the phasor diagram shown in (a).
Simpli-Vmax
φ
X L – X C Z
R
Figure 33.12 An impedance
tri-angle for a series RLC circuit gives
the relationship
Z√R2 (X L X C) 2
Trang 12Label each part of Figure 33.13 as being X L X C,X L X C,or X L X C.
Quick Quiz 33.3
Finding L from a Phasor Diagram
EXAMPLE33.5
circuit contains an inductor whose inductance can be varied,
a 200- resistor, and a 4.00- F capacitor What value of L
In a series RLC circuit, the applied voltage has a maximum
value of 120 V and oscillates at a frequency of 60.0 Hz The
TABLE 33.1 Impedance Values and Phase Angles for Various
a In each case, an ac voltage (not shown) is applied across the elements.
C R R
L
Trang 13Analyzing a Series RLC Circuit
EXAMPLE 33.6
(d) Find both the maximum voltage and the neous voltage across each element.
instanta-Solution The maximum voltages are
Using Equations 33.19, 33.20, and 33.21, we find that we can write the instantaneous voltages across the three elements as
Comments The sum of the maximum voltages across the
much greater than the maximum voltage of the generator,
150 V As we saw in Quick Quiz 33.2, the sum of the mum voltages is a meaningless quantity because when sinu-
maxi-soidally varying quantities are added, both their amplitudes and
their phases must be taken into account We know that the
A series RLC ac circuit has , H,
inductive reactance, the capacitive reactance, and the
imped-ance of the circuit.
Solution The reactances are and
Because the circuit is more capacitive than inductive,
negative and the current leads the applied voltage.
should an engineer analyzing the circuit choose such that the
voltage across the capacitor lags the applied voltage by 30.0°?
Solution The phase relationships for the drops in voltage
across the elements are shown in Figure 33.14 From the
fig-ure we see that the phase angle is This is because
the phasors representing Imaxand V Rare in the same
direc-tion (they are in phase) From Equadirec-tion 33.25, we find that
Substituting Equations 33.10 and 33.17 (with 2f ) into
this expression gives
Substituting the given values into the equation gives L
Trang 14POWER IN AN AC CIRCUIT
No power losses are associated with pure capacitors and pure inductors in
an ac circuit To see why this is true, let us first analyze the power in an ac circuitcontaining only a generator and a capacitor
When the current begins to increase in one direction in an ac circuit, chargebegins to accumulate on the capacitor, and a voltage drop appears across it Whenthis voltage drop reaches its maximum value, the energy stored in the capacitor is
However, this energy storage is only momentary The capacitor ischarged and discharged twice during each cycle: Charge is delivered to the capacitorduring two quarters of the cycle and is returned to the voltage source during the re-maining two quarters Therefore, the average power supplied by the source iszero In other words, no power losses occur in a capacitor in an ac circuit.Similarly, the voltage source must do work against the back emf of the induc-tor When the current reaches its maximum value, the energy stored in the induc-tor is a maximum and is given by When the current begins to decrease inthe circuit, this stored energy is returned to the source as the inductor attempts tomaintain the current in the circuit
In Example 28.1 we found that the power delivered by a battery to a dc circuit
is equal to the product of the current and the emf of the battery Likewise, the stantaneous power delivered by an ac generator to a circuit is the product of the
in-generator current and the applied voltage For the RLC circuit shown in Figure
33.9a, we can express the instantaneous power as
(33.26)
Clearly, this result is a complicated function of time and therefore is not very ful from a practical viewpoint What is generally of interest is the average powerover one or more cycles Such an average can be computed by first using thetrigonometric identity sin(
use-into Equation 33.26 gives
(33.27)
We now take the time average of over one or more cycles, noting that Imax,
Vmax,
in Equation 33.27 involves the average value of sin2t, which is (as shown in
footnote 2) The time average of the second term on the right is identically zerobecause sin t cos t sin 2t, and the average value of sin 2t is zero There-fore, we can express the average power as
1 2
ᏼ
ᏼ ImaxVmax sin2 maxVmax sin
ImaxVmax sin
ᏼ i v Imax sin( max sin t
maximum voltages across the various elements occur at
dif-ferent times That is, the voltages must be added in a way that
takes account of the different phases When this is done,
Equation 33.22 is satisfied You should verify this result.
Exercise Construct a phasor diagram to scale, showing the voltages across the elements and the applied voltage From your diagram, verify that the phase angle is 34.0°.
Trang 15where the quantity cos
we see that the maximum voltage drop across the resistor is given by
Using Equation 33.5 and the fact that cos
we find that we can express as
After making the substitution from Equation 33.4, we have
(33.30)
In words, the average power delivered by the generator is converted to
inter-nal energy in the resistor, just as in the case of a dc circuit No power loss
oc-curs in an ideal inductor or capacitor When the load is purely resistive, then
Equation 33.29 shows that the power delivered by an ac source to any circuit
depends on the phase, and this result has many interesting applications For
exam-ple, a factory that uses large motors in machines, generators, or transformers has a
large inductive load (because of all the windings) To deliver greater power to
such devices in the factory without using excessively high voltages, technicians
in-troduce capacitance in the circuits to shift the phase
Calculate the average power delivered to the series RLC
cir-cuit described in Example 33.6.
Solution First, let us calculate the rms voltage and rms
cur-rent, using the values of Vmaxand Imaxfrom Example 33.6:
RESONANCE IN A SERIES RLC CIRCUIT
A series RLC circuit is said to be in resonance when the current has its maximum
value In general, the rms current can be written
Trang 16Because the impedance depends on the frequency of the source, the current in
the RLC circuit also depends on the frequency The frequency 0 at which
is called the resonance frequency of the circuit To find 0, we usethe condition from which we obtain , or
(33.33)
Note that this frequency also corresponds to the natural frequency of oscillation of
an LC circuit (see Section 32.5) Therefore, the current in a series RLC circuit
reaches its maximum value when the frequency of the applied voltage matches the
natural oscillator frequency — which depends only on L and C Furthermore, at
this frequency the current is in phase with the applied voltage
What is the impedance of a series RLC circuit at resonance? What is the current in the
cir-cuit at resonance?
A plot of rms current versus frequency for a series RLC circuit is shown in
Fig-ure 33.15a The data assume a constant that H, andthat The three curves correspond to three values of R Note that in
each case the current reaches its maximum value at the resonance frequency 0.Furthermore, the curves become narrower and taller as the resistance decreases
By inspecting Equation 33.32, we must conclude that, when the currentbecomes infinite at resonance Although the equation predicts this, real circuits al-ways have some resistance, which limits the value of the current
7 6 5 4 3 2 1
Figure 33.15 (a) The rms current versus frequency for a series RLC circuit, for three values of
R The current reaches its maximum value at the resonance frequency 0 (b) Average power
versus frequency for the series RLC circuit, for two values of R
... lags behind the current by 90° Usingthese phase relationships, we can express the instantaneous voltages across thethree elements as(33. 19) (33. 20)
(33. 21)
where...
in Figure 33. 11b, we see that
(33. 22)
Therefore, we can express the maximum current as
The impedance Z of the circuit is defined as
(33. 23)
where...
Section 33. 7
Table 33. 1 gives impedance values and phase angles for various series circuits
containing different combinations of elements
Figure 33. 11