OpenStax Staging4 module: m10418
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Resonance in an AC Circuit
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Words Numbers
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Abstract
By the end of the section, you will be able to:
• Determine the peak ac resonant angular frequency for a RLC circuit
• Explain the width of the average power versus angular frequency curve and its signicance using
terms like bandwidth and quality factor
In the
RLC series circuit of , the current amplitude is, from ,
I0 = q
V0
.
(1)
2
R2 + (ωL − 1/ωC)
If we can vary the frequency of the ac generator while keeping the amplitude of its output voltage constant,
then the current changes accordingly. A plot of
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I0
versus
ω
is shown in Figure 1.
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Figure 1: At an
maximum value.
RLC
2
circuit's resonant frequency, ω0 =
p
1/LC, the current amplitude is at its
In Oscillations, we encountered a similar graph where the amplitude of a damped harmonic oscillator
was plotted against the angular frequency of a sinusoidal driving force (see [link]Figure 15_07_ForcDmpAmp[/link]). This similarity is more than just a coincidence, as shown earlier by the application of Kirchho 's loop rule to the circuit of . This yields
di
q
+ iR +
= V0 sin ωt,
dt
C
(2)
d2 q
dq
1
+ R + q = V0 sin ωt,
dt2
dt
C
(3)
L
or
L
where we substituted
dq(t)/dt for i (t). A comparison of (2) and, from Oscillations, [link]Equation_15_05_01[/link]
RLC series circuit is the electrical analog
for damped harmonic motion clearly demonstrates that the driven
of the driven damped harmonic oscillator.
The
resonant frequencyf0 of the RLC circuit is the frequency at which the amplitude of the current is
a maximum and the circuit would oscillate if not driven by a voltage source. By inspection, this corresponds
to the angular frequency
ω0 = 2πf0
at which the impedance
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Z in (1) is a minimum, or when
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ω0 L =
1
ω0 C
(4)
and
note:
r
ω0 =
1
.
LC
(5)
This is the resonant angular frequency of the circuit. Substituting
ϕ = tan−1 (0) = 0, I0 = V0 /R,
Therefore, at resonance, an
ω0
into , , and , we nd that at resonance,
Z = R.
and
(6)
RLC circuit is purely resistive, with the applied emf and current in phase.
What happens to the power at resonance? tells us how the average power transferred from an ac generator
to the
RLC combination varies with frequency. In addition, P
on the frequency, is a minimum, that is, when
output of the source in an
RLC series circuit is a maximum.
Figure 2 is a typical plot of
Pave
versus
ω
ave
reaches a maximum when
XL = XC and Z = R.
From , this maximum is
quantity known as the
quality factorQ
2
/R.
Vrms
in the region of maximum power output. The
of the resonance peak is dened as the range of angular frequencies
greater than one-half the maximum value of
Z, which depends
Thus, at resonance, the average power
Pave .
ω
bandwidth∆ω
over which the average power
Pave
is
The sharpness of the peak is described by a dimensionless
of the circuit. By denition,
note:
Q=
where
ω0
is the resonant angular frequency. A high
ω0
,
∆ω
Q indicates a sharp resonance peak.
(7)
We can give
Q in
terms of the circuit parameters as
note:
Q=
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ω0 L
.
R
(8)
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Figure 2: Like the current, the average power transferred from an ac generator to an RLC circuit peaks
at the resonant frequency.
Resonant circuits are commonly used to pass or reject selected frequency ranges. This is done by adjusting the
value of one of the elements and hence tuning the circuit to a particular resonant frequency. For example,
in radios, the receiver is tuned to the desired station by adjusting the resonant frequency of its circuitry
to match the frequency of the station. If the tuning circuit has a high
Q, it will have a small bandwidth,
so signals from other stations at frequencies even slightly dierent from the resonant frequency encounter a
high impedance and are not passed by the circuit. Cell phones work in a similar fashion, communicating
with signals of around 1 GHz that are tuned by an inductor-capacitor circuit.
One of the most common
applications of capacitors is their use in ac-timing circuits, based on attaining a resonant frequency. A metal
detector also uses a shift in resonance frequency in detecting metals (Figure 3).
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Figure 3: When a metal detector comes near a piece of metal, the self-inductance of one of its coils
changes. This causes a shift in the resonant frequency of a circuit containing the coil. That shift is
detected by the circuitry and transmitted to the diver by means of the headphones.
Example 1
Resonance in an
RLC Series Circuit
(a) What is the resonant frequency of the circuit of ? (b) If the ac generator is set to this frequency
without changing the amplitude of the output voltage, what is the amplitude of the current?
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Strategy
The resonant frequency for a
RLC
circuit is calculated from (5), which comes from a balance
between the reactances of the capacitor and the inductor.
Since the circuit is at resonance, the
impedance is equal to the resistor. Then, the peak current is calculated by the voltage divided by
the resistance.
Solution
a. The resonant frequency is found from (5):
f0 =
1
2π
q
1
LC
=
= 1.03 × 10
2
1
2π
q
1
(3.00 × 10−3
H)(8.00
× 10−4
F)
(9)
Hz.
b. At resonance, the impedance of the circuit is purely resistive, and the current amplitude is
I0 =
0.100 V
= 2.50 × 10−2 A.
4.00Ω
(10)
Signicance
If the circuit were not set to the resonant frequency, we would need the impedance of the entire
circuit to calculate the current.
Example 2
Power Transfer in an RLC Series Circuit at Resonance
(a) What is the resonant angular frequency of an
and
C = 2.00 × 10
−6
RLC circuit with R = 0.200Ω,L = 4.00 × 10−3 H,
F? (b) If an AC source of constant amplitude 4.00 V is set to this frequency,
what is the average power transferred to the circuit? (c) Determine
Q and the bandwidth of this
circuit.
Strategy
The resonant angular frequency is calculated from (5). The average power is calculated from the
rms voltage and the resistance in the circuit.
The quality factor is calculated from (8) and by
knowing the resonant frequency. The bandwidth is calculated from (7) and by knowing the quality
factor.
Solution
a. The resonant angular frequency is
ω0 =
q
1
LC
=
q
1
(4.00 × 10−3
4
= 1.12 × 10
H)(2.00
× 10−6
F)
(11)
rad/s.
b. At this frequency, the average power transferred to the circuit is a maximum. It is
Pave
V2
= rms =
R
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√ 2
1/ 2 (4.00 V)
= 40.0 W.
0.200Ω
(12)
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c. The quality factor of the circuit is
1.12 × 104 rad/s 4.00 × 10−3 H
ω0 L
Q=
=
= 224.
R
0.200Ω
(13)
We then nd for the bandwidth
∆ω =
ω0
1.12 × 104 rad/s
=
= 50.0 rad/s.
Q
224
(14)
Signicance
If a narrower bandwidth is desired, a lower resistance or higher inductance would help.
How-
ever, a lower resistance increases the power transferred to the circuit, which may not be desirable,
depending on the maximum power that could possibly be transferred.
Exercise 1
Check Your Understanding In the circuit of , L = 2.0
(Solution on p. 10.)
note:
and
R = 40Ω.
(a) What is the resonant frequency?
circuit at resonance?
−3
× 10
H,C
= 5.0 × 10−4 F,
(b) What is the impedance of the
(c) If the voltage amplitude is 10 V, what is
i (t)
at resonance?
(d) The frequency of the AC generator is now changed to 200 Hz. Calculate the phase
dierence between the current and the emf of the generator.
Exercise 2
(Solution on p. 10.)
Check Your Understanding What happens to the resonant frequency of an RLC series
note:
circuit when the following quantities are increased by a factor of 4: (a) the capacitance,
(b) the self-inductance, and (c) the resistance?
Exercise 3
(Solution on p. 10.)
Check Your Understanding The resonant angular frequency of an RLC series circuit
note:
4.0 × 102 rad/s. An ac source operating at this frequency transfers an average power of
2.0 × 10−2 W to the circuit. The resistance of the circuit is 0.50Ω. Write an expression
is
for the emf of the source.
1 Summary
ˆ
ˆ
At the resonant frequency, inductive reactance equals capacitive reactance.
The average power versus angular frequency plot for a
RLC circuit has a peak located at the resonant
frequency; the sharpness or width of the peak is known as the bandwidth.
ˆ
The bandwidth is related to a dimensionless quantity called the quality factor. A high quality factor
value is a sharp or narrow peak.
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2 Problems
Exercise 4
(a) Calculate the resonant angular frequency of an
75
C = 4.0µF.
mH, and
(b) If
R
is changed to
RLC
300Ω,
(Solution on p. 10.)
series circuit for which
R = 20Ω, L =
what happens to the resonant angular
frequency?
Exercise 5
(Solution on p. 10.)
The resonant frequency of an
RLC series circuit is 2.0 × 103 Hz. If the self-inductance in the circuit
is 5.0 mH, what is the capacitance in the circuit?
Exercise 6
(a) What is the resonant frequency of an
C = 4.0µF?
RLC
(Solution on p. 10.)
series circuit with
R = 20Ω, L = 2.0 mH,
Exercise 7
For an
and
(b) What is the impedance of the circuit at resonance?
RLC
(Solution on p. 10.)
series circuit,
R = 100Ω, L = 150 mH,
and
C = 0.25µF.
(a) If an ac source of
variable frequency is connected to the circuit, at what frequency is maximum power dissipated in
the resistor? (b) What is the quality factor of the circuit?
Exercise 8
An ac source of voltage amplitude 100 V and variable frequency
with
R = 10Ω, L = 2.0 mH,
f
and
C = 25µF.
f
(Solution on p. 10.)
drives an
RLC
series circuit
(a) Plot the current through the resistor as a function
of the frequency . (b) Use the plot to determine the resonant frequency of the circuit.
Exercise 9
(Solution on p. 10.)
(a) What is the resonant frequency of a resistor, capacitor, and inductor connected in series if
R = 100Ω, L = 2.0 H,
and
C = 5.0µF?
(b) If this combination is connected to a 100-V source
operating at the constant frequency, what is the power output of the source? (c) What is the
Q of
the circuit? (d) What is the bandwidth of the circuit?
Exercise 10
(Solution on p. 10.)
Suppose a coil has a self-inductance of 20.0 H and a resistance of
200Ω.
What (a) capacitance and
(b) resistance must be connected in series with the coil to produce a circuit that has a resonant
frequency of 100 Hz and a
Exercise 11
Q of 10?
(Solution on p. 10.)
An ac generator is connected to a device whose internal circuits are not known. We only know
current and voltage outside the device, as shown below. Based on the information given, what can
you infer about the electrical nature of the device and its power usage?
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Solutions to Exercises in this Module
Solution to Exercise (p. 7)
a. 160 Hz; b.
40Ω;
(0.25 A)
c.
Solution to Exercise (p. 7)
sin
103 t;
d. 0.023 rad
a. halved; b. halved; c. same
Solution to Exercise (p. 7)
v (t) = (0.14 V)
4.0 × 102 t
sin
Solution to Exercise (p. 8)
a.
1.8 × 103 rad/s;
b.
1.83 × 103 rad/s,
Solution to Exercise (p. 8)
because it is independent of
1.3 × 10−7 F
Solution to Exercise (p. 8)
a. 1780 Hz; b.
20Ω
Solution to Exercise (p. 8)
a. 820 Hz; b. 7.8
Solution to Exercise (p. 8)
a.
b. 712 Hz
Solution to Exercise (p. 8)
a. 50 Hz; b. 50 W; c. 13; d. 25 rad/s
Solution to Exercise (p. 8)
a.
1.27 × 10−7 F;
b.
1060Ω
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R
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Solution to Exercise (p. 8)
The reactance of the capacitor is larger than the reactance of the inductor because the current leads the
voltage. The power usage is 30 W.
Glossary
Denition 1: bandwidth
range of angular frequencies over which the average power is greater than one-half the maximum
value of the average power
Denition 2: quality factor
dimensionless quantity that describes the sharpness of the peak of the bandwidth; a high quality
factor is a sharp or narrow resonance peak
Denition 3: resonant frequency
frequency at which the amplitude of the current is a maximum and the circuit would oscillate if
not driven by a voltage source
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