Electric Circuits II Resonance
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Dr. Firas Obeidat
Dr. Firas Obeidat β Philadelphia University
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β’ Series Resonance
Table of Contents
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Series Resonance
Resonance is a condition in an RLC circuit in which the capacitive and inductive
reactances are equal in magnitude, thereby resulting in a purely resistive
impedance (voltage and current at the circuit input terminals are in phase).
Resonant circuits (series or parallel) are useful for constructing filters. They
are used in many applications such as selecting the desired stations in radio
and TV receivers.
For the series RLC circuit. The input
impedance in frequency domain is
Resonance results when the imaginary part of the transfer function is zero
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Series Resonance The value of that satisfies this condition is called the resonant frequency. Thus, the
resonance condition is
Since
or
At resonance:-
1. The impedance is purely resistive, thus Z=R, In other words, the LC series
combination acts like a short circuit, and the entire voltage is across R.
2. The voltage Vs and the current I are in phase, so that the power factor is unity.
3. The magnitude of the transfer function H(π)=Z(π) is minimum.
4. The maximum current for the circuit for an applied voltage Vs since Z is a
minimum value.
5. The inductor voltage and capacitor voltage can be much more than the source
voltage.
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Series Resonance
At resonance, when ΞΈ=0, the voltage across each of the elements in the circuit can
be written as
I =π
πβ ΞΈ
π β 0=
ππ
π β ΞΈ
ππ = πΌπ β 0π
ππΏ = πΌππΏβ 90π
ππΆ = πΌππΆβ β 90π
Because πΏπ³ = πΏC , The magnitude of VL equals to the magnitude of VC but 180Β° out
of phase
The average power dissipated by the resistor and the reactive powers of the
inductor and capacitor as follows
ππ =1
2πΌ2π (w)
ππΏ =1
2πΌ2ππΏ
(VAR)
ππΆ =1
2πΌ2ππΆ
(VAR)
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Series Resonance
The frequency response of the circuitβs current magnitude is as in the figure
The magnitude of the current at resonance is
I =ππ
π
π =1
2πΌ2π =
1
2
ππ
2
π
The maximum power dissipated by the series resonant
circuit is given as
The bandwidth, BW, of the resonant circuit is the
difference between the frequencies at which the circuit
delivers half of the maximum power. The frequencies π1
and π2 are called the half-power frequencies, the cutoff
frequencies, or the band frequencies.
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Series Resonance
For the series resonant circuit the power at any frequency is determined as
π =1
2πΌ2π =
1
2
ππ
|π|
2π =
1
2
ππ
2
π 2+ ππΏβ1
ππΆ2
π
At the half-power frequencies, the power must be
1
2Γ
1
2
ππ2
π =
1
2
ππ2
π 2 + ππΏ β1
ππΆ2
π β 1
2π =
π
π 2 + ππΏ β1
ππΆ2
2π 2 = π 2 +π2πΏπΆ β 1
ππΆ
2
β π 2 =π2πΏπΆ β 1
ππΆ
2
Β±π =π2πΏπΆβ1
ππΆ β Β±π ππΆ = π2πΏπΆ β 1
βπ ππΆ = π2πΏπΆ β 1 β π2πΏπΆ + π ππΆ β 1 =0
The solution of the above equation yields four values for the cutoff frequency.
Only two of these values are positive and have physical significance
+π ππΆ = π2πΏπΆ β 1 β π2πΏπΆ β π ππΆ β 1 =0 To find π2 take +πΉππͺ
To find π1 take βπΉππͺ
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Series Resonance
π1 is called the lower half power frequency and π2 is called upper half
power frequency.
ππ =π
π³πͺ= ππππ
π1π2 = βπ
2πΏ+
π
2πΏ
2
+1
πΏπΆ
π
2πΏ+
π
2πΏ
2
+1
πΏπΆ=
1
πΏπΆ
ππ = βπΉ
ππ³+
πΉ
ππ³
π
+π
π³πͺ ππ =
πΉ
ππ³+
πΉ
ππ³
π
+π
π³πͺ
π1 and π2 are in general not symmetrical around the resonant frequency, because
the frequency response is not generally symmetrical
The height of the curve is determined by R. The width of the response curve
depends on the bandwidth B, which is defined as the difference between the two
half-power frequencies.
Solving the above two equations gives
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Series Resonance
The βsharpnessβ of the resonance in a resonant circuit is measured quantitatively
by the quality factor Q.
πΈ = ππ Peak energy stored in the circuit
Energy dissipated by the circuit in one period at resonance
π2 β π1 =π
2πΏ+
π
2πΏ
2+
1
πΏπΆβ β
π
2πΏ+
π
2πΏ
2+
1
πΏπΆ =
π
πΏ
π© = ππ β ππ =πΉ
π³
The relationship between the bandwidth B and the quality factor Q is given by
πΈ =πππ³
πΉ=
π
πππͺπΉ
The quality factor of a resonant circuit is the ratio of its resonant frequency to its
bandwidth.
πΈ =ππ
π©
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Series Resonance
A resonant circuit is designed to operate at or near its resonant frequency. It is
said to be a high-Q circuit when its quality factor is equal to or greater than 10.
For high-Q circuits, the approximate expressions for Ο1 and Ο2 are
The cutoff frequencies can be written in terms of the center frequency and the
bandwidth as:
ππ = βπ©
π+
π©
π
π
+ πππ ππ =
π©
π+
π©
π
π
+ πππ
The cutoff frequencies can be written in terms of the quality factor and the center
frequency as:
ππ = ππ βπ
ππΈ+ π +
π
ππΈ
π
ππ = ππ
π
ππΈ+ π +
π
ππΈ
π
ππ β ππ βπ©
π ππ β ππ +
π©
π
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Series Resonance
Example: In the circuit R=2Ξ©, L= 1 mH and C=0.4 ΞΌF. (a) Find the resonant frequency
and the half-power frequencies. (b) Calculate the quality factor and bandwidth.
(c) Determine the amplitude of the current at πo, π1 and π2.
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Series Resonance
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Series Resonance Example:
a. For the series resonant circuit, find I, VR, VL, and
VC at resonance.
b. What is the Qs of the circuit?
c. If the resonant frequency is 5000 Hz, find the
bandwidth.
d. What is the power dissipated in the circuit at the
half-power frequencies?.
π) π = 2Ξ©
πΌ =ππ
π =
10β 0π
2= 5β 0π
π΄
π£π = ππ = 10β 0πA
π£πΏ = ππΏπΌ = 10β 90π Γ 5β 0π= 50β 90π
π£πΆ = ππΆπΌ = 10β β 90π Γ 5β 0π= 50β β 90π
π) ππ =πππΏ
π =
10Ξ©
2Ξ©= 5
π) π΅ =ππ
ππ
=2 Γ Ο Γ 5000
5= 6283.19
π) ππ»ππΉ =1
2ππππ₯ =
1
2Γ
1
2πΌπππ₯
2R
=1
2Γ
1
2Γ 52 Γ 2 =12.5w
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Series Resonance Example: The bandwidth of a series resonant circuit is 2513.27.
a. If the resonant frequency is 4000 Hz, what is the value of Qs?
b. If R=10 Ξ©, what is the value of XL at resonance?
c. Find the inductance L and capacitance C of the circuit.
π) ππ =π
π
π΅ =
2ΓΟΓ4000
2513.27 =10
π) πΏ =ππΏ
2Οππ
=100
2Ο Γ 4000π»π§= 3.98 ππ»
π) ππ =π
ππΏ
π β πππΏ = ππΏ = πππ =10Γ10=100Ξ©
At resonant XL=XC.
πΆ =1
2ΟππππΆ
=1
2Ο Γ 4000π»π§ Γ 100Ξ©= 0.398 ΞΌπΉ
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Series Resonance Example: A band pass circuit able to select inputs within the 1-10 kHz frequency band
(cutoff frequencies). If C=1 ΞΌF, find the value of L and R.
π1 = 2ππ1 = 2π Γ 1000 = 6283.19 πππ/π
π2 = 2ππ2 = 2π Γ 10000 = 62831.9 πππ/π
ππ= π1π2 = 6283.19 Γ 62831.9 = 19869.19 πππ/π
ππ = 3162.28 Hz
π΅ = π2 β π1 = 62831.9 β 6283.19 = 56548.71
π =ππ
π΅=
19869.19
56548.71 = 0.3514
π =1
πππΆπ=
1
19869.19 Γ 1 Γ 10β6 Γ 0.3514= 143.24 Ξ©
πΏ =1
ππ2πΆ
=1
19869.192 Γ 1 Γ 10β6 = 2.533 ππ»
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Series Resonance Example: A series RLC network has R=2 kΞ©, L=40 mH and C=1ΞΌF. Calculate the
impedance at resonance and at one-fourth, one-half, twice, and four times the resonant
frequency.
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Series Resonance
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Series Resonance
Example: Design a series RLC resonant circuit with R=10Ξ©, πo=1000 rad/s and B=20 rad/s. then find circuitβs Q.
Example: Design a series RLC resonant circuit with πo=40 rad/s and B=10 rad/s.
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