Resonance In AC Circuits
Dr Miguel Cavero
August 22, 2014
Resonance
August 22, 2014
1 / 19
AC Circuit Example
A current in a series RL circuit is given by
I = 8 cos(1000t) A
where t is in seconds and R = 40 Ω and L = 30 mH.
Calculate the
(a) reactance and impedance,
(b) the peak voltage across the RL combination,
(c) the instantaneous voltage across the RL combination,
(d) and the rms voltage across L.
Resonance
Recap
August 22, 2014
3 / 19
AC Circuit Example
A current in a series RL circuit is given by
I = 8 cos(1000t) A
where t is in seconds and R = 40 Ω and L = 30 mH.
(a) Calculate the reactance and impedance.
XL = ωL = (1000) × (30 × 10−3 )
= 30 Ω
p
p
Z =
R2 + (XL − XC )2 = 402 + 302
= 50 Ω
Resonance
Recap
August 22, 2014
4 / 19
AC Circuit Example
A current in a series RL circuit is given by
I = 8 cos(1000t) A
where t is in seconds and R = 40 Ω and L = 30 mH.
(b) Calculate the the peak voltage across the RL combination.
The peak voltage is
V0 = I0 Z = 8 × 50 = 400 V
Resonance
Recap
August 22, 2014
5 / 19
AC Circuit Example
A current in a series RL circuit is given by
I = 8 cos(1000t) A
where t is in seconds and R = 40 Ω and L = 30 mH.
(c) Calculate the the instantaneous voltage across the RL
combination.
The instantaneous voltage across the combination is
V = V0 cos(ωt + α)
What is α?
tan α =
Resonance
XL − XC
30
=
R
40
Recap
August 22, 2014
6 / 19
AC Circuit Example
A current in a series RL circuit is given by
I = 8 cos(1000t) A
where t is in seconds and R = 40 Ω and L = 30 mH.
(d) Calculate the rms voltage across L.
What is the peak voltage across L?
V0L = I0 XL = 8 × 30 = 240 V
The rms voltage is then
VL
240
L
Vrms
= √0 = √
2
2
Resonance
Recap
August 22, 2014
7 / 19
Frequency Depedence
Recall that in an RLC circuit, the impedance Z depends on the
frequency.
s
p
1 2
2
2
2
Z = R + (XL − XC ) = R + ωL −
ωC
The peak current and peak voltage in a circuit is given by the
impedance:
V0 = I0 Z
V
I
√0 = √0 Z
2
2
Vrms = Irms Z
Changing the angular frequency results in a change in the current.
Resonance
Resonace
August 22, 2014
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Frequency Depedence
From the equation
Vrms
Z
the current reaches a maximum value when the impedance is at a
minimum.
Since the impedance is the combination of resistance and reactance
p
Z = R2 + (XL − XC )2
Irms =
it is a minimum when the inductive reactance is equal to the
capacitative reactance.
XL = XC ⇒ Z = R
Resonance
Resonace
August 22, 2014
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Resonant Frequency
When the inductive reactance is equal to the capacitative reactance,
ω0 L =
ω02 =
∴ ω0 =
1
ωC
1
LC
1
√
LC
where ω0 is the resonant frequency, which results in the minimum
impedance.
At resonance, the circuit is purely resistive (since Z = R).
Clearly, the current in the circuit will be a maximum when the
impedance is a minimum.
Resonance
Resonace
August 22, 2014
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Resonant Frequency
A graph of current versus frequency for different resistors R1 and R2 (L
and C are kept constant) shows that the current reaches a maximum
when the frequency is ω0 .
Resonance
Resonace
August 22, 2014
12 / 19
Resonant Frequency
Since XL = ωL and XC = 1/ωC, when XL increases then XC
decreases, and vice versa.
At low frequencies, the circuit is more capacitative and at high
frequencies the circuit is more inductive.
Resonance
Resonace
August 22, 2014
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Resonant Frequency
In this graph R1 < R2 since, at every frequency, the current is greater
through R1 .
Resonance
Resonace
August 22, 2014
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Resonant Frequency
Irms =
Vrms
Vrms
=q
Z
R2 + ωL −
1 2
ωC
What happens at resonance if the resistance is zero?
Resonance
Resonace
August 22, 2014
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Metal Detectors
A metal detector (for example in an airport) is essentially a coil with
many turns.
The frequency of the circuit is set to be the resonant frequency when
there is no metal in the coil.
Any metal that is placed inside the detector changes the effective
inductance of the coil, therefore it changes the current in the circuit. An
alarm is set to go off when a change in current is detected.
Resonance
Resonace
August 22, 2014
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Capacitor In An AC Circuit
A series RLC circuit has resistance R = 150 Ω, inductance
L = 20.0 mH, capacitance C and is connected to an alternating emf
source of 20.0 V rms at 796 Hz.
(a) Find the value for C for which the current is a maximum.
(b) Hence, find the maximum rms current in the circuit.
Resonance
Resonace
August 22, 2014
17 / 19
Capacitor In An AC Circuit
A series RLC circuit has resistance R = 150 Ω, inductance
L = 20.0 mH, capacitance C and is connected to an alternating emf
source of 20.0 V rms at 796 Hz.
(a) Find the value for C for which the current is a maximum.
The resonant (angular) frequency is given by
1
ω0 = 2πf0 = √
LC
The current is maximum at resonance, therefore the capacitance
C at resonance is
1
√
f0 =
2π LC
1
C =
2
4π f02 L
1
=
= ... F
2
2
4π (796) (20.0 × 10−3 )
Resonance
Resonace
August 22, 2014
18 / 19
Capacitor In An AC Circuit
A series RLC circuit has resistance R = 150 Ω, inductance
L = 20.0 mH, capacitance C and is connected to an alternating emf
source of 20.0 V rms at 796 Hz.
(b) Hence, find the maximum rms current in the circuit.
What is the value of the impedance at resonance?
At resonance, the circuit is purely resistive (there is no reactance).
The rms current is
Irms =
Resonance
Vrms
Vrms
20.0
=
=
= ... A
Z
R
150
Resonace
August 22, 2014
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