32.5.3 Resonance

Suppose we vary the emf frequency ω while keeping everything else constant. There is very little current at very low frequencies because the capacitive reactance XC=1/ωC (and thus Z) is very large. Similarly, there is very little current at very high frequencies because the inductive reactance XL=ωL becomes very large.

If I approaches zero at very low and very high frequencies, there should be some intermediate frequency where I is a maximum. Indeed, you can see from Equation 32.24 that the denominator will be a minimum, making I a maximum, when X​L=X​C, or

ωL=1ωC (32.29)

The frequency ω0 that satisfies Equation 32.29 is called the resonance frequency:

ω0=1LC (32.30)

This is the frequency for maximum current in the series RLC circuit. The maximum current

Imax=ε​0R (32.31)

is that of a purely resistive circuit because the impedance is Z=R at resonance.

You’ll recognize ω0 as the oscillation frequency of the LC circuit we analyzed in Chapter 30. The current in an ideal LC circuit oscillates forever as energy is transferred back and forth between the capacitor and the inductor. This is analogous to an ideal, frictionless simple harmonic oscillator in which the energy is transformed back and forth between kinetic and potential.

Adding a resistor to the circuit is like adding damping to a mechanical oscillator. The emf is then a sinusoidal driving force, and the series RLC circuit is directly analogous to the driven, damped oscillator that you studied in Chapter 15. A mechanical oscillator exhibits resonance by having a large-amplitude response when the driving frequency matches the system’s natural frequency. Equation 32.30 is the natural frequency of the series RLC circuit, the frequency at which the current would like to oscillate. Consequently, the circuit has a large current response when the oscillating emf matches this frequency.

FIGURE 32.18 shows the peak current I of a series RLC circuit as the emf frequency ω is varied. Notice how the current increases until reaching a maximum at frequency ω0, then decreases. This is the hallmark of a resonance.

FIGURE 32.18 A graph of the current I versus emf frequency for a series RLC circuit.

The figure illustrates the Current versus e m f frequency graph for three values of R in a series R L C circuit.
FIGURE 32.18 Full Alternative Text

As R decreases, causing the damping to decrease, the maximum current becomes larger and the curve in Figure 32.18 becomes narrower. You saw exactly the same behavior for a driven mechanical oscillator. The emf frequency must be very close to ω0 in order for a lightly damped system to respond, but the response at resonance is very large.

For a different perspective, FIGURE 32.19 graphs the instantaneous emf ε=ε​0 cos ωt and current i=I cos(ωt−ϕ) for frequencies below, at, and above ω0. The current and the emf are in phase at resonance (ϕ=0 rad) because the capacitor and inductor essentially cancel each other to give a purely resistive circuit. Away from resonance, the current decreases and begins to get out of phase with the emf. You can see, from Equation 32.27, that the phase angle ϕ is negative when XL<XC (i.e., the frequency is below resonance) and positive when XL>XC (the frequency is above resonance).

FIGURE 32.19 Graphs of the emf ε and the current i at frequencies below, at, and above the resonance frequency ω0.

The figure consists of three parts that illustrate the graphs of e m f versus time.
FIGURE 32.19 Full Alternative Text

Resonance circuits are widely used in radio, television, and communication equipment because of their ability to respond to one particular frequency (or very narrow range of frequencies) while suppressing others. The selectivity of a resonance circuit improves as the resistance decreases, but the inherent resistance of the wires and the inductor coil keeps R from being 0 Ω.

EXAMPLE 32.6 Designing a radio receiver

An AM radio antenna picks up a 1000 kHz signal with a peak voltage of 5.0 mV. The tuning circuit consists of a 60 μH inductor in series with a variable capacitor. The inductor coil has a resistance of 0.25 Ω, and the resistance of the rest of the circuit is negligible.

  1. To what value should the capacitor be tuned to listen to this radio station?

  2. What is the peak current through the circuit at resonance?

  3. A stronger station at 1050 kHz produces a 10 mV antenna signal. What is the current at this frequency when the radio is tuned to 1000 kHz?

MODEL The inductor’s 0.25 Ω resistance can be modeled as a resistance in series with the inductance, hence we have a series RLC circuit. The antenna signal at ω=2π×1000 kHz is the emf.

VISUALIZE The circuit looks like Figure 32.16.

SOLVE

  1. The capacitor needs to be tuned to where it and the inductor are resonant at ω0=2π×1000 kHz. The appropriate value is

    C=1Lω02=1(60×10−6 H)(6.28×106 rad/s)2=4.2×10−10 F=420 pF
  2. XL=XC at resonance, so the peak current is

    I=ε​0R=5.0×10−3 V0.25 Ω=0.020 A=20 mA
  3. The 1050 kHz signal is “off resonance,” so we need to compute XL=ωL=396 Ω and XC=1/ωC=361 Ω at ω=2π×1050 kHz. The peak voltage of this signal is ε0=10 mV. With these values, Equation 32.24 for the peak current is

    I=ε​0R2+(XL−XC )2=0.28 mA

REVIEW These are realistic values for the input stage of an AM radio. You can see that the signal from the 1050 kHz station is strongly suppressed when the radio is tuned to 1000 kHz.