15.5 Resonance in an AC Circuit

In the RLC series circuit of Figure 15.11, the current amplitude is, from Equation 15.10,

$I_{0} = \frac{V_{0}}{\sqrt{R^{2} + {(ω L - 1 / ω C)}^{2}}} .$

(15.15)

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 $I sub 0$ versus $ω$ is shown in Figure 15.17.

Function graph showing y = 100/sqrt(16 + (0.3*x - 10000/(x*Cu))^2) on x in [0.4, 48] and the parametric curve (6.455, t) for t in [0, 27]. Adjustable parameter: Tuning capacitance C (the worked example uses 800 μF) (Cu) = 800 μF. Viewing window: x from -0.01 to 48.01, y from -1.35 to 28.35. — The current amplitude this section plots, I₀ = V₀/√(R² + (ωL − 1/ωC)²), for the circuit the example Resonance in an RLC Series Circuit uses: V₀ = 0.100 V, R = 4.00 Ω, L = 3.00 × 10⁻³ H and C = 8.00 × 10⁻⁴ F. The horizontal axis is the driving angular frequency in hundreds of rad/s and the vertical axis is the current amplitude in milliamperes. At the default the peak sits on the dashed line at ω₀ = 1/√(LC) = 645 rad/s — the example's f₀ = 1.03 × 10² Hz — and reaches I₀ = V₀/R = 25.0 mA, because at resonance the two reactances cancel and the circuit is purely resistive. The slider is the tuning capacitor, and it does exactly what the section says tuning does: the peak slides along the frequency axis as 1/√(LC), from 408 rad/s at 2000 μF out to 2582 rad/s at 50 μF, while its height never changes. That is the point worth taking away — C picks the frequency the circuit selects, R alone fixes how much current it passes there, and the dashed line stays behind to show how far you have tuned away from the example.

Figure shows a graph of I0 versus omega. The curve ascends gradually, has one blunt peak at the centre and then gradually descends to its original value. The y-value at the peak is V0 by R and the x-value is omega 0. — **Figure 15.17** At an *RLC* circuit’s resonant frequency, $ω_{0} = \sqrt{1 / L C},$ the current amplitude is at its maximum value.

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 Forced Oscillations). This similarity is more than just a coincidence, as shown earlier by the application of Kirchhoff’s loop rule to the circuit of Figure 15.11. This yields

$L \frac{d i}{d t} + i R + \frac{q}{C} = V_{0} sin ω t,$

(15.16)

$L \frac{d^{2} q}{d t^{2}} + R \frac{d q}{d t} + \frac{1}{C} q = V_{0} sin ω t,$

where we substituted dq(t)/dt for i(t). A comparison of Equation 15.16 and, from Oscillations, Damped Oscillations for damped harmonic motion clearly demonstrates that the driven RLC series circuit is the electrical analog of the driven damped harmonic oscillator.

The resonant frequency $f sub 0$ 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 $ω sub 0 equals 2 π f sub 0$ at which the impedance Z in Equation 15.15 is a minimum, or when

$ω_{0} L = \frac{1}{ω_{0} C}$

and

$ω_{0} = \sqrt{\frac{1}{L C}} .$

(15.17)

This is the resonant angular frequency of the circuit. Substituting $ω sub 0$ into Equation 15.9, Equation 15.10, and Equation 15.11, we find that at resonance,

$ϕ = {tan}^{−1} (0) = 0, I_{0} = V_{0} / R, and Z = R .$

Therefore, at resonance, an RLC circuit is purely resistive, with the applied emf and current in phase.

What happens to the power at resonance? Equation 15.14 tells us how the average power transferred from an ac generator to the RLC combination varies with frequency. In addition, $P_{ave}$ reaches a maximum when Z, which depends on the frequency, is a minimum, that is, when $X_{L} = X_{C} and Z = R .$ Thus, at resonance, the average power output of the source in an RLC series circuit is a maximum. From Equation 15.14, this maximum is $V rms 2 / R.$

Figure 15.18 is a typical plot of $P_{ave}$ versus $ω$ in the region of maximum power output. The bandwidth $Δ ω$ of the resonance peak is defined as the range of angular frequencies $ω$ over which the average power $P_{ave}$ is greater than one-half the maximum value of $P_{ave} .$ The sharpness of the peak is described by a dimensionless quantity known as the quality factor Q of the circuit. By definition,

$Q = \frac{ω_{0}}{Δ ω},$

(15.18)

where $ω sub 0$ is the resonant angular frequency. A high Q indicates a sharp resonance peak. We can give Q in terms of the circuit parameters as

$Q = \frac{ω_{0} L}{R} .$

(15.19)

Figure shows a graph of P bar versus omega. The curve ascends gradually, has one blunt peak at the centre and then gradually descends to its original value. The y-value at the peak is V squared subscript rms by R and the x-value is omega 0. The y value near the middle of the curve is V squared subscript rms by 2R. The width of the curve near the middle is labeled delta omega. — **Figure 15.18** 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 different 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 15.19).

Photograph of an underwater diver using a metal detector. — **Figure 15.19** 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.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. (credit: modification of work by Eric Lippmann, U.S. Navy)

Example 1 Resonance in an RLC Series Circuit

(a) What is the resonant frequency of a circuit using the voltage and LRC values all wired in series from Example 1 in Section 15.3? (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?

Strategy The resonant frequency for a RLC circuit is calculated from Equation 15.17, 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

The resonant frequency is found from Equation 15.17:
$\begin{matrix} f_{0} & = \frac{1}{2 π} \sqrt{\frac{1}{L C}} = \frac{1}{2 π} \sqrt{\frac{1}{(3.00 \times 10^{−3} H) (8.00 \times 10^{−4} F)}} \\ = 1.03 \times 10^{2} Hz . \end{matrix}$
At resonance, the impedance of the circuit is purely resistive, and the current amplitude is
$I_{0} = \frac{0.100 V}{4.00 Ω} = 2.50 \times 10^{−2} A .$

Significance 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 RLC circuit with $R equals 0.200 Ω,$ $L = 4.00 \times 10^{−3} H,$ and $C = 2.00 \times 10^{−6} F?$ (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 Equation 15.17. The average power is calculated from the rms voltage and the resistance in the circuit. The quality factor is calculated from Equation 15.19 and by knowing the resonant frequency. The bandwidth is calculated from Equation 15.18 and by knowing the quality factor.

Solution

The resonant angular frequency is
$\begin{matrix} ω_{0} & = \sqrt{\frac{1}{L C}} = \sqrt{\frac{1}{(4.00 \times 10^{−3} H) (2.00 \times 10^{−6} F)}} \\ = 1.12 \times 10^{4} rad/s . \end{matrix}$
At this frequency, the average power transferred to the circuit is a maximum. It is
$P_{ave} = \frac{V_{rms}^{2}}{R} = \frac{{[(1 / \sqrt{2}) (4.00 V)]}^{2}}{0.200 Ω} = 40.0 W .$
The quality factor of the circuit is
$Q = \frac{ω_{0} L}{R} = \frac{(1.12 \times 10^{4} rad/s) (4.00 \times 10^{−3} H)}{0.200 Ω} = 224 .$

We then find for the bandwidth

$Δ ω = \frac{ω_{0}}{Q} = \frac{1.12 \times 10^{4} rad/s}{224} = 50.0 rad/s .$

Significance If a narrower bandwidth is desired, a lower resistance or higher inductance would help. However, 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.

Summary

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.
The bandwidth is related to a dimensionless quantity called the quality factor. A high quality factor value is a sharp or narrow peak.

Problems

(a) Calculate the resonant angular frequency of an RLC series circuit for which $R equals 20 Ω$ , $L equals 7 5 mH$ , and $C equals 4.0 μ F.$ (b) If R is changed to $300 Ω,$ what happens to the resonant angular frequency?

The resonant frequency of an RLC series circuit is $2.0 times 10 cubed Hz.$ If the self-inductance in the circuit is 5.0 mH, what is the capacitance in the circuit?

$1.3 \times 10^{−6} F$

(a) What is the resonant frequency of an RLC series circuit with $R equals 20 Ω$ , $L equals 2.0 mH$ , and $C equals 4.0 μ F$ ? (b) What is the impedance of the circuit at resonance?

For an RLC series circuit, $R equals 100 Ω$ , $L equals 150 mH$ , and $C equals 0.25 μ F.$ (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?

a. 820 Hz; b. 7.8

An ac source of voltage amplitude 100 V and variable frequency f drives an RLC series circuit with $R equals 10 Ω$ , $L equals 2.0 mH$ , and $C equals 25 μ F.$ (a) Plot the current through the resistor as a function of the frequency f. (b) Use the plot to determine the resonant frequency of the circuit.

(a) What is the resonant frequency of a resistor, capacitor, and inductor connected in series if $R equals 100 Ω,$ $L equals 2.0 H$ , and $C equals 5.0 μ F$ ? (b) If this combination is connected to a 100-V source operating at the resonant 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?

a. 50 Hz; b. 50 W; c. 6.32; d. 50 rad/s

Suppose a coil has a self-inductance of 20.0 H and a resistance of $200 Ω$ . 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 Q of 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?

Figure shows an AC source connected to a box labeled Z. The source is 170V, cos 120 pi t. The current through the circuit is 0.5 Amp, cos parentheses 120 pi t plus pi by 4 parentheses.

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.