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  5. Resonance in an ac circuit

15.5 Resonance in an ac circuit  (Page 2/5)

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Photograph of an underwater diver using a metal detector.
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.

Resonance in an RLC Series circuit

(a) What is the resonant frequency of the circuit of [link] ? (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 [link] , 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

  1. The resonant frequency is found from [link] :
    f 0 = 1 2 π 1 L C = 1 2 π 1 ( 3.00 × 10 −3 H ) ( 8.00 × 10 −4 F ) = 1.03 × 10 2 Hz .
  2. At resonance, the impedance of the circuit is purely resistive, and the current amplitude is
    I 0 = 0.100 V 4.00 Ω = 2.50 × 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.

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Power transfer in an RLC Series circuit at resonance

(a) What is the resonant angular frequency of an RLC circuit with R = 0.200 Ω , L = 4.00 × 10 −3 H, and C = 2.00 × 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 [link] . The average power is calculated from the rms voltage and the resistance in the circuit. The quality factor is calculated from [link] and by knowing the resonant frequency. The bandwidth is calculated from [link] and by knowing the quality factor.

Solution

  1. The resonant angular frequency is
    ω 0 = 1 L C = 1 ( 4.00 × 10 −3 H ) ( 2.00 × 10 −6 F ) = 1.12 × 10 4 rad/s .
  2. At this frequency, the average power transferred to the circuit is a maximum. It is
    P ave = V rms 2 R = [ ( 1 / 2 ) ( 4.00 V ) ] 2 0.200 Ω = 40.0 W .
  3. The quality factor of the circuit is
    Q = ω 0 L R = ( 1.12 × 10 4 rad/s ) ( 4.00 × 10 −3 H ) 0.200 Ω = 224 .

We then find for the bandwidth

Δ ω = ω 0 Q = 1.12 × 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.

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Check Your Understanding In the circuit of [link] , L = 2.0 × 10 −3 H, C = 5.0 × 10 −4 F, and R = 40 Ω . (a) What is the resonant frequency? (b) What is the impedance of the circuit at resonance? (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 difference between the current and the emf of the generator.

a. 160 Hz; b. 40 Ω ; c. ( 0.25 A ) sin 10 3 t ; d. 0.023 rad

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Check Your Understanding What happens to the resonant frequency of an RLC series circuit when the following quantities are increased by a factor of 4: (a) the capacitance, (b) the self-inductance, and (c) the resistance?

a. halved; b. halved; c. same

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Check Your Understanding The resonant angular frequency of an RLC series circuit is 4.0 × 10 2 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 0.50 Ω . Write an expression for the emf of the source.

v ( t ) = ( 0.14 V ) sin ( 4.0 × 10 2 t )

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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 = 20 Ω , L = 7 5 mH , and C = 4.0 μ F . (b) If R is changed to 300 Ω , what happens to the resonant angular frequency?

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The resonant frequency of an RLC series circuit is 2.0 × 10 3 Hz . If the self-inductance in the circuit is 5.0 mH, what is the capacitance in the circuit?

1.3 × 10 −7 F

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

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For an RLC series circuit, R = 100 Ω , L = 150 mH , and C = 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

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An ac source of voltage amplitude 100 V and variable frequency f drives an RLC series circuit with R = 10 Ω , L = 2.0 mH , and C = 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.

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(a) What is the resonant frequency of a resistor, capacitor, and inductor connected in series if R = 100 Ω, L = 2.0 H , and C = 5.0 μ F ? (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?

a. 50 Hz; b. 50 W; c. 13; d. 25 rad/s

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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?

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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.

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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