Lesson Explainer: Resonance in Alternating Current Circuits Physics

In this explainer, we will learn how to calculate the resonant frequency of simple resistive-capacitive-inductive circuits.

A circuit containing a resistor (R), inductor (L), and capacitor (C) connected to an alternating potential difference source is shown in the following figure.

The alternating potential difference source alternates a particular frequency, 𝑓.

The impedance of the circuit is affected by the frequency of the alternating potential difference. Increasing the frequency of the alternating potential difference increases the inductive reactance and decreases the capacitive reactance (but not equally). The impedance of a circuit depends on the absolute difference between the inductive and capacitive reactance of the circuit.

The impedance of a series alternating current circuit with inductance and capacitance is given by a formula.

Formula: Impedance

The impedance, 𝑍, of a circuit is given by 𝑍=𝑅+(π‘‹βˆ’π‘‹), where 𝑅 is the resistance of the circuit, π‘‹οŒ« is the inductive reactance of the circuit, and π‘‹οŒ’ is the capacitive reactance of the circuit.

The resonant frequency of a circuit is the frequency of an applied alternating potential difference that generates the greatest current in the circuit.

Alternating potential difference frequencies near the resonant frequency generate currents close to the magnitude of the resonant frequency current, while frequencies further from the resonant frequency generate lesser-magnitude currents.

The following figure shows how the maximum current in a circuit varies with frequency.

From the formula for impedance, 𝑍=𝑅+(π‘‹βˆ’π‘‹), we can see that when π‘‹βˆ’π‘‹=0, it must be the case that 𝑍=𝑅, which corresponds to the minimum impedance of the circuit. The current is greatest for this impedance.

The value of capacitive reactance is given by a formula.

Formula: Capacitive Reactance

The capacitive reactance, π‘‹οŒ’, of a circuit with capacitance 𝐢 that is carrying an alternating current with a frequency 𝑓 is given by 𝑋=12πœ‹π‘“πΆ.

The value of inductive reactance is given by a formula.

Formula: Inductive Reactance

The inductive reactance, π‘‹οŒ«, of a circuit with inductance 𝐿 that is carrying an alternating current with a frequency 𝑓 is given by 𝑋=2πœ‹π‘“πΏ.

For minimum impedance, it must be the case that 2πœ‹π‘“πΏβˆ’12πœ‹π‘“πΆ=0.

This equation can be rearranged as follows: 2πœ‹π‘“πΏ=12πœ‹π‘“πΆ2πœ‹π‘“=12πœ‹π‘“πΏπΆ(2πœ‹π‘“)=1𝐿𝐢2πœ‹π‘“=ο„ž1𝐿𝐢, where 𝑓 is the resonant frequency of the circuit.

We hence obtain the formula for resonant frequency.

Formula: Resonant Frequency

The resonant frequency, 𝑓, of a circuit with inductance 𝐿 and capacitance 𝐢 is given by 2πœ‹π‘“=ο„ž1𝐿𝐢.

Let us now look at some examples.

Example 1: Determining the Resonant Frequency of a Circuit

What is the resonant frequency of the circuit shown in the diagram? Give your answer to one decimal place.

Answer

The resonant frequency, 𝑓, is given by 2πœ‹π‘“=ο„ž1𝐿𝐢.

Substituting the values given in the question, we obtain 2πœ‹π‘“=ο„ž17.5Γ—3.5Γ—10.HFοŠͺ

To one decimal place, 𝑓 is 3.1 Hz.

Example 2: Determining the Peak Current in a Circuit Oscillating at Its Resonant Frequency

A circuit containing a resistor, a capacitor, and an inductor in series has a resonant frequency of 372 Hz. The resistor has a resistance of 440 Ξ© and the capacitor has a capacitance of 112 mF. The peak voltage across the circuit is 28 V. What is the peak current in the circuit when an alternating current in the circuit has a frequency of 372 Hz? Give your answer to two decimal places.

Answer

The resonant frequency of the circuit is stated to be 372 Hz.

The question requires the peak current to be determined.

The peak applied potential difference is stated to be 28 V. To determine the peak current, the impedance of the circuit must be determined.

The impedance, 𝑍, of a circuit is given by 𝑍=𝑅+(π‘‹βˆ’π‘‹), where 𝑅 is the resistance of the circuit, π‘‹οŒ« is the inductive reactance of the circuit, and π‘‹οŒ’ is the capacitive reactance of the circuit.

The values of π‘‹οŒ’ and π‘‹οŒ« are not stated. We can find these values from the formulas for the capacitive and inductive reactance of a circuit, 𝑋=12πœ‹π‘“πΆ,𝑋=2πœ‹π‘“πΏ, and the resonant frequency of the circuit, 2πœ‹π‘“=ο„ž1𝐿𝐢.

However, as the circuit is oscillating at its resonant frequency, it must be the case that π‘‹βˆ’π‘‹=0 and, hence, that 𝑍=𝑅.

To find the greatest current, we need only divide the greatest potential difference across it by 𝑅: 𝐼=28440.VΞ©

To two decimal places, 𝐼 is 0.06 A.

Example 3: Determining the Capacitance of a Circuit Oscillating at Its Resonant Frequency

A circuit containing a capacitor and an inductor in series has a resonant frequency of 575 kHz. The inductor in the circuit has an inductance of 1.25 H. What is the capacitance of the capacitor? Give your answer in scientific notation to two decimal places.

Answer

The capacitance of the capacitor is related to the resonant frequency of the circuit, which is given by 2πœ‹π‘“=ο„ž1𝐿𝐢, where 𝐿 is stated to be 1.25 H and 𝑓 is stated to be 575 kHz, which is 5.75Γ—10 Hz.

The formula must be rearranged to make 𝐢 the subject, as follows: (2πœ‹π‘“)=1𝐿𝐢𝐢=1(2πœ‹π‘“)𝐿.

Substituting the known values, we obtain 𝐢=1(2πœ‹Γ—5.75Γ—10)Γ—1.25.HzH

In scientific notation, to two decimal places, 𝐢 is 6.13Γ—10οŠͺ F.

Example 4: Determining the Inductive Reactance of a Circuit Oscillating at Its Resonant Frequency

A circuit containing a capacitor and an inductor in series has a resonant frequency of 155 kHz. The capacitor in the circuit has a capacitance of 215 Β΅F. What is the inductive reactance of the circuit? Give your answer in scientific notation to two decimal places.

Answer

The inductive reactance of the circuit is given by 𝑋=2πœ‹π‘“πΏ, where 𝑓 is stated to be 155 kHz, which is 1.55Γ—10 Hz, and 𝐿 is the inductance of the inductor.

The value of 𝐿 is not stated. It can, however, be found using the resonant frequency of the circuit, which is given by 2πœ‹π‘“=ο„ž1𝐿𝐢, where 𝐢 is stated to be 215 Β΅F, which is 2.15Γ—10οŠͺ F.

The formula for resonant frequency must be rearranged to make 𝐿 the subject, as follows: (2πœ‹π‘“)=1𝐿𝐢𝐿=1(2πœ‹π‘“)𝐢.

Substituting known values, we obtain 𝐿=1(2πœ‹Γ—1.55Γ—10)Γ—2.15Γ—10πΏβ‰ˆ4.9Γ—10.οŠͺHzFH

We can substitute this value into 𝑋=2πœ‹π‘“πΏ.

In scientific notation, to two decimal places, π‘‹οŒ« is 4.78Γ—10 Ξ©.

This question can also be solved by recalling that at the resonant frequency of a circuit, the capacitive and inductive reactances are equal, and so the capacitive reactance could have been calculated instead of the inductive reactance. The capacitive reactance can be determined simply from the frequency and capacitance of the circuit, by rearrangement of the formula 𝑋=12πœ‹π‘“πΆ, as the values of 𝑓 and 𝐢 are both given in the question.

Let us now summarize what has been learned in this explainer.

Key Points

  • The resonant frequency of an alternating current series circuit that has inductance and capacitance is the frequency at which an applied alternating potential difference generates the greatest current.
  • At the resonant frequency of an alternating current circuit, the impedance of the circuit equals the resistance of the circuit.
  • The resonant frequency, 𝑓, of an alternating current circuit is given by 2πœ‹π‘“=ο„ž1𝐿𝐢, where 𝐿 is the inductance of the circuit and 𝐢 is the capacitance of the circuit.

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