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Alternating Current - Resonance and Q-factor

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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In a series LCRLCR circuit, resonance occurs when the inductive reactance XLX_L is equal to the capacitive reactance XCX_C, leading to the condition XL=XCX_L = X_C.

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At the resonant frequency ω0\omega_0, the impedance ZZ of the circuit reaches its minimum value, which is equal to the resistance (Z=RZ = R).

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Since the impedance is minimum at resonance, the current amplitude I0I_0 in the circuit reaches its maximum value: Imax=V0RI_{max} = \frac{V_0}{R}.

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The phase difference ϕ\phi between the voltage and the current at resonance is zero, meaning the power factor cos⁡ϕ=1\cos \phi = 1. The circuit behaves as a purely resistive circuit.

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The Quality Factor (QQ-factor) is a dimensionless parameter that describes how 'sharp' the resonance is. A higher QQ indicates a narrower bandwidth and higher selectivity.

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The Bandwidth is defined as the difference between the two frequencies ω1\omega_1 and ω2\omega_2 at which the power dissipated in the circuit is half the maximum power (where current falls to 12\frac{1}{\sqrt{2}} of its maximum value).

📐Formulae

ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}}

f0=12πLCf_0 = \frac{1}{2\pi\sqrt{LC}}

Z=R2+(XL−XC)2Z = \sqrt{R^2 + (X_L - X_C)^2}

Q=ω0LR=1ω0CRQ = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 CR}

Q=1RLCQ = \frac{1}{R} \sqrt{\frac{L}{C}}

Δω=ω2−ω1=RL\Delta \omega = \omega_2 - \omega_1 = \frac{R}{L}

Q=ω0ΔωQ = \frac{\omega_0}{\Delta \omega}

💡Examples

Problem 1:

A series LCRLCR circuit with L=0.12 HL = 0.12 \, H, C=480 nFC = 480 \, nF, and R=23 ΩR = 23 \, \Omega is connected to a 230 V230 \, V variable frequency supply. (a) What is the source frequency for which current amplitude is maximum? (b) What is the QQ-factor of this circuit?

Solution:

(a) Maximum current occurs at resonance. The resonant frequency is ω0=1LC=10.12×480×10−9=15.76×10−8=12.4×10−4≈4166.67 rad/s\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.12 \times 480 \times 10^{-9}}} = \frac{1}{\sqrt{5.76 \times 10^{-8}}} = \frac{1}{2.4 \times 10^{-4}} \approx 4166.67 \, rad/s. To find frequency in HzHz: f0=ω02π≈663 Hzf_0 = \frac{\omega_0}{2\pi} \approx 663 \, Hz. (b) Q=ω0LR=4166.67×0.1223≈21.74Q = \frac{\omega_0 L}{R} = \frac{4166.67 \times 0.12}{23} \approx 21.74.

Explanation:

We used the resonance condition ω0=1/LC\omega_0 = 1/\sqrt{LC} for the peak current and applied the standard definition of QQ-factor using the calculated resonant frequency.

Problem 2:

For an LCRLCR circuit, R=20 ΩR = 20 \, \Omega, L=1.5 HL = 1.5 \, H, and C=35 μFC = 35 \, \mu F. If the circuit is in resonance, calculate the power factor and the bandwidth.

Solution:

At resonance, the phase angle ϕ=0\phi = 0, so the power factor is cos⁡ϕ=cos⁡(0)=1\cos \phi = \cos(0) = 1. The bandwidth is given by Δω=RL=201.5≈13.33 rad/s\Delta \omega = \frac{R}{L} = \frac{20}{1.5} \approx 13.33 \, rad/s.

Explanation:

At resonance, XL=XCX_L = X_C, making the circuit purely resistive, which yields a power factor of 11. The bandwidth is determined solely by the ratio of resistance to inductance.