What is the resonance frequency of ac circuit?

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Examoogle · Accepted Answer

1/√LC Explanation: The resonance frequency of an RLC circuit is the specific frequency at which the inductive reactance equals the capacitive reactance. At this point, the impedance of the circuit is purely resistive and reaches its minimum value. Working Principle: In a series RLC circuit, resonance occurs when $X_L = X_C$, where inductive reactance $X_L = 2\pi f L$ and capacitive reactance $X_C = \frac{1}{2\pi f C}$. By equating these two, we solve for frequency $f$: $2\pi f L = \frac{1}{2\pi f C}$, which simplifies to $f^2 = \frac{1}{4\pi^2 LC}$. Taking the square root, we get the resonant frequency $f = \frac{1}{2\pi\sqrt{LC}}$. Note that the provided option refers to angular frequency $\omega_0$ in some contexts, but specifically denotes the frequency constant for tuning. Key Formulas: $f_0 = \frac{1}{2\pi\sqrt{LC}}$ — Resonant frequency in Hertz; $\omega_0 = \frac{1}{\sqrt{LC}}$ — Angular resonant frequency in rad/s Key Points: At resonance, the circuit current is maximum because impedance is minimum ($Z = R$).; The power factor of an RLC series circuit at resonance is unity.; The phase angle between voltage and current is zero at resonance. Advantages: Used in tuning circuits for radio and TV receivers.; Allows selection of a specific frequency while rejecting others. Disadvantages: Can lead to high voltages across L and C (magnification) which may damage components.; Requires high-quality factor (Q) components for sharp frequency selection. Applications: Radio frequency (RF) tuning circuits.; Oscillators and filter design. Additional Info: The option provided in the question, 1/√LC, represents the angular resonant frequency $\omega_0$ if the $2\pi$ factor is omitted, which is common in simplified engineering notation.. Option C, $\sqrt{LC}$, is the reciprocal of the angular frequency. Answer: A is correct — The resonance frequency is defined by the condition where inductive and capacitive reactances cancel each other out, resulting in the expression $f = \frac{1}{2\pi\sqrt{LC}}$. Exam Tip: Always verify if the question asks for frequency in Hertz ($f$) or angular frequency in radians/second ($\omega$); the $2\pi$ factor is the most common trap in resonance problems.

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