Derive an expression for resonant frequency of series resonant circuit.
Q.16 Derive an expression for resonant frequency of series resonant circuit.
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Answer: \(f_r = \dfrac{1}{2\pi\sqrt{LC}}\)
In a series LCR circuit connected to an a.c. source \(e = e_0\sin\omega t\), the impedance is
\[Z = \sqrt{R^2 + (X_L - X_C)^2},\qquad X_L = \omega L,\ X_C = \frac{1}{\omega C}\]and the peak current is \(i_0 = \dfrac{e_0}{Z}\). As the frequency is changed, \(X_L\) increases and \(X_C\) decreases. At one particular frequency they become equal; then \(Z = R\) is minimum and the current is maximum. This condition is called resonance and the circuit is a series resonant (acceptor) circuit.
At resonance, \(X_L = X_C\):
\[\omega_r L = \frac{1}{\omega_r C} \Rightarrow \omega_r^2 = \frac{1}{LC} \Rightarrow \omega_r = \frac{1}{\sqrt{LC}}\]Since \(\omega_r = 2\pi f_r\),
\[\boxed{f_r = \frac{1}{2\pi\sqrt{LC}}}\]At resonance the voltage and current are in phase (\(\phi = 0\)) and the power factor is 1.
Doubtly AI
· 27 Oct 2026
AI-verified · Oct 2026
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