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(C) At resonance,the inductive reactance $(X_L)$ and capacitive reactance $(X_C)$ are equal:$X_L = X_C$$\omega L = \frac{1}{\omega C}$$\omega^2 = \fra

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$\frac{250}{\pi} Hz$

Vedclass · Answer

(C) At resonance,the inductive reactance $(X_L)$ and capacitive reactance $(X_C)$ are equal:$X_L = X_C$$\omega L = \frac{1}{\omega C}$$\omega^2 = \frac{1}{LC}$$\omega = \frac{1}{\sqrt{LC}}$Since $\omega = 2 \pi f$,the resonant frequency $f$ is given by:$f = \frac{1}{2 \pi \sqrt{LC}}$Given values: $L = 0.16 H$ and $C = 25 	imes 10^{-6} F$.Substituting these values:$f = \frac{1}{2 \pi \sqrt{0.16 	imes 25 	imes 10^{-6}}}$$f = \frac{1}{2 \pi \sqrt{4 	imes 10^{-6}}}$$f = \frac{1}{2 \pi 	imes 2 	imes 10^{-3}}$$f = \frac{1}{4 \pi 	imes 10^{-3}}$$f = \frac{1000}{4 \pi} Hz = \frac{250}{\pi} Hz$.

$A$ circuit of negligible resistance has an inductor of $0.16 H$ and a capacitor of $25 \mu F$ connected in series with an alternating voltage source. The resonant frequency of the circuit is:

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