A 10 Omega resistance,5 mH coil,and 10 mu F | vedclass

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(D) At resonance in an $LCR$ series circuit,the inductive reactance $(X_L)$ is equal to the capacitive reactance $(X_C)$.$X_L = X_C$$2 \pi f L = \frac

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remains unchanged.

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(D) At resonance in an $LCR$ series circuit,the inductive reactance $(X_L)$ is equal to the capacitive reactance $(X_C)$.$X_L = X_C$$2 \pi f L = \frac{1}{2 \pi f C}$Rearranging for the resonant frequency $(f)$:$f^2 = \frac{1}{4 \pi^2 LC}$$f = \frac{1}{2 \pi \sqrt{LC}}$From this formula,it is clear that the resonant frequency $(f)$ depends only on the inductance $(L)$ and the capacitance $(C)$.It is independent of the resistance $(R)$.Therefore,if the resistance is halved,the resonant frequency remains unchanged.

$A$ $10 \Omega$ resistance,$5 \ mH$ coil,and $10 \ \mu F$ capacitor are joined in series. When an alternating current source of suitable frequency is joined to this combination,the circuit resonates. If the resistance is halved,the resonant frequency is

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