An LCR circuit contains R = 50 , Omega,L = 1 | vedclass

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Question

(A) The impedance of an $LCR$ circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.For the impedance to be minimum,the circuit must be in resonance,wh

Vedclass · Accepted Answer

$\frac{10^5}{2\pi} \, 	ext{s}^{-1}$

Vedclass · Answer

(A) The impedance of an $LCR$ circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.For the impedance to be minimum,the circuit must be in resonance,where $X_L = X_C$.The resonant frequency $
u_0$ is given by the formula $
u_0 = \frac{1}{2\pi \sqrt{LC}}$.Given $L = 1 \, 	ext{mH} = 10^{-3} \, 	ext{H}$ and $C = 0.1 \, \mu	ext{F} = 0.1 	imes 10^{-6} \, 	ext{F} = 10^{-7} \, 	ext{F}$.Substituting the values: $
u_0 = \frac{1}{2\pi \sqrt{10^{-3} 	imes 10^{-7}}} = \frac{1}{2\pi \sqrt{10^{-10}}}$.$
u_0 = \frac{1}{2\pi 	imes 10^{-5}} = \frac{10^5}{2\pi} \, 	ext{Hz}$ (or $	ext{s}^{-1}$).Thus,the correct option is $A$.

An $LCR$ circuit contains $R = 50 \, \Omega$,$L = 1 \, \text{mH}$,and $C = 0.1 \, \mu\text{F}$. The impedance of the circuit will be minimum for a frequency of:

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