In a series LCR circuit,a resistor of 300 Om | vedclass

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Question

(A) For maximum current in a series $LCR$ circuit,the circuit must be in resonance.At resonance,the angular frequency $\omega$ is given by the formula

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) For maximum current in a series $LCR$ circuit,the circuit must be in resonance.At resonance,the angular frequency $\omega$ is given by the formula $\omega = \frac{1}{\sqrt{LC}}$.Given: $L = 100 \ 	ext{mH} = 100 	imes 10^{-3} \ 	ext{H} = 0.1 \ 	ext{H}$ and $C = 25 \ 	ext{nF} = 25 	imes 10^{-9} \ 	ext{F}$.Substituting the values:$\omega = \frac{1}{\sqrt{0.1 	imes 25 	imes 10^{-9}}}$$\omega = \frac{1}{\sqrt{25 	imes 10^{-10}}}$$\omega = \frac{1}{5 	imes 10^{-5}}$$\omega = 0.2 	imes 10^5 \ 	ext{rad s}^{-1} = 2 	imes 10^4 \ 	ext{rad s}^{-1}$.Thus,the value is $2$.

In a series $LCR$ circuit,a resistor of $300 \ \Omega$,a capacitor of $25 \ \text{nF}$ and an inductor of $100 \ \text{mH}$ are used. For maximum current in the circuit,the angular frequency of the ac source is $. . . . \times 10^4 \ \text{rad s}^{-1}$.

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