(A) The bandwidth $(BW)$ of a series $RLC$ circuit is given by $BW = \frac{f_r}{Q}$,where $Q = \frac{X_L}{R}$.Substituting the values: $Q = \frac{300

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$11900 \text{ Hz}, 12100 \text{ Hz}$

(A) The bandwidth $(BW)$ of a series $RLC$ circuit is given by $BW = \frac{f_r}{Q}$,where $Q = \frac{X_L}{R}$.Substituting the values: $Q = \frac{300 \text{ } \Omega}{5 \text{ } \Omega} = 60$.Now,$BW = \frac{12000 \text{ Hz}}{60} = 200 \text{ Hz}$.The half-power frequencies are $f_1 = f_r - \frac{BW}{2}$ and $f_2 = f_r + \frac{BW}{2}$.$f_1 = 12000 - 100 = 11900 \text{ Hz}$.$f_2 = 12000 + 100 = 12100 \text{ Hz}$.

Class 12 Physics Alternating Current Half Power Frequency , Quality Factor ,Resonance in AC Circuit

Difficult

In a series $RLC$ circuit,the resonance frequency is $12 \text{ kHz}$. If $R = 5 \text{ } \Omega$ and $X_L$ at resonance is $300 \text{ } \Omega$,the half-power frequencies will be:

A
$11900 \text{ Hz}, 12100 \text{ Hz}$
B
$9000 \text{ Hz}, 10000 \text{ Hz}$
C
$200 \text{ Hz}, 210 \text{ Hz}$
D
None of the above

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Half Power Frequency , Quality Factor ,Resonance in AC Circuit

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In the circuit shown, the ratio of the quality factor $(Q)$ and the band width $(BW)$ is (Given: band width = $R/L$)

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The power in an $AC$ circuit is given by $P = E_{rms}I_{rms} \cos \phi$. The value of $\cos \phi$ in a series $LCR$ circuit at resonance is:

Medium

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$A$ resistor of $100 \Omega$, an inductor of $\frac{25}{\pi^2} \text{ mH}$ and a capacitor of $0.1 \mu\text{F}$ are connected in series to an $AC$ source. The impedance of the circuit is minimum for a frequency of (in $\text{ kHz}$)

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$A$ series $LCR$ circuit is designed to resonate at an angular frequency $\omega_{0} = 10^{5} \, rad/s$. The circuit draws $16 \, W$ power from a $120 \, V$ source at resonance. The value of resistance $R$ in the circuit is ...... $\Omega$.

MediumJEE MAIN 2021

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In a series $RLC$ circuit,the resonance frequency is $12 \text{ kHz}$. If $R = 5 \text{ } \Omega$ and $X_L$ at resonance is $300 \text{ } \Omega$,the half-power frequencies will be:

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