In a series LCR resonant circuit,the quality | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(NONE) The quality factor $Q$ of a series $LCR$ circuit is given by the formula: $Q = \frac{1}{R} \sqrt{\frac{L}{C}}$.Initially,$Q = 100$.When the ind

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(NONE) The quality factor $Q$ of a series $LCR$ circuit is given by the formula: $Q = \frac{1}{R} \sqrt{\frac{L}{C}}$.Initially,$Q = 100$.When the inductance $L$ is increased by two fold $(L' = 2L)$ and the resistance $R$ is decreased by two fold $(R' = R/2)$,the new quality factor $Q'$ becomes:$Q' = \frac{1}{R'} \sqrt{\frac{L'}{C}} = \frac{1}{(R/2)} \sqrt{\frac{2L}{C}} = 2 	imes \sqrt{2} 	imes \left( \frac{1}{R} \sqrt{\frac{L}{C}} ight)$.Substituting the initial value of $Q$:$Q' = 2 \sqrt{2} 	imes Q = 2 	imes 1.414 	imes 100 = 282.8$.

In a series $LCR$ resonant circuit,the quality factor is measured as $100$. If the inductance is increased by two fold and resistance is decreased by two fold,then the quality factor after this change will be .........

Similar Questions

In a series $LCR$ circuit,the frequency of the source is half of the resonance frequency. The nature of the circuit will be:

An inductance of $1\, mH$,a capacitor of $10\, \mu F$,and a resistance of $50\, \Omega$ are connected in series. The reactances of the inductor and the capacitor are the same. The reactance of either of them will be........$\Omega$.

For an $R-L-C$ circuit,driven with voltage of amplitude $V_m$ and frequency $\omega_0 = \frac{1}{\sqrt{LC}}$,the current exhibits resonance. The quality factor $Q$ is

$A$ resistor $R$,an inductor $L$,and a capacitor $C$ are connected in series to an oscillator of frequency $n$. If the resonant frequency is $n_r$,then the current lags behind the voltage when:

The frequency at resonance for the circuit shown in the figure is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module