Figure (a) shows the plot of voltage across t | vedclass

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

Question

(C) The quality factor $Q$ is defined as $Q = \frac{\omega_0 L}{R} = \frac{1}{R} \sqrt{\frac{L}{C}}$.In figure $(a)$,the sharpness of the resonance pe

Vedclass · Accepted Answer

The circuit of graph $1$ has a high quality factor.

Vedclass · Answer

(C) The quality factor $Q$ is defined as $Q = \frac{\omega_0 L}{R} = \frac{1}{R} \sqrt{\frac{L}{C}}$.In figure $(a)$,the sharpness of the resonance peak indicates the quality factor. $A$ sharper peak corresponds to a higher $Q$ value.Graph $(1)$ is the sharpest,so it has the highest $Q$. Graph $(3)$ is the flattest,so it has the lowest $Q$.In figure $(b)$,the phase angle $\phi$ changes more rapidly near resonance $(\omega = \omega_0)$ for higher $Q$ values. The curve that changes most steeply is $a$,which corresponds to the highest $Q$ (graph $1$). The curve that changes least steeply is $c$,which corresponds to the lowest $Q$ (graph $3$).Therefore,the matching is: $1 \rightarrow a$,$2 \rightarrow b$,and $3 \rightarrow c$.Thus,graph $1$ has the highest quality factor,making option $(C)$ correct.

Similar Questions

$A$ sinusoidal voltage of peak value $250\, V$ is applied to a series $LCR$ circuit,in which $R = 8\, \Omega$,$L = 24\, mH$ and $C = 60\, \mu F$. The value of power dissipated at resonant condition is $'x'\, kW$. The value of $x$ to the nearest integer is .............

The $Q$ factor of a series $LCR$ circuit with $L = 2 \ H$,$C = 32 \ \mu F$,and $R = 10 \ \Omega$ is:

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 $. . . . \times 10^4 \ \text{rad s}^{-1}$.

In the circuit shown,the readings of voltmeters $V_1, V_2$ and $V_3$ at resonance are given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module