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

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(B) The quality factor $Q$ is defined as $Q = \frac{\omega_0 L}{R} = \frac{1}{R} \sqrt{\frac{L}{C}}$.Higher $Q$ implies a sharper resonance peak in th

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Graph $(1)$ corresponds to graph $(c)$

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(B) The quality factor $Q$ is defined as $Q = \frac{\omega_0 L}{R} = \frac{1}{R} \sqrt{\frac{L}{C}}$.Higher $Q$ implies a sharper resonance peak in the voltage vs frequency graph (figure $a$) and a steeper change in the phase angle $\phi$ near resonance (figure $b$).In figure $(a)$,graph $(1)$ is the sharpest,indicating the highest $Q$ value. Graph $(3)$ is the broadest,indicating the lowest $Q$ value.In figure $(b)$,graph $(c)$ shows the steepest change in phase angle near $\omega / \omega_0 = 1$,corresponding to the highest $Q$. Graph $(a)$ shows the shallowest change,corresponding to the lowest $Q$.Therefore,graph $(1)$ matches graph $(c)$,graph $(2)$ matches graph $(b)$,and graph $(3)$ matches graph $(a)$.Thus,the statement 'Graph $(1)$ corresponds to graph $(c)$' is correct.

Similar Questions

$A$ series $LCR$ circuit with $R = 20 \ \Omega$, $L = 1.6 \ \text{H}$ and $C = 40 \ \mu\text{F}$ is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is . . . . . . $\Omega$.

In a series $LCR$ circuit,the capacitance is changed from $C$ to $4C$. To keep the resonance frequency unchanged,the new inductance should be:

For the series $LCR$ circuit shown in the figure,what is the resonance angular frequency and the amplitude of the current at the resonating frequency? The source voltage is given as $220 \, V$ ($RMS$ value).

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