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

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(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.

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(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.

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