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**Hint:**In order to solve this question one should be aware of the concept of resonance and the conditions of resonance in an LCR circuit. The condition of the resonance depends on a particular frequency, also known as resonance frequency. When the frequency of the LCR is equal to the resonance frequency only then the condition of resonance is attained.

**Complete step by step solution:**

The condition for resonance in the LCR circuit is given by,

${\omega _o} = \dfrac{1}{{\sqrt {LC} }}$

Here, we can easily observe that the amplitude of the current is maximum at the resonance frequency ${\omega _o}$ . Since, ${i_m} = {V_m}IR$ at resonance, the amplitude of current for case ${R_2}$ is sharper to that for case ${R_1}$ . The quality factor or can be said as the Q-factor of a resonance LCR circuit is defined as the ratio of voltage drop across the capacitor to that of applied voltage.

It is given by,

$Q = \dfrac{1}{R}\sqrt {\dfrac{L}{C}} $

The Q-factor or Quality factor determines the sharpness of the resonance curve. If the resonance curve is less sharp, then the maximum current decreases and also the circuit is close to the resonance for a larger range $\Delta \omega $ of frequencies and the regulation of the circuit then will not be good. Therefore, the sharp resonance curve is better.

**Note:**Less sharp the resonance is better, as for less sharp resonance curve, less is the selectivity of the circuit while the Q-Factor will be higher, sharper is the resonance curve and lesser will be the loss in energy of the circuit.

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