Easy MCQ +4 / -1 PYQ · JEE Mains 2022

In series RLC resonator, if the self inductance and capacitance become double, the new resonant frequency (f2) and new quality factor (Q2) will be :

(f1 = original resonant frequency, Q1 = original quality factor)

  1. A ${f_2} = {{{f_1}} \over 2}$ and ${Q_2} = {Q_1}$ Correct answer
  2. B ${f_2} = {f_1}$ and ${Q_2} = {{{Q_1}} \over {{Q_2}}}$
  3. C ${f_2} = 2{f_1}$ and ${Q_2} = {Q_1}$
  4. D ${f_2} = {f_1}$ and ${Q_2} = 2{Q_1}$

Solution

<p>We know,</p> <p>Quality factor (Q factor)</p> <p>${Q_1} = {{{w_1}} \over {\Delta w}}$</p> <p>$= {1 \over {\sqrt {LC} }} \times {L \over R}$</p> <p>$= {1 \over R}\sqrt {{L \over C}}$</p> <p>Now, when $L' = 2L$ and $C' = 2C$ then $${Q_2} = {1 \over R}\sqrt {{{2L} \over {2C}}} = {1 \over R}\sqrt {{L \over C}} = {Q_1}$$</p> <p>$\therefore$ Q<sub>2</sub> remains same as Q<sub>1</sub>.</p> <p>Also, as ${w_1} = {1 \over {\sqrt {LC} }}$</p> <p>$\Rightarrow 2\pi {f_1} = {1 \over {\sqrt {LC} }}$</p> <p>$\Rightarrow {f_1} = {1 \over {2\pi \sqrt {LC} }}$</p> <p>$\therefore$ When $L' = 2L$ and $C' = 2C$ then new resonating frequency</p> <p>$${f_2} = {1 \over {2\pi \sqrt {2L \times 2C} }} = {1 \over {2\pi \times 2\sqrt {LC} }} = {1 \over 2} \times {f_1}$$</p>

About this question

Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C

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