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)
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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