JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 15)

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)

$${f_2} = {{{f_1}} \over 2}$$ and $${Q_2} = {Q_1}$$

$${f_2} = {f_1}$$ and $${Q_2} = {{{Q_1}} \over {{Q_2}}}$$

$${f_2} = 2{f_1}$$ and $${Q_2} = {Q_1}$$

$${f_2} = {f_1}$$ and $${Q_2} = 2{Q_1}$$

We know,

Quality factor (Q factor)

$${Q_1} = {{{w_1}} \over {\Delta w}}$$

$$ = {1 \over {\sqrt {LC} }} \times {L \over R}$$

$$ = {1 \over R}\sqrt {{L \over C}} $$

Now, when $$L' = 2L$$ and $$C' = 2C$$ then $${Q_2} = {1 \over R}\sqrt {{{2L} \over {2C}}} = {1 \over R}\sqrt {{L \over C}} = {Q_1}$$

$$\therefore$$ Q₂ remains same as Q₁.

Also, as $${w_1} = {1 \over {\sqrt {LC} }}$$

$$ \Rightarrow 2\pi {f_1} = {1 \over {\sqrt {LC} }}$$

$$ \Rightarrow {f_1} = {1 \over {2\pi \sqrt {LC} }}$$

$$\therefore$$ When $$L' = 2L$$ and $$C' = 2C$$ then new resonating frequency

$${f_2} = {1 \over {2\pi \sqrt {2L \times 2C} }} = {1 \over {2\pi \times 2\sqrt {LC} }} = {1 \over 2} \times {f_1}$$