Question 175: AC fundamentals

Question

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)

Accepted Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

AC fundamentals

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)

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