JEE MAIN - Physics (2021 - 18th March Morning Shift - No. 11)
In a series LCR resonance circuit, if we change the resistance only, from a lower to higher value :
The bandwidth of resonance circuit will increase.
The resonance frequency will increase.
The quality factor will increase.
The quality factor and the resonance frequency will remain constant.
Explanation
$${\omega } = {1 \over {\sqrt {LC} }}$$
$$ \Rightarrow $$ 2$$\pi $$f = $${1 \over {\sqrt {LC} }}$$
$$ \Rightarrow $$ f = $${1 \over {2\pi \sqrt {LC} }}$$
f does not depends on resistance(R).
Quality factor, $$Q = {{\omega L} \over R}$$
$$ \Rightarrow $$ $$Q \propto {1 \over R}$$
So if R increase then Q will decrease.
Also, $$Q = {{\omega L} \over R} = {\omega \over {\Delta \beta}}$$
where $$\Delta \beta$$ = bandwidth
$$\Delta \beta = {R \over L}$$
So if R increase then $$\Delta \beta$$ will increase too.
$$ \Rightarrow $$ 2$$\pi $$f = $${1 \over {\sqrt {LC} }}$$
$$ \Rightarrow $$ f = $${1 \over {2\pi \sqrt {LC} }}$$
f does not depends on resistance(R).
Quality factor, $$Q = {{\omega L} \over R}$$
$$ \Rightarrow $$ $$Q \propto {1 \over R}$$
So if R increase then Q will decrease.
Also, $$Q = {{\omega L} \over R} = {\omega \over {\Delta \beta}}$$
where $$\Delta \beta$$ = bandwidth
$$\Delta \beta = {R \over L}$$
So if R increase then $$\Delta \beta$$ will increase too.
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