In a series LCR resonance circuit, if we change the resistance only, from a lower to higher value :
Solution
${\omega } = {1 \over {\sqrt {LC} }}$
<br><br>$\Rightarrow$ 2$\pi$f = ${1 \over {\sqrt {LC} }}$
<br><br>$\Rightarrow$ f = ${1 \over {2\pi \sqrt {LC} }}$
<br><br>f does not depends on resistance(R).
<br><br>Quality factor, $Q = {{\omega L} \over R}$
<br><br>$\Rightarrow$ $Q \propto {1 \over R}$
<br><br>So if R increase then Q will decrease.
<br><br>Also, $Q = {{\omega L} \over R} = {\omega \over {\Delta \beta}}$
<br><br>where $\Delta \beta$ = bandwidth
<br><br>$\Delta \beta = {R \over L}$
<br><br>So if R increase then $\Delta \beta$ will increase too.
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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