A) \[\frac{R}{2\pi L}\]
B) \[\frac{R}{\pi L}\]
C) \[\frac{2R}{\pi L}\]
D) \[\frac{3R}{2\pi L}\]
Correct Answer: A
Solution :
[a] At resonance \[{{I}_{R}}=\frac{{{E}_{0}}}{R}\Rightarrow \frac{{{I}_{R}}}{\sqrt{2}}=\frac{{{E}_{0}}}{\sqrt{2}R}=\frac{{{E}_{0}}}{\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}}\] \[\Rightarrow \,\,{{\omega }_{1}}L-\frac{1}{{{\omega }_{1}}C}=-R\] and \[{{\omega }_{2}}L-\frac{1}{{{\omega }_{2}}C}=+R\] \[\Rightarrow \,\,\,L({{\omega }_{1}}+{{\omega }_{2}})=\left( \frac{{{\omega }_{1}}+{{\omega }_{2}}}{{{\omega }_{1}}{{\omega }_{2}}} \right)\frac{1}{C}\Rightarrow {{\omega }_{1}}{{\omega }_{2}}=\frac{1}{LC}\] and \[L({{\omega }_{2}}-{{\omega }_{1}})+\left( \frac{{{\omega }_{2}}-{{\omega }_{1}}}{{{\omega }_{1}}{{\omega }_{2}}} \right)\frac{1}{C}=2R\] Putting value of \[{{\omega }_{1}}\]and \[{{\omega }_{2}}\]we get \[{{\omega }_{2}}-{{\omega }_{1}}=\frac{R}{L}\Rightarrow {{f}_{2}}-{{f}_{1}}=\frac{R}{2\pi L}\]
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