Electro Magnetic Induction And Alternating Currents Question 158
Question: In a series LCR circuit, the difference of the frequencies at which current amplitude falls to $ \frac{1}{\sqrt{2}} $ of the current amplitude at resonance is
Options:
A) $ \frac{R}{2\pi L} $
B) $ \frac{R}{\pi L} $
C) $ \frac{2R}{\pi L} $
D) $ \frac{3R}{2\pi L} $
Show Answer
Answer:
Correct Answer: A
Solution:
- 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}+{{( \omega L-\frac{1}{\omega C} )}^{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}})=( \frac{{\omega_{1}}+{\omega_{2}}}{{\omega_{1}}{\omega_{2}}} )\frac{1}{C}\Rightarrow {\omega_{1}}{\omega_{2}}=\frac{1}{LC} $
and $ L({\omega_{2}}-{\omega_{1}})+( \frac{{\omega_{2}}-{\omega_{1}}}{{\omega_{1}}{\omega_{2}}} )\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} $
