SATHEE: Electro Magnetic Induction And Alternating Currents Question 158

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) $\frac{R}{2 π L}$

B) $\frac{R}{π L}$

C) $\frac{2 R}{π L}$

D) $\frac{3 R}{2 π L}$

Show Answer

Answer:

Correct Answer: A

Solution:

At resonance

$I_{R} = \frac{E_{0}}{R} \Rightarrow \frac{I_{R}}{\sqrt{2}} = \frac{E_{0}}{\sqrt{2} R} = \frac{E_{0}}{\sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}}$
$\Rightarrow ω_{1} L - \frac{1}{ω_{1} C} = - R$

and $ω_{2} L - \frac{1}{ω_{2} C} = + R$
$\Rightarrow, L (ω_{1} + ω_{2}) = (\frac{ω_{1} + ω_{2}}{ω_{1} ω_{2}}) \frac{1}{C} \Rightarrow ω_{1} ω_{2} = \frac{1}{L C}$

and $L (ω_{2} - ω_{1}) + (\frac{ω_{2} - ω_{1}}{ω_{1} ω_{2}}) \frac{1}{C} = 2 R$

Putting value of $ω_{1}$ and $ω_{2}$

we get $ω_{2} - ω_{1} = \frac{R}{L} \Rightarrow f_{2} - f_{1} = \frac{R}{2 π L}$

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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 12 of the current amplitude at resonance is

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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