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Question: Answered & Verified by Expert
An alternating e.m.f. of frequency $v\left(=\frac{1}{2 \pi \sqrt{L C}}\right)$ is applied to a series $L C R$ circuit. For this frequency of the applied e.m.f.
PhysicsAlternating CurrentJEE Main
Options:
  • A The circuit is at resonance and its impedance is made up only of a reactive part
  • B The current in the circuit is in phase with the applied e.m.f. and the voltage across $R$ equals this applied emf
  • C The sum of the p.d.'s across the inductance and capacitance equals the applied e.m.f. which is $180^{\circ}$ ahead of phase of the current in the circuit
  • D The quality factor of the circuit is $\omega L / R$ or $1 / \omega C R$ and this is a measure of the voltage magnification (produced by the circuit at resonance) as well as the sharpness of resonance of the circuit
Solution:
1112 Upvotes Verified Answer
The correct answer is: The circuit is at resonance and its impedance is made up only of a reactive part
As we know resonating angular frequency is
$\omega=\frac{1}{\sqrt{\mathrm{LC}}}$
so resonating frequency is
$\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$
so circuit is in resonance and impedence is made up of only resistance part

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