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PhysicsEMI/ACSeries AC CircuitsEasy2 minPYQ_2021

PhysicsEasynumerical

In a series LCR resonant circuit, the quality factor is measured as $100$ . If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be _______

Answer:

282.84

Solution:

$Q = \frac{X_{L}}{R} = \frac{ω L}{R} = \frac{1}{\sqrt{L C}} \frac{L}{R} = \frac{\sqrt{L}}{R \sqrt{C}}$
$Q^{'} = \frac{\sqrt{2 L}}{(\frac{R}{2}) \sqrt{C}} = 2 \sqrt{2} Q = 2 \sqrt{2} (100)$
$= 282.84$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Series AC Circuits

⏱ 2mℹ️ Source: PYQ_2021

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