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Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACQuality and Power factorEasy2 minQB

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:

$Q = \frac{X _{L}}{R} = \frac{ω L}{R} = \frac{1}{L C} \frac{L}{R} = \frac{L}{R C}$
$Q^{^{'}} = \frac{2 L}{( \frac{R}{2} ) C} = 22 Q = 22 (100)$
$= 282.84$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Quality and Power factor

⏱ 2mℹ️ Source: QB

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