Alternating Current

JEE Main level, numerical answer, medium

118 of 190

Question 118 of 190

In a series LCR resonant circuit, the quality factor is measured as 100100. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be _______

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Answer: 282.84

Q=XLR=ωLR=1LCLR=LRCQ=\dfrac{{X}_{L}}{R}=\dfrac{\omega L}{R}=\dfrac{1}{\sqrt{LC}}\dfrac{L}{R}=\dfrac{\sqrt{L}}{R\sqrt{C}}
Q′=2L(R2)C=22Q=22(100){Q}^{'}=\dfrac{\sqrt{2L}}{\left(\dfrac{R}{2}\right)\sqrt{C}}=2\sqrt{2}Q=2\sqrt{2}\left(100\right)
=282.84=282.84

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