A) \[\sqrt{2}A\]
B) 1A
C) 2 A
D) 3 A
Correct Answer: C
Solution :
Idea This question is based on the resonant frequency in an AC circuit. At resonance\[{{X}_{C}}={{X}_{L}}\] At\[{{\omega }_{0}}/2,X{{'}_{C}}=2{{x}_{C}}\] and\[X{{'}_{L}}=\frac{{{X}_{L}}}{2}\] \[[\because {{X}_{C}}=\frac{1}{\omega C}i.e.,{{X}_{C}}\propto \frac{1}{\omega }\text{and}{{\text{X}}_{L}}=\omega Li.e.,{{X}_{L}}\propto \omega ]\]Then, \[Z'=\sqrt{X'_{C}^{2}'+X'_{L}^{2}}\] \[=\sqrt{4X_{C}^{2}+\frac{X_{L}^{2}}{4}}=\frac{\sqrt{17}}{2}.{{X}_{C}}\]\[\left[ \because {{X}_{C}}={{X}_{L}} \right]\] i.e., \[Z'=\frac{\sqrt{17}}{2}{{X}_{C}}\] Now, when frequency is\[2{{\omega }_{0}}\]then\[X_{C}^{''}=\frac{{{X}_{C}}}{2}\]and \[X_{L}^{''}=2{{X}_{L}}\] So. \[Z''=\sqrt{X_{C}^{''2}+X_{L}^{''2}}=\frac{\sqrt{17}}{2}{{X}_{C}}=Z'\] Since impedance remains the same at\[\omega =\frac{{{\omega }_{0}}}{2}\]and at\[\omega =2{{\omega }_{0}}\] So, current shall be 2 A. TEST Edge A question on quality factor could be asked from the same tonic.
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