MCQMediumJEE Main 2024 · 30 January, Shift 1LCR Circuits & Resonance

Physics Question from JEE Main 2024 · 30 January, Shift 1

A series LRL-R circuit connected to an AC source E=25sin(1000t)VE = 25 \sin(1000t) \, \text{V} has a power factor of 12\frac{1}{\sqrt{2}}. If the source of emf is changed to E=20sin(2000t)VE = 20 \sin(2000t) \, \text{V}, the new power factor of the circuit will be:

  • A

    12\frac{1}{\sqrt{2}}

  • B

    13\frac{1}{\sqrt{3}}

  • C

    15\frac{1}{\sqrt{5}}

  • D

    17\frac{1}{\sqrt{7}}

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