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JEE Main 201322 AprilOnline
Exam
JEE Main
Year
2013
Date
22 April
Mode
Online
Subject
Physics
Chapter
Alternating Current

JEE Main 2013 · 22 April · Online · Physics · Alternating Current

In a series L-C-R circuit, C=10^ -11 Farad, L=10^ - ^5 Henry and...

In a series L−C−R\mathrm{L}-\mathrm{C}-\mathrm{R} circuit, C=10−11\mathrm{C}=10^{-11} Farad, L=10−\mathrm{L}=10^{-} 5{ }^5 Henry and R=100Ohm\mathrm{R}=100 \mathrm{Ohm}, when a constant D.C. voltage E\mathrm{E} is applied to the circuit, the capacitor acquires a charge 10−910^{-9} C. The D.C. source is replaced by a sinusoidal voltage source in which the peak voltage E0\mathrm{E}_0 is equal to the constant D.C. voltage E. At resonance the peak value of the charge acquired by the capacitor will be :

Options

  1. A10−15C10^{-15} \mathrm{C}
  2. B10−6C10^{-6} \mathrm{C}
  3. C10−10C10^{-10} \mathrm{C}
  4. D10−8C10^{-8} \mathrm{C}

Correct answer

D. 10−8C10^{-8} \mathrm{C}

Step-by-step solution

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