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JEE Main 202126 AugustMorning Shift
Exam
JEE Main
Year
2021
Date
26 August
Shift
Morning Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2021 · 26 August · Morning Shift · Physics · Alternating Current

A series LCR circuit driven by 300V at a frequency of 50Hz contains...

A series LCR circuit driven by 300V300V at a frequency of 50Hz50Hz contains a resistance R=3kΩR = 3kΩ, an inductor of inductive reactance XL=250πΩ{X}_{L} = 250\piΩ and an unknown capacitor. The value of capacitance to maximise the average power should be:
(takeπ2=10)\left(\text{take}{\pi}^{2} = 10\right)

Options

  1. A400μF400\mu F
  2. B4\muF4\muF
  3. C40\muF40\muF
  4. D25\muF25\muF

Correct answer

B. 4\muF4\muF

Step-by-step solution

Angular frequency, ω=2\pif=2π×50=100πrads−1\omega = 2\pif = 2\pi \times 50 = 100\pi rad{s}^{- 1}

At resonance,

XC=XL⇒1ωC=ωL⇒1ωC=250π⇒C=1100π×250πF=4μF{X}_{C} = {X}_{L} ⇒ \dfrac{1}{\omega C} = \omega L ⇒ \dfrac{1}{\omega C} = 250\pi ⇒ C = \dfrac{1}{100\pi \times 250\pi}F = 4\mu F

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