Verified Solution • 2025–26 Session moderate

QJEEM21Ph886d: JEE Main 2021 Physics - Class 12 JEEM21Ph886d Solution: A series LCR circuit driven by 300 V at a frequency of 50 Hz…

A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance R = 3 k Ω\Omega , an inductor of inductive reactance XL = 250 π\pi Ω\Omega and an unknown capacitor. The value of capacitance to maximize the average power should be : (Take π\pi 2 = 10)

Multiple Choice Options

  • A 4 μ\mu F
  • B 25 μ\mu F
  • C 400 μ\mu F
  • D 40 μ\mu F

Step-by-Step Verified Solution

From maximum average power

XL = XC

250 π\pi = 12π(50)C{1 \over {2\pi (50)C}}

⇒\Rightarrow C = 4 ×\times 10 −- 6

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