Easy MCQ +4 / -1 PYQ · JEE Mains 2021

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

  1. A 4 $\mu$F Correct answer
  2. B 25 $\mu$F
  3. C 400 $\mu$F
  4. D 40 $\mu$F

Solution

From maximum average power<br><br>X<sub>L</sub> = X<sub>C</sub><br><br>250$\pi$ = ${1 \over {2\pi (50)C}}$<br><br>$\Rightarrow$ C = 4 $\times$ 10<sup>$-$6</sup>

About this question

Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C

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