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

(Take π2 = 10)

Option 1 -

 400  μ F

Option 2 -

4 μ F

Option 3 -

40 μ F

Option 4 -

25 μ F

0 2 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago
    Correct Option - 2


    Detailed Solution:

    Power is maximum at resonance

    So, X L = X C = 1 ω C

    ⇒ C = 1 ω × X L

    = 1 2 5 0 × 1 0 0 × π 2

    = 4 μ F [ π 2 = 1 0 ]

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