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AP EAMCET · PHYSICS · Alternating Current

In a series resonant LCR circuit, for the power dissipated to become half of the maximum power dissipated, the current amplitude is

  1. A \(\frac{1}{\sqrt{2}}\) times its maximum value.
  2. B \(\frac{1}{2}\) times its maximum value.
  3. C twice its maximum value.
  4. D \(\sqrt{2}\) times its maximum value.
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{\sqrt{2}}\) times its maximum value.

Step-by-step Solution

Detailed explanation

\(P_{max} = \frac{1}{2} I_{max}^2 R\) \(P' = \frac{1}{2} P_{max} \implies \frac{1}{2} I'^2 R = \frac{1}{2} \left( \frac{1}{2} I_{max}^2 R \right)\) \(I'^2 = \frac{1}{2} I_{max}^2\) \(I' = \frac{1}{\sqrt{2}} I_{max}\)