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
- A \(\frac{1}{\sqrt{2}}\) times its maximum value.
- B \(\frac{1}{2}\) times its maximum value.
- C twice its maximum value.
- D \(\sqrt{2}\) times its maximum value.
Answer & Solution
Correct Answer
(A) \(\frac{1}{\sqrt{2}}\) times its maximum value.
Step-by-step Solution
Detailed explanation
\(P_{max} = \frac{1}{2} I_{max}^2 R\) \(P' = \frac{1}{2} P_{max} \implies \frac{1}{2} I'^2 R = \frac{1}{2} \left( \frac{1}{2} I_{max}^2 R \right)\) \(I'^2 = \frac{1}{2} I_{max}^2\) \(I' = \frac{1}{\sqrt{2}} I_{max}\)
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