CUET · PHYSICS · PYQ PAPER 2025
For a series LCR circuit, \( R = X_L = 3X_C \). The impedance of the circuit and the phase difference between voltage V and current I, respectively, will be:
- A \( 3R\sqrt{13} \text{ } \Omega \text{ and } \tan^{-1}\left(\frac{3}{2}\right) \)
- B \( 3R\sqrt{13} \text{ } \Omega \text{ and } \tan^{-1}\left(\frac{2}{3}\right) \)
- C \( \frac{R\sqrt{13}}{3} \text{ } \Omega \text{ and } \tan^{-1}\left(\frac{2}{3}\right) \)
- D \( \frac{R\sqrt{13}}{3} \text{ } \Omega \text{ and } \tan^{-1}\left(\frac{3}{2}\right) \)
Answer & Solution
Correct Answer
(C) \( \frac{R\sqrt{13}}{3} \text{ } \Omega \text{ and } \tan^{-1}\left(\frac{2}{3}\right) \)
Step-by-step Solution
Detailed explanation
\( X_C = \frac{X_L}{3} = \frac{R}{3} \) \( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + \left(R - \frac{R}{3}\right)^2} = \sqrt{R^2 + \left(\frac{2R}{3}\right)^2} = \sqrt{R^2 + \frac{4R^2}{9}} = \sqrt{\frac{13R^2}{9}} = \frac{R\sqrt{13}}{3} \)…
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