CUET · PHYSICS · PYQ PAPER 2025
In a series L, C and R circuit with an ac source of frequency \( \omega \), the current leads the voltage by \( \pi/4 \). The value of capacitance C will be
- A \( \omega(\omega L - R)^{-1} \)
- B \( \omega^{-1}(\omega L - R) \)
- C \( (\omega^2 L)^{-1} \)
- D \( (\omega^2 L + \omega R)^{-1} \)
Answer & Solution
Correct Answer
(D) \( (\omega^2 L + \omega R)^{-1} \)
Step-by-step Solution
Detailed explanation
\(\tan \phi = \frac{X_L - X_C}{R}\) Current leads voltage by \(\pi/4 \implies \phi = -\pi/4\) \(\tan(-\pi/4) = \frac{\omega L - \frac{1}{\omega C}}{R}\) \(-1 = \frac{\omega L - \frac{1}{\omega C}}{R}\) \(-R = \omega L - \frac{1}{\omega C}\) \(\frac{1}{\omega C} = \omega L + R\)…
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