Given:
- Components: Resistor \( R \), inductor \( L \), capacitor \( C \) in series.
- AC voltage source: \( V(t) = V_0 \sin(\omega t) \).
- Phase difference between current and voltage: \( \phi = \frac{\pi}{4} \).
- Impedance: \( Z = \sqrt{R^2 + \left( \omega L - \frac{1}{\omega C} \right)^2} \).
At resonance, the inductive reactance \( \omega L \) equals the capacitive reactance \( \frac{1}{\omega C} \). This makes the net reactance zero, and the impedance of the circuit is purely resistive, minimizing the overall impedance and maximizing the current.
The condition for resonance in a series RLC circuit is:
\[ \omega L = \frac{1}{\omega C} \]
This means that the inductive reactance \( \omega L \) should exactly cancel out the capacitive reactance \( \frac{1}{\omega C} \).
To check this with respect to the phase difference:
1. Calculate \( \tan(\phi) \):
Since the phase difference \( \phi \) is \( \frac{\pi}{4} \):
\[ \phi = \tan^{-1}\left( \frac{\omega L - \frac{1}{\omega C}}{R} \right) \]
2. Using the phase difference \( \phi = \frac{\pi}{4} \):
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]
Thus,
\[ 1 = \frac{\omega L - \frac{1}{\omega C}}{R} \]
3. Simplifying the equation:
Since \( \tan(\frac{\pi}{4}) = 1 \),
\[ R = \omega L - \frac{1}{\omega C} \]
For resonance, the reactances should be equal:
\[ \omega L = \frac{1}{\omega C} \]
Thus, the correct condition for resonance is:
A. \( \omega L = \frac{1}{\omega C} \)
So, the correct answer is A. \( \omega L = \frac{1}{\omega C} \).
