Solution:
In an inductive-resistive (RL) series circuit, the phase angle \( \phi \) is given by \( \tan \phi = \frac{X_L}{R} \). Since \( \phi = 60^\circ \), \( \tan 60^\circ = \frac{X_L}{R} \implies X_L = \sqrt{3}R \).
In an inductive-resistive (RL) series circuit, the phase angle \( \phi \) is given by \( \tan \phi = \frac{X_L}{R} \). Since \( \phi = 60^\circ \), \( \tan 60^\circ = \frac{X_L}{R} \implies X_L = \sqrt{3}R \).
Leave a Reply