(b) $Z = \sqrt{R^{2} + X_{L}^{2}},\ X_{L} = \omega L$ and $\omega = 2\pi f$ $\therefore Z = \sqrt{R^{2} + 4\pi^{2}f^{2}L^{2}}$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.