Q 12-07-060JEE MainJEE Main 2024 (8 Apr, Shift 2)Medium
A coil of negligible resistance is connected in series with a $90\ \Omega$ resistor across a $120\ \text{V}$, $60\ \text{Hz}$ supply. A voltmeter reads $36\ \text{V}$ across the resistance. Inductance of the coil is:
Answer: (B) $0.76\ \text{H}$
Current: $I = \dfrac{36}{90} = 0.4\ \text{A}$.
The voltages across $R$ and $L$ are $90^\circ$ out of phase:
$$V_L = \sqrt{120^2 - 36^2} = \sqrt{13104} \approx 114.5\ \text{V}$$
$$X_L = \frac{114.5}{0.4} \approx 286\ \Omega,\qquad L = \frac{X_L}{2\pi f} = \frac{286}{2\pi\times60} \approx 0.76\ \text{H}$$
Solution by Sreeraj P, M.Sc Physics