A coil of inductance 0.50 H and resistance 100 Ω is connected to 240 V, 50 Hz supply. The maximum current in the coil is:

1.82 A

$ Z = \sqrt{R^2+ \omega^2L^2} = \sqrt{100^2+ 4\pi^2\nu^2L^2}=  \sqrt{100^2+ 4\pi^2 \times 50^2 \times 0.5^2} = 186 \Omega $

$ I = \frac{V}{Z} = \frac{240\sqrt{2}}{186} = 1.82A$

The correct answer is Option (3) → 1.82 A