A coil of negligible resistance is connected in series with $90 \Omega$ resistor across $120 \mathrm{~V}, 60…
- $0.286 \mathrm{H}$
- $0.76 \mathrm{H}$
- $2.86 \mathrm{H}$
- $0.91 \mathrm{H}$
Solution

$\begin{aligned} & 36=I_{\mathrm{rms}} \mathrm{R} \\ & 36=\frac{120}{\sqrt{\mathrm{X}_{\mathrm{L}}^2+\mathrm{R}^2}} \times \mathrm{R} \\ & \mathrm{R}=90 \Omega \Rightarrow 36=\frac{120 \times 90}{\sqrt{\mathrm{X}_{\mathrm{L}}^2+90^2}} \\ & \sqrt{\mathrm{X}_{\mathrm{L}}^2+90^2}=300 \\ & \mathrm{X}_{\mathrm{L}}^2=81900 \\ & \mathrm{X}_{\mathrm{L}}=286.18 \\ & \omega \mathrm{L}=286.18 \\ & \mathrm{~L}=\frac{286.18}{376.8} \\ & \mathrm{~L}=0.76 \mathrm{H}\end{aligned}$
Asked in: JEE Main 2024 (08 Apr Shift 2)