Exam

JEE Main

Year

2024

Date

8 April

Shift

Evening Shift

Subject

Physics

Chapter

Alternating Current

JEE Main 2024 · 8 April · Evening Shift · Physics · Alternating Current

A coil of negligible resistance is connected in series with 90...

A coil of negligible resistance is connected in series with $90 \Omega$ resistor across $120 \mathrm{~V}, 60 \mathrm{~Hz}$ supply. A voltmeter reads $36 \mathrm{~V}$ across resistance. Inductance of the coil is :

Step-by-step solution

$$\begin{aligned} & 36=I_{\mathrm{rms}} \mathrm{R} \\ & 36=\frac{120}{\sqrt{\mathrm{X}_{\mathrm{L}}^2+\mathrm{R}^2}} \times \mathrm{R} \\ & \mathrm{R}=90 \Omega \Rightarrow 36=\frac{120 \times 90}{\sqrt{\mathrm{X}_{\mathrm{L}}^2+90^2}} \\ & \sqrt{\mathrm{X}_{\mathrm{L}}^2+90^2}=300 \\ & \mathrm{X}_{\mathrm{L}}^2=81900 \\ & \mathrm{X}_{\mathrm{L}}=286.18 \\ & \omega \mathrm{L}=286.18 \\ & \mathrm{~L}=\frac{286.18}{376.8} \\ & \mathrm{~L}=0.76 \mathrm{H}\end{aligned}$$