ExamBro
ExamBro
JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2024
  • Hard
  • MCQ

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 _______.

Pick an option to check your answer.
Verified Solution

Answer & Solution

Correct Answer

(A) \(0.76 \mathrm{H}\)

Step-by-step Solution

Detailed explanation

\(36=I_{\text {ms }} R\)

\(36=\frac{120}{\sqrt{\mathrm{X}_{\mathrm{L}}^2+\mathrm{R}^2}} \times \mathrm{R}\)

\(\mathrm{R}=90 \Omega \Rightarrow 36=\frac{120 \times 90}{\sqrt{\mathrm{X}_{\mathrm{L}}^2+90^2}}\)

\(\sqrt{\mathrm{X}_{\mathrm{L}}^2+90^2}=300\)

\(\mathrm{X}_{\mathrm{L}}^2=81900\)

\(\mathrm{X}_{\mathrm{L}}=286.18\)

\(\omega \mathrm{L}=286.18\)

\(\mathrm{~L}=\frac{286.18}{376.8}\)

\(\mathrm{~L}=0.76 \mathrm{H}\)

Same subject
Explore more questions on app