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.
Answer & Solution
Correct Answer
(A) \(0.76 \mathrm{H}\)
Step-by-step Solution
Detailed explanation
\(36=I_{\text {ms }} R\)
\(36=\frac{120}{\sqrt{\mathrm{X}_{\mathrm{L}}^2+\mathrm{R}^2}} \times \mathrm{R}\)
\(\mathrm{R}=90 \Omega \Rightarrow 36=\frac{120 \times 90}{\sqrt{\mathrm{X}_{\mathrm{L}}^2+90^2}}\)
\(\sqrt{\mathrm{X}_{\mathrm{L}}^2+90^2}=300\)
\(\mathrm{X}_{\mathrm{L}}^2=81900\)
\(\mathrm{X}_{\mathrm{L}}=286.18\)
\(\omega \mathrm{L}=286.18\)
\(\mathrm{~L}=\frac{286.18}{376.8}\)
\(\mathrm{~L}=0.76 \mathrm{H}\)
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