JEE Mains · Physics · STD 12 -7. Alternating current
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 _______.
- A \(0.76 \mathrm{H}\)
- B \(2.86 \mathrm{H}\)
- C \(0.286 \mathrm{H}\)
- D \(0.91 \mathrm{H}\)
Answer & Solution
Correct Answer
(A) \(0.76 \mathrm{H}\)
Step-by-step Solution
Detailed explanation
\(36=I_{\text {ms }} R\) \(36=\frac{120}{\sqrt{\mathrm{X}_{\mathrm{L}}^2+\mathrm{R}^2}} \times \mathrm{R}\) \(\mathrm{R}=90 \Omega \Rightarrow 36=\frac{120 \times 90}{\sqrt{\mathrm{X}_{\mathrm{L}}^2+90^2}}\) \(\sqrt{\mathrm{X}_{\mathrm{L}}^2+90^2}=300\)…
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