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JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2020
  • Hard
  • MCQ

In \(LC\) circuit the inductance \(\mathrm{L}=40\; \mathrm{mH}\) and capacitance \(\mathrm{C}=100\; \mu \mathrm{F} .\) If a voltage \(\mathrm{V}(\mathrm{t})=10 \sin (314 \mathrm{t})\) is applied to the circuit, the current in the circuit is given as 

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Answer & Solution

Correct Answer

(A) \(0.52 \cos 314 t\)

Step-by-step Solution

Detailed explanation

\(\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=314 \times 40 \times 10^{-3}=12.56 \Omega\)

\(\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{314 \times 100 \times 10^{-6}}\)

\(=\frac{10^{4}}{314}=31.84 \Omega\)

\(\mathrm{V}_{\mathrm{m}}=\mathrm{I}_{\mathrm{m}}\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)\)

\(10=\mathrm{I}_{\mathrm{m}}(31.84-12.56)\)

\(I_{m}=\frac{10}{19.28}=0.52 A\)

\(I=0.52 \sin \left(314 t+\frac{\pi}{2}\right)\)

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