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7.Alternating Current
hard
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
A
$0.52 \cos 314 t$
B
$0.52 \sin 314 t$
C
$10 \cos 314 t$
D
$5.2 \cos 314 t$
(JEE MAIN-2020)
Solution
$\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=314 \times 40 \times 10^{-3}=12.56 \Omega$
$\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{314 \times 100 \times 10^{-6}}$
$=\frac{10^{4}}{314}=31.84 \Omega$
$\mathrm{V}_{\mathrm{m}}=\mathrm{I}_{\mathrm{m}}\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)$
$10=\mathrm{I}_{\mathrm{m}}(31.84-12.56)$
$I_{m}=\frac{10}{19.28}=0.52 A$
$I=0.52 \sin \left(314 t+\frac{\pi}{2}\right)$
Standard 12
Physics
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