In L C circuit the inductance L = 40   mH and capacitance C = 100   μF . If a voltage V t =…

1. $0.52\cos \left(314t\right)$
2. $10\cos \left(314t\right)$
3. $5.2\cos \left(314t\right)$
4. $0.52\sin \left(314t\right)$

Solution

$z=\sqrt{R^2+{\left(X_C-X_L\right)}^2}$
$R=0$
$Z=X_C-X_L$
$=\frac{1}{\omega C}-\omega L$
$=\frac{1}{314\times 100\times {10}^{-6}}-314\times 40\times {10}^{-3}$
$=31.84-12.56$
$=19.28 \mathrm{\Omega }$

$i=\frac{V_0}{Z}\sin \left(314t+\frac{\pi }{2}\right)$
$\therefore i=\frac{V_0}{Z}\cos \left(314t\right)\Rightarrow i=\frac{10}{19.28}\cos \left(314t\right)$
$\Rightarrow i=0.52\cos \left(314t\right)$

Asked in: JEE Main 2020 (09 Jan Shift 2)

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