Question 128 of 190

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

A$0.52\cos ⁡\left(314t\right)$B$10\cos ⁡\left(314t\right)$C$5.2\cos ⁡\left(314t\right)$D$0.52\sin ⁡\left(314t\right)$

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Answer: (A) $0.52\cos ⁡\left(314t\right)$

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

$i=\dfrac{{V}_{0}}{Z}\sin ⁡\left(314t+\dfrac{\pi }{2}\right)$
$∴i=\dfrac{{V}_{0}}{Z}\cos ⁡\left(314t\right)\Rightarrow i=\dfrac{10}{19.28}\cos ⁡\left(314t\right)$
$\Rightarrow i=0.52\cos ⁡\left(314t\right)$