Question:
In LC circuit the inductance $\mathrm{L}=40 \mathrm{mH}$ and capacitance
$\mathrm{C}=100 \mu \mathrm{F}$. If a voltage $\mathrm{V}(t)=10 \sin (314 t)$ is applied to
the circuit, the current in the circuit is given as:
Correct Option: 1
Solution:
(1)
Given, Inductance, $L=40 \mathrm{mH}$
Capacitance, $C=100 \mu F$
Impedance, $Z=X_{C}-X_{L}$
$\Rightarrow Z=\frac{1}{\omega C}-\omega L \quad\left(\because X_{c}=\frac{1}{\omega C}\right.$ and $\left.X_{L}=\omega L\right)$
$=\frac{1}{314 \times 100 \times 10^{-6}}-314 \times 40 \times 10^{-3}$
$=19.28 \Omega$
Current, $\quad i=\frac{V_{0}}{Z} \sin (\omega t+\pi / 2)$
$\Rightarrow i=\frac{10}{19.28} \cos \omega t=0.52 \cos (314 \mathrm{t})$