Question

In $LC$ circuit the inductance $L=40\text{mH}$ and capacitance $C=100\mu \text{F}.$ If a voltage $V\left(t\right)=10\sin \left(314t\right)$ 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. $0.52\sin \left(314t\right)$

Answer 1

The correct option is A $0.52\cos \left(314t\right)$

Given:
Inductance, $L=40\text{mH}$
Capacitance, $C=100\mu \text{F}$
Angular frequency, $\omega =314\text{rad/s}$

Now, Impedance of the circuit,

$Z=X_C-X_L$

$\Rightarrow Z=\frac{1}{\omega C}-\omega L$

$=\frac{1}{314\times 100\times {10}^{-6}}-314\times 40\times {10}^{-3}$

$\Rightarrow Z=19.28\Omega$


Here, $X_C>X_L$. So, current lead by $\frac{\pi }{2}$.

Current, $i=\frac{V_0}{Z}\sin (\omega t+\frac{\pi }{2})$

$\Rightarrow i=\frac{10}{19.28}\cos \omega t=0.52\cos \left(314t\right)$

Hence, option $\left(\text{A}\right)$ is correct.