Question

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

• A. $10\, cos\, 314\,t$
• B. $0.52\, cos\, 314\, t$
• C. $5.2\, cos\, 314\, t$
• D. $0.52 \,sin \,314\,t$

Answer 1

Answer: (B)

$X_{L} = \omega L = 314 × 40 × 10^{-3} = 12.56 \Omega$
$X_{C} = \frac{1}{\omega C} = \frac{1}{314\times 100\times 10^{-6}}$
$= \frac{10^{4}}{314} = 31.84 \Omega$
Phasor
$V_{m} = I_{m}\left(X_{C} - X_{L}\right)$
$10 = I_{m}\left(31.84 - 12.56\right)$
$I_{m} = \frac{10}{19.28} = 0.52A$
$I = 0.52\, sin \left(314t+\frac{\pi}{2}\right)$