# In LC circuit the inductance L=40 mH and capacitance

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:

1. (1) $0.52 \cos 314 \mathrm{t}$

2. (2) $10 \cos 314 t$

3. (3) $5.2 \cos 314 \mathrm{t}$

4. (4) $0.52 \sin 314 \mathrm{t}$

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, $\quad i=\frac{V_{0}}{Z} \sin (\omega t+\pi / 2)$

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

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