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

We have an ideal series $$LC$$ circuit fed by the sinusoidal voltage

$$v(t)=10\sin(314t)\;{\rm volt}$$

Comparing with the standard form $$V_0\sin(\omega t)$$, we identify the angular frequency

$$\omega = 314\;{\rm rad\,s^{-1}}\; , \qquad V_0 = 10\;{\rm V}.$$

The given component values are

$$L = 40\;{\rm mH}=40\times10^{-3}\;{\rm H}=0.04\;{\rm H},$$

$$C = 100\;\mu{\rm F}=100\times10^{-6}\;{\rm F}=1.0\times10^{-4}\;{\rm F}.$$

For a series circuit containing only $$L$$ and $$C$$, the impedance is purely reactive. We must therefore find the inductive reactance $$X_L$$ and the capacitive reactance $$X_C$$ and then combine them.

Formulae stated first:

Inductive reactance: $$X_L = \omega L.$$

Capacitive reactance: $$X_C = \dfrac{1}{\omega C}.$$

Substituting the numerical values

$$X_L = 314 \times 0.04 = 12.56\;\Omega,$$

$$X_C = \dfrac{1}{314 \times 1.0\times10^{-4}} = \dfrac{1}{3.14\times10^{-2}} = 31.85\;\Omega.$$

Because the inductor and capacitor are in series, the combined reactance is

$$X = X_L - X_C.$$

So,

$$X = 12.56 - 31.85 = -19.29\;\Omega.$$

The negative sign shows that the net reactance is capacitive. Its magnitude (which is the impedance, since there is no resistance) is

$$Z = |X| = 19.29\;\Omega.$$

Current amplitude

Ohm’s law for AC gives $$I_0 = \dfrac{V_0}{Z}.$$ Substituting,

$$I_0 = \dfrac{10}{19.29} \approx 0.52\;{\rm A}.$$

Phase of the current

For a purely capacitive reactance, current leads voltage by $$\dfrac{\pi}{2}$$ (90°). Mathematically, if the voltage is $$\sin(\omega t)$$, the current will be $$\sin(\omega t + \dfrac{\pi}{2}) = \cos(\omega t).$$

Hence the complete time‐domain expression for the current is

$$i(t) = 0.52\cos(314t)\;{\rm A}.$$

Matching this with the options provided, we see that it corresponds to Option A.

Hence, the correct answer is Option A.