1. Given data

• Inductance: $L = 40\text{ mH} = 0.04\text{ H}$
• Capacitance: $C = 100\,\mu\text{F} = 100 \times 10^{-6}\text{ F} = 10^{-4}\text{ F}$
• Applied voltage: $V(t)=10\sin(314t)$

So the angular frequency is $\omega = 314\ \text{rad s}^{-1}$

2. Find reactances

For the inductor, $X_L=\omega L=314 \times 0.04=12.56\,\Omega$

For the capacitor, $X_C=\frac{1}{\omega C}=\frac{1}{314\times 10^{-4}}\approx 31.85\,\Omega$

3. Net reactance of LC series circuit

For a series $LC$ circuit, impedance is purely reactive: $Z = j\left(X_L-X_C\right)$

Magnitude of impedance: $|Z|=|X_L-X_C|=|12.56-31.85|=19.29\,\Omega$

4. Current amplitude

Given voltage amplitude $V_0=10$ V, current amplitude is $I_0=\frac{V_0}{|Z|}=\frac{10}{19.29}\approx 0.52\text{ A}$

5. Phase relation

Since $X_C > X_L$, the circuit is capacitive. Therefore current leads the voltage by $\frac{\pi}{2}$.

Given $V(t)=10\sin(314t)$ so current is $I(t)=0.52\sin\left(314t+\frac{\pi}{2}\right)=0.52\cos(314t)$

6. Check options

• A: $0.52\cos 314t$ ✅
• B: $5.2\cos 314t$ ❌ amplitude incorrect
• C: $0.52\sin 314t$ ❌ wrong phase
• D: $10\cos 314t$ ❌ wrong amplitude

Therefore, the correct current is $\boxed{I(t)=0.52\cos(314t)}$

So, Option A is correct.