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Step-by-Step Solution
Step 1: Identify the Given Quantities
Inductance:
$L = 40\,\text{mH} = 40 \times 10^{-3}\,\text{H}$
Capacitance:
$C = 100\,\mu\text{F} = 100 \times 10^{-6}\,\text{F}$
Applied voltage:
$V(t) = 10\,\sin(314t)$
Step 2: Calculate Inductive Reactance $X_L$
The inductive reactance is given by
$$
X_L = \omega L
$$
where $\omega = 314\,\text{rad/s}$. Thus:
$$
X_L = 314 \times 40 \times 10^{-3} = 314 \times 0.04 = 12.56\,\Omega
$$
Step 3: Calculate Capacitive Reactance $X_C$
The capacitive reactance is given by
$$
X_C = \frac{1}{\omega C}
$$
Substituting $\omega = 314\,\text{rad/s}$ and
$C = 100 \times 10^{-6}\,\text{F}$:
$$
X_C = \frac{1}{314 \times 100 \times 10^{-6}}
= \frac{1}{314 \times 10^{-4}}
\approx 31.85\,\Omega
$$
Step 4: Find the Net Reactance of the LC Circuit
Since the circuit is purely reactive (no resistance), the net reactance is
$$
X = \bigl|X_C - X_L\bigr|
$$
From the calculations:
$$
X = |31.85 - 12.56| = 19.29\,\Omega
$$
Step 5: Determine the Current Amplitude
The maximum (peak) current $I_0$ is given by
$$
I_0 = \frac{V_0}{X}
$$
where $V_0$ is the peak value of the voltage (10 V). Thus,
$$
I_0 = \frac{10}{19.29} \approx 0.52\,\text{A}
$$
Step 6: Determine the Phase of the Current
Since $X_C > X_L$, the circuit behaves predominantly like a capacitor, and the current leads the voltage by $90^\circ$.
A leading current can be written as a cosine function when the voltage is a sine function at the same angular frequency. Hence, the current is
$$
i(t) = 0.52 \cos(314t).
$$
Final Answer
The current in the circuit is:
$$
i(t) = 0.52\,\cos(314t).
$$
