To solve this problem, let's go through the steps to determine the value of the capacitor $C$ that makes the current $i(t)$ in phase with the source voltage $e(t)$, and then find $i(t)$.

Step 1: Analyzing the Circuit and Phasor Conditions

The given circuit consists of:

1. A resistor $R = 10 \ \Omega$.
2. An inductor with inductance $L = 20 \ \text{mH}$.
3. A capacitor $C$ (whose value we need to determine).

The source voltage $e(t) = 30 \cos(377t + 30^\circ) \, \text{V}$ has an angular frequency $\omega = 377 \ \text{rad/s}$.

For the current $i(t)$ to be in phase with $e(t)$, the overall impedance of the circuit must be purely resistive. This means that the reactive components (inductive and capacitive reactances) must cancel each other out, resulting in zero net reactance.

Step 2: Calculate the Inductive and Capacitive Reactances

1. Inductive Reactance $X_L$: $X_L = \omega L = 377 \times 0.02 = 7.54 \ \Omega$

2. Capacitive Reactance $X_C$: $X_C = \frac{1}{\omega C}$

For the current to be in phase with the voltage, we need $X_L = X_C$.

Step 3: Set Up the Equation for $C$

Equating the inductive and capacitive reactances: $X_L = X_C \Rightarrow 7.54 = \frac{1}{377C}$

Solving for $C$: $C = \frac{1}{377 \times 7.54} = \frac{1}{2845.38} \approx 351.5 \ \mu\text{F}$

Thus, the value of $C$ should be approximately $351.5 \ \mu\text{F}$.

Step 4: Determine the Current $i(t)$

Now that we know the impedance is purely resistive, the total impedance $Z$ of the circuit is just the resistance $R = 10 \ \Omega$.

1. Calculate the RMS Value of the Voltage: The peak voltage $V_{\text{peak}} = 30 \ \text{V}$, so the RMS voltage $V_{\text{rms}}$ is: $V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} = \frac{30}{\sqrt{2}} \approx 21.21 \ \text{V}$

2. Calculate the RMS Current: Using Ohm's Law, $I_{\text{rms}} = \frac{V_{\text{rms}}}{R}$: $I_{\text{rms}} = \frac{21.21}{10} = 2.121 \ \text{A}$

3. Express $i(t)$ in Time Domain: Since $i(t)$ is in phase with $e(t)$, it has the same angular frequency and phase shift. The peak current $I_{\text{peak}} = I_{\text{rms}} \cdot \sqrt{2}$: $I_{\text{peak}} = 2.121 \times \sqrt{2} \approx 3 \ \text{A}$

   Therefore: $i(t) = 3 \cos(377t + 30^\circ) \ \text{A}$

Final Answers

1. Capacitance: $C \approx 351.5 \ \mu\text{F}$
2. Current: $i(t) = 3 \cos(377t + 30^\circ) \ \text{A}$

Would you like more details or have any questions?

Here are some related questions for further exploration:

1. What would happen to the phase of $i(t)$ if $C$ were increased or decreased from this calculated value?
2. How does the presence of both an inductor and a capacitor impact the resonance of the circuit?
3. How would you determine the power dissipated in this circuit?
4. What would be the effect on $i(t)$ if the source frequency were changed?
5. How can you calculate the quality factor of this circuit?

Tip: In AC circuits, to ensure in-phase conditions, always balance inductive and capacitive reactances to make the net reactance zero.