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To solve for \( v_C(t) \), the voltage across the capacitor in this RLC circuit, let’s analyze the circuit components and use phasor analysis. The circuit consists of a resistor (\(R = 10 \ \Omega\)), an inductor (\(L = 0.5 \ \text{H}\)), and a capacitor (\(C = 0.05 \ \text{F}\)) in series with a sinusoidal AC source. The source voltage is given as:

\[
V_{\text{source}}(t) = 10 \cos(10t + 75^\circ) \ \text{V}
\]

### Step 1: Convert Source Voltage to Phasor Form
The phasor form of the source voltage \(V_{\text{source}}\) is:
\[
V_{\text{source}} = 10\angle 75^\circ \ \text{V}
\]

### Step 2: Determine Angular Frequency
The angular frequency \( \omega \) can be identified from the cosine function as:
\[
\omega = 10 \ \text{rad/s}
\]

### Step 3: Calculate Impedances
1. **Impedance of the Resistor (\( R \))**:
   \[
   Z_R = R = 10 \ \Omega
   \]

2. **Impedance of the Inductor (\( L \))**:
   \[
   Z_L = j\omega L = j \cdot 10 \cdot 0.5 = j5 \ \Omega
   \]

3. **Impedance of the Capacitor (\( C \))**:
   \[
   Z_C = \frac{1}{j\omega C} = \frac{1}{j \cdot 10 \cdot 0.05} = -j2 \ \Omega
   \]

### Step 4: Total Impedance of the Series RLC Circuit
The total impedance \( Z_{\text{total}} \) is the sum of the individual impedances:
\[
Z_{\text{total}} = Z_R + Z_L + Z_C = 10 + j5 - j2 = 10 + j3 \ \Omega
\]

### Step 5: Calculate the Current in Phasor Form
Using Ohm’s law in phasor form:
\[
I = \frac{V_{\text{source}}}{Z_{\text{total}}} = \frac{10\angle 75^\circ}{10 + j3}
\]

Convert \(10 + j3\) to polar form to find the magnitude and phase:
\[
|Z_{\text{total}}| = \sqrt{10^2 + 3^2} = \sqrt{100 + 9} = \sqrt{109} \approx 10.44
\]
\[
\theta_Z = \arctan\left(\frac{3}{10}\right) \approx 16.7^\circ
\]
Thus,
\[
Z_{\text{total}} \approx 10.44 \angle 16.7^\circ \ \Omega
\]

The current phasor \( I \) is:
\[
I = \frac{10 \angle 75^\circ}{10.44 \angle 16.7^\circ} = \frac{10}{10.44} \angle (75^\circ - 16.7^\circ)
\]
\[
I \approx 0.958 \angle 58.3^\circ \ \text{A}
\]

### Step 6: Voltage Across the Capacitor
The phasor voltage across the capacitor \( V_C \) is given by \( V_C = I \cdot Z_C \):
\[
V_C = 0.958 \angle 58.3^\circ \cdot (-j2)
\]
Calculating the multiplication:
\[
|V_C| = 0.958 \times 2 = 1.916
\]
The angle of \( -j2 \) is \( -90^\circ \), so:
\[
\angle V_C = 58.3^\circ - 90^\circ = -31.7^\circ
\]
Thus,
\[
V_C = 1.916 \angle -31.7^\circ \ \text{V}
\]

### Step 7: Convert \( V_C \) Back to Time Domain
The voltage across the capacitor in the time domain is:
\[
v_C(t) = 1.916 \cos(10t - 31.7^\circ) \ \text{V}
\]

### Final Answer
\[
v_C(t) = 1.916 \cos(10t - 31.7^\circ) \ \text{V}
\]

Would you like a detailed breakdown of any step, or do you have any further questions?

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### Related Questions
1. How would the solution change if the values of \( L \) or \( C \) were different?
2. What would happen to \( v_C(t) \) if the source frequency were increased?
3. How does the phase angle of the capacitor voltage relate to the source voltage in a series RLC circuit?
4. Could you solve this problem using differential equations instead of phasor analysis?
5. What is the resonance frequency for this RLC circuit, and how would the response change at resonance?

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### Tip
For AC circuits, phasor analysis simplifies calculations by transforming time-domain sinusoidal functions into frequency-domain phasors. This technique is especially useful for circuits with sinusoidal sources.

For the circuit shown below, calculate v_C(t), the voltage across the capacitor.

Explore the solution to calculating v_C(t), the voltage across the capacitor, in an RLC circuit with a 10 cos(10t + 75°) V AC source. Learn about AC circuit analysis, impedance calculations, and phasor transformations. Suitable for undergraduate electrical engineering students.

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