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This is a series RLC circuit without resistance (pure LC circuit), and we are given the voltage across the capacitor, \( V_C(t) = 4 \cos(105t) \, \text{V} \). We need to find the source voltage \( V_S(t) \).

### Step 1: Angular Frequency
We are given \( V_C(t) = 4 \cos(105t) \), where 105 represents the angular frequency \( \omega \) in radians per second. Thus, \( \omega = 105 \, \text{rad/s} \).

### Step 2: Impedances of Inductor and Capacitor
For an inductor, the impedance is given by:
\[
Z_L = j\omega L
\]
where \( L = 2 \, \text{mH} = 2 \times 10^{-3} \, \text{H} \), and \( \omega = 105 \, \text{rad/s} \). So,
\[
Z_L = j(105)(2 \times 10^{-3}) = j0.21 \, \Omega
\]

For a capacitor, the impedance is:
\[
Z_C = \frac{1}{j\omega C}
\]
where \( C = 80 \, \text{nF} = 80 \times 10^{-9} \, \text{F} \), and \( \omega = 105 \, \text{rad/s} \). So,
\[
Z_C = \frac{1}{j(105)(80 \times 10^{-9})} = \frac{1}{j(8.4 \times 10^{-6})} = -j119.05 \, \Omega
\]

### Step 3: Total Impedance
The total impedance of the LC circuit is:
\[
Z_{\text{total}} = Z_L + Z_C = j0.21 - j119.05 = -j118.84 \, \Omega
\]

### Step 4: Voltage Relationship
The voltage across the entire circuit (i.e., \( V_S(t) \)) is related to the capacitor voltage by the voltage divider rule. Since the capacitor voltage is given, we can find the source voltage using:
\[
V_S(t) = V_C(t) \times \frac{Z_{\text{total}}}{Z_C}
\]

Substitute the values:
\[
V_S(t) = 4 \cos(105t) \times \frac{-j118.84}{-j119.05}
\]
\[
V_S(t) = 4 \cos(105t) \times \frac{118.84}{119.05} \approx 4 \cos(105t) \times 0.9982
\]
\[
V_S(t) \approx 3.993 \cos(105t) \, \text{V}
\]

Thus, the source voltage \( V_S(t) \approx 3.993 \cos(105t) \, \text{V} \).

Would you like further details on any step or have any questions?

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Here are 5 related questions you might explore:
1. How would the result change if resistance was introduced into the circuit?
2. What happens if the frequency of \( V_C(t) \) changes?
3. How does the impedance of the capacitor and inductor vary with frequency?
4. What is the resonant frequency of this LC circuit?
5. How does energy transfer between the inductor and capacitor in an LC circuit?

**Tip:** In LC circuits, resonance occurs when the inductive and capacitive reactances cancel each other out, leading to maximum voltage across the components.

If Vc(t) = 4Cos(105t) V in the circuit given below, find Vs(t).

Learn how to calculate the source voltage Vs(t) in a series LC circuit, where Vc(t) = 4Cos(105t) V. This solution covers AC circuit analysis, impedance calculations for inductors and capacitors, and applies the voltage divider rule. Suitable for university-level electrical engineering students.

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