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We are given a simple series RLC circuit with the following information:

- \( V_C(t) = 4 \cos(105t) \) V (capacitor voltage)
- Inductance \( L = 2 \, \text{mH} = 2 \times 10^{-3} \, \text{H} \)
- Capacitance \( C = 80 \, \text{nF} = 80 \times 10^{-9} \, \text{F} \)
- We are asked to find \( V_S(t) \), the source voltage.

### Step 1: Analyze the circuit

Since this is an RLC series circuit, the total voltage \( V_S(t) \) is the sum of the voltage across the inductor \( V_L(t) \), the capacitor \( V_C(t) \), and any resistor (though there doesn't seem to be a resistor in this case, we assume the resistance is negligible).

The relationship between the current and voltages in an RLC circuit is given by:
- Voltage across the inductor: \( V_L(t) = L \frac{dI(t)}{dt} \)
- Voltage across the capacitor: \( V_C(t) = \frac{1}{C} \int I(t) \, dt \)
- The current through the inductor and capacitor is the same, as they are in series.

### Step 2: Obtain the angular frequency

From the given capacitor voltage \( V_C(t) = 4 \cos(105t) \), we see that the angular frequency \( \omega = 105 \, \text{rad/s} \).

### Step 3: Find the impedance of the inductor and capacitor

- The impedance of the inductor \( Z_L = j\omega L = j(105)(2 \times 10^{-3}) = j0.21 \, \Omega \)
- The impedance of the capacitor \( Z_C = \frac{1}{j\omega C} = \frac{1}{j(105)(80 \times 10^{-9})} \approx -j119.05 \, \Omega \)

### Step 4: Calculate the total impedance

The total impedance \( Z_{total} \) is the sum of the inductor and capacitor impedances:

\[
Z_{total} = Z_L + Z_C = j0.21 - j119.05 = -j118.84 \, \Omega
\]

### Step 5: Relate the

If Vc(t) = 4Cos105t V in the circuit given below then find Vs(t)?

Explore how to determine the source voltage in an RLC series circuit given the capacitor voltage. This guide covers complex impedance calculations, differential equations, and the application of the superposition theorem in AC circuits. Suitable for electrical engineering students at the university level.

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