A sinusoidal voltage of peak value `283 V` and frequency `50 Hz` is applied to a series `LCR` circuit in which `R = 3 Omega, L = 25.48 mH`, and `C = 796 mu F`. Find the impdedance of the circuit.

Explanation

To find the impedance of the series LCR circuit, we can follow these steps:

### Step 1: Identify the given values
- Peak voltage (V₀) = 283 V
- Frequency (f) = 50 Hz
- Resistance (R) = 3 Ω
- Inductance (L) = 25.48 mH = 25.48 × 10⁻³ H
- Capacitance (C) = 796 μF = 796 × 10⁻⁶ F

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of frequency:
\[ \omega = 2\pi \times 50 \approx 314.16 \, \text{rad/s} \]

### Step 3: Calculate the inductive reactance (Xₗ)
The inductive reactance (Xₗ) is calculated using the formula:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 314.16 \times 25.48 \times 10^{-3} \approx 8.00 \, \Omega \]

### Step 4: Calculate the capacitive reactance (Xᶜ)
The capacitive reactance (Xᶜ) is calculated using the formula:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values:
\[ X_C = \frac{1}{314.16 \times 796 \times 10^{-6}} \approx 3.98 \, \Omega \]

### Step 5: Calculate the total reactance (X)
The total reactance (X) in the circuit is given by:
\[ X = X_L - X_C \]
Substituting the values:
\[ X = 8.00 - 3.98 \approx 4.02 \, \Omega \]

### Step 6: Calculate the impedance (Z)
The impedance (Z) of the circuit can be calculated using the formula:
\[ Z = \sqrt{R^2 + X^2} \]
Substituting the values:
\[ Z = \sqrt{3^2 + 4.02^2} \approx \sqrt{9 + 16.16} \approx \sqrt{25.16} \approx 5.02 \, \Omega \]

### Final Answer
The impedance of the circuit is approximately **5.02 Ω**.

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