A sinusoidal voltage of peak value 293 V and frequency 50 Hz is applie to a series LCR circuit in which `R=6 Omega, L=25 mH` and `C=750mu F`. The impedance of the circuit is

Options

A. `7.0 Omega`
B. `8.9 Omega`
C. `9.9 Omega`
D. `10.0 Omega`

Explanation

To find the impedance of the given LCR circuit, we will follow these steps:

### Step 1: Identify the given values
- Peak voltage (V_peak) = 293 V (not needed for impedance calculation)
- Frequency (f) = 50 Hz
- Resistance (R) = 6 Ω
- Inductance (L) = 25 mH = 25 × 10^(-3) H
- Capacitance (C) = 750 μF = 750 × 10^(-6) F

### Step 2: Calculate the capacitive reactance (X_C)
The formula for capacitive reactance is:
\[ X_C = \frac{1}{2 \pi f C} \]
Substituting the values:
\[ X_C = \frac{1}{2 \pi (50)(750 \times 10^{-6})} \]
Calculating this gives:
\[ X_C = \frac{1}{2 \pi (50)(0.00075)} \approx 4.24 \, \Omega \]

### Step 3: Calculate the inductive reactance (X_L)
The formula for inductive reactance is:
\[ X_L = 2 \pi f L \]
Substituting the values:
\[ X_L = 2 \pi (50)(25 \times 10^{-3}) \]
Calculating this gives:
\[ X_L = 2 \pi (50)(0.025) \approx 7.85 \, \Omega \]

### Step 4: Calculate the impedance (Z)
The formula for impedance in a series LCR circuit is:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the values:
\[ Z = \sqrt{6^2 + (7.85 - 4.24)^2} \]
Calculating this step by step:
1. \( R^2 = 6^2 = 36 \)
2. \( X_L - X_C = 7.85 - 4.24 = 3.61 \)
3. \( (X_L - X_C)^2 = (3.61)^2 \approx 13.0321 \)
4. Now, add these values:
\[ Z = \sqrt{36 + 13.0321} = \sqrt{49.0321} \approx 7.0 \, \Omega \]

### Final Answer
The impedance of the circuit is approximately \( Z \approx 7.0 \, \Omega \).

---