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 series LCR circuit, we will follow these steps:

### Step 1: Identify the given values
- Peak Voltage (V₀) = 293 V
- 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 inductive reactance (Xₗ)
The formula for inductive reactance is:
\[ Xₗ = 2 \pi f L \]
Substituting the values:
\[ Xₗ = 2 \pi (50) (25 \times 10^{-3}) \]
Calculating:
\[ Xₗ = 2 \times 3.14 \times 50 \times 0.025 \]
\[ Xₗ = 7.85 \, \Omega \]

### Step 3: Calculate the capacitive reactance (Xₜ)
The formula for capacitive reactance is:
\[ Xₜ = \frac{1}{\omega C} \]
Where \( \omega = 2 \pi f \). First, calculate \( \omega \):
\[ \omega = 2 \pi (50) = 100 \pi \, \text{rad/s} \]
Now substituting into the formula for \( Xₜ \):
\[ Xₜ = \frac{1}{(100 \pi)(750 \times 10^{-6})} \]
Calculating:
\[ Xₜ = \frac{1}{0.00075 \times 100 \pi} \]
\[ Xₜ \approx 4.25 \, \Omega \]

### Step 4: Calculate the net reactance (X)
The net reactance \( X \) is given by:
\[ X = Xₗ - Xₜ \]
Substituting the values:
\[ X = 7.85 - 4.25 = 3.6 \, \Omega \]

### Step 5: Calculate the impedance (Z)
The impedance \( Z \) is given by:
\[ Z = \sqrt{R^2 + X^2} \]
Substituting the values:
\[ Z = \sqrt{6^2 + 3.6^2} \]
Calculating:
\[ Z = \sqrt{36 + 12.96} \]
\[ Z = \sqrt{48.96} \]
\[ Z \approx 7.0 \, \Omega \]

### Final Answer
The impedance of the circuit is approximately **7.0 Ω**.

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