A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R = 5 `Omega` L = 25 mH and C = 1000 `mu` F. The total impedance, and phse difference between the voltage across the source and the current will respectively be

Options

A. 10 `Omega and ((5)/(3)).`
B. ` 7 Omega and 45^(@)`.
C. `10 Omega and ((8)/(3))`
D. `7 Omega and ((5)/(3))`.

Explanation

To solve the problem, we need to find the total impedance (Z) and the phase difference (φ) between the voltage across the source and the current in a series LCR circuit.

### Given Data:
- Peak voltage (V₀) = 283 V
- Angular frequency (ω) = 320 rad/s
- Resistance (R) = 5 Ω
- Inductance (L) = 25 mH = 25 × 10⁻³ H
- Capacitance (C) = 1000 μF = 1000 × 10⁻⁶ F = 10⁻³ F

### Step 1: Calculate Inductive Reactance (Xₗ)
Inductive reactance (Xₗ) is given by the formula:
\[ Xₗ = ωL \]

Substituting the values:
\[ Xₗ = 320 \, \text{rad/s} \times 25 \times 10^{-3} \, \text{H} \]
\[ Xₗ = 320 \times 0.025 = 8 \, \Omega \]

### Step 2: Calculate Capacitive Reactance (Xₑ)
Capacitive reactance (Xₑ) is given by the formula:
\[ Xₑ = \frac{1}{ωC} \]

Substituting the values:
\[ Xₑ = \frac{1}{320 \, \text{rad/s} \times 10^{-3} \, \text{F}} \]
\[ Xₑ = \frac{1}{0.32} = 3.125 \, \Omega \]

### Step 3: Calculate Total Impedance (Z)
The total impedance (Z) in a series LCR circuit is given by:
\[ Z = \sqrt{R^2 + (Xₗ - Xₑ)^2} \]

Calculating \( Xₗ - Xₑ \):
\[ Xₗ - Xₑ = 8 - 3.125 = 4.875 \]

Now substituting into the impedance formula:
\[ Z = \sqrt{5^2 + (4.875)^2} \]
\[ Z = \sqrt{25 + 23.765625} \]
\[ Z = \sqrt{48.765625} \]
\[ Z \approx 6.99 \, \Omega \]

### Step 4: Calculate Phase Difference (φ)
The phase difference (φ) is given by:
\[ \tan(φ) = \frac{Xₗ - Xₑ}{R} \]

Substituting the values:
\[ \tan(φ) = \frac{4.875}{5} \]
\[ φ = \tan^{-1}(0.975) \]

Calculating φ:
Using a calculator,
\[ φ \approx 44.75^\circ \]

### Final Answers:
- Total Impedance (Z) ≈ 6.99 Ω (or approximately 7 Ω)
- Phase Difference (φ) ≈ 44.75° (or approximately 45°)