An sinusoidal voltage of peak value 300 V and an argular frequency `omega` = 400 `rads^(-1) ` is applied to series L-C-R circuit , in which R = ` 3 Omega` , L = 20 mH and C = 625 `mu`F .The peak current in the circuit is

Options

A. `30sqrt(2)`A
B. 60 A
C. 100 A
D. `60sqrt(2)`A

Explanation

To find the peak current in the given series L-C-R circuit, we will follow these steps:

### Step 1: Identify the given values
- Peak voltage (V₀) = 300 V
- Angular frequency (ω) = 400 rad/s
- Resistance (R) = 3 Ω
- Inductance (L) = 20 mH = 20 × 10^(-3) H
- Capacitance (C) = 625 μF = 625 × 10^(-6) F

### Step 2: Calculate the inductive reactance (X_L)
The inductive reactance (X_L) is given by the formula:
\[ X_L = ωL \]
Substituting the values:
\[ X_L = 400 \, \text{rad/s} \times 20 \times 10^{-3} \, \text{H} \]
\[ X_L = 400 \times 0.02 = 8 \, \Omega \]

### Step 3: Calculate the capacitive reactance (X_C)
The capacitive reactance (X_C) is given by the formula:
\[ X_C = \frac{1}{ωC} \]
Substituting the values:
\[ X_C = \frac{1}{400 \, \text{rad/s} \times 625 \times 10^{-6} \, \text{F}} \]
\[ X_C = \frac{1}{400 \times 625 \times 10^{-6}} \]
Calculating the denominator:
\[ 400 \times 625 = 250000 \]
Thus,
\[ X_C = \frac{1}{0.25} = 4 \, \Omega \]

### Step 4: Calculate the impedance (Z)
The impedance (Z) in a series L-C-R circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the values:
\[ Z = \sqrt{3^2 + (8 - 4)^2} \]
Calculating:
\[ Z = \sqrt{9 + 4} = \sqrt{13} \]
Thus,
\[ Z \approx 3.61 \, \Omega \]

### Step 5: Calculate the peak current (I₀)
The peak current (I₀) is given by:
\[ I₀ = \frac{V₀}{Z} \]
Substituting the values:
\[ I₀ = \frac{300 \, V}{\sqrt{13}} \]
Calculating:
\[ I₀ \approx \frac{300}{3.61} \approx 83.3 \, A \]

### Conclusion
The peak current in the circuit is approximately 83.3 A.